Proof. Suppose P ₀ is not cyclic. Then lemma 2.2 implies that $G/P_0$ is Dedekind. Therefore $P/P_0 \unlhd G/P_0$, which implies that $P \unlhd G$. The proof of the rest of the statement is completely as the proof of [Reference Liu and Passman33, Lemma 2.5. (1)].

Proof. Note that $\text{dim}_F(eFG) = k^2[D:F]$ and $\chi(1) = k[D:F]$. So the primitive central idempotent e corresponding to this component by Theorem 2.6 has the form

\begin{equation*}e = \frac{\chi(1)}{[D:F]|G|}\sum_{g \in G} \chi(g^{-1})g = \frac{k}{|G|}\sum_{g \in G} \chi(g^{-1})g. \end{equation*}

Hence for the product we have

\begin{align*} ne &= \frac{k}{|G|}\sum_{g \in G} \sum_{h \in G} \alpha(g) \chi(h^{-1})gh \\ &= \frac{k}{|G|}\sum_{g \in G} \left(\sum_{h \in G} \alpha(gh^{-1}) \chi(h^{-1})\right)g = \frac{k}{|G|}\sum_{g \in G} \left(\sum_{h \in G} \alpha(h) \chi(g^{-1}h)\right)g. \end{align*}

So the coefficient of ne at g is

\begin{equation*}\frac{k}{|G|} \sum_{h \in G} \alpha(gh^{-1}) \chi(h^{-1}) = \frac{k}{|G|} \sum_{h \in G} \alpha(h) \chi(g^{-1}h). \end{equation*}

Proof. Assume P and Q are a non-cyclic p-Sylow and a q-Sylow subgroup of G respectively. Then $P \unlhd G$ and so by lemma 2.2 we know that Q is a Dedekind group. But as this argument applies to every prime, this means that G is Dedekind, contradicting the assumption.

Proof of lemma 2.11

Assume N is a normal subgroup of G contained in Q. If N is not cyclic, then $G/N$ is Dedekind by lemma 2.2, so that $Q/N \unlhd G/N$, which implies the contradiction $Q \unlhd G$. So part (a) follows.

The proof of part (b) is by two ‘iterations’. First assume N with the described properties exists. Then Q is cyclic or a generalized quaternion group: indeed, if $n \in N$ has order p and $q \in Q$ is an element of order p not lying in $\langle n \rangle$, then $\langle n \rangle \langle q \rangle \unlhd G$, as G is a group with SN. But this contradicts part (a) as $\langle n \rangle \langle q \rangle$ is not cyclic. So Q contains exactly one subgroup of order p, implying Q is cyclic or generalized quaternion by lemma 2.12. Next we claim that, independently from the existence of N, the group Q is cyclic, elementary abelian or generalized quaternion. For this assume Q is maximal non-normal and let M be a subgroup of G containing Q such that $[M:Q] = p$. By the maximality of Q, we get $M \unlhd G$ and so $\Phi(M) \unlhd G$, where $\Phi(M)$ denotes the Frattini subgroup of M. As Q is a maximal subgroup of M, it contains $\Phi(M)$ and so $\Phi(M)$ is cyclic by part (a). If $\Phi(M) = 1$, then Q is elementary abelian. If $\Phi(M) \neq 1$, then Q is cyclic or generalized quaternion by the first claim proved in this paragraph.

It remains to show that in the two claims proven in the previous paragraph we can replace generalized quaternion groups by quaternion groups of order 8. Assume first N exists and Q is a generalized quaternion with $n \in Q$ the unique involution. Then $Q/\langle n \rangle$ is a dihedral group of order $|Q|/2$. If $|Q|/2 \geq 8$, this implies, by the claim proven in the previous paragraph and the fact that the SN property is inherited by quotients, that $Q/\langle n \rangle \unlhd G/\langle n \rangle$. This would imply $Q \unlhd G$. So $|Q|/2 \leq 4$ which means that Q is the quaternion group of order 8. Now again ignore the existence of N, let Q be again maximal non-normal, M a normal subgroup of G containing Q such that $[M:Q] = p$ and assume that Q is generalized quaternion. Then as before $\Phi(M) \unlhd G$ and $\Phi(M) \neq 1$, as Q is not elementary abelian. It follows that the unique involution of Q is central in G and so the same argument as before can be used to show $|Q| = 8$.

Proof. If p = 2 let Q be the elementary abelian subgroup of G which is not normal. When p is odd, by lemma 2.10, G contains a non-cyclic subgroup Q which is not normal in G. Then Q is elementary abelian by lemma 2.11. We choose Q maximal with these properties, in particular for $Q \lneq M \leq G$ we have $M \unlhd G$. By lemma 2.11, if Q contains a subgroup N which is normal in G, then N = 1 (note that this is trivial if $|Q| = 2$). In particular, we have $\mathcal{Z}(G) \cap Q = 1$.

Claim 1: $\mathcal{Z}(G)$ is cyclic.

Assume first that $\mathcal{Z}(G)$ contains an elementary abelian subgroup $\langle z_1 \rangle \times \langle z_2 \rangle$. Then, by our choice of Q and the fact that $\mathcal{Z}(G) \cap Q = 1$, we have $Q\langle z_1 \rangle \unlhd G$ and $Q\langle z_2 \rangle \unlhd G$. This implies that $[G,Q] \leq Q \langle z_1 \rangle$ and $[G,Q] \leq Q \langle z_2 \rangle$, respectively. So $[G,Q] \leq Q \langle z_1 \rangle \cap Q \langle z_2 \rangle = Q$, which would imply that $Q \unlhd G$. Hence $\mathcal{Z}(G)$ contains at most one subgroup of order p. As $\mathcal{Z}(G)$ cannot be generalized quaternion, because such a group is not abelian, the claim follows from lemma 2.12.

We denote an element of order p in $\mathcal{Z}(G)$ by z.

Claim 2: $G' = \langle z \rangle$. Moreover, if $g\ \in G$ such that $z \notin \langle g \rangle$, then $g^p = 1$.

Let $g,h \in G$ such that $[g,h] \neq 1$. Assume first that $z \notin \langle g \rangle$. Note that we can assume this without changing the value of $[g,h]$ when p is odd. As G is a group with SN, this implies $\langle g, z \rangle = \langle g \rangle \times \langle z \rangle \unlhd G$. As z is central and G is a p-group we hence get $[G,g] \subseteq \langle g^p \rangle \times \langle z \rangle$. If $g^p \neq 1$, then we have $[G, g^p] \subseteq \langle g^{p^2} \rangle$, $[G, g^{p^2}] \subseteq \langle g^{p^3} \rangle$, …, $[G, g^{\circ(g)/p}] = 1$, implying $g^{\circ(g)/p} \subseteq \mathcal{Z}(G)$ which contradicts Claim 1, as $z \notin \langle g \rangle$ by assumption. Hence $g^p = 1$ and $[G,g] \subseteq \langle z \rangle$, in particular $[g,h] \in \langle z \rangle$. Now suppose that z lies in every non-trivial subgroup of $\langle g, h \rangle$, i.e. $\langle g,h \rangle$ is a generalized quaternion group. If $\langle g,h \rangle$ has order 8, then $[g,h] = z$ and there is nothing more to prove. So assume $\langle g,h \rangle \cong Q_{2^m}$ for some $m \geq 4$. Say $g^{2^{m-1}} = h^4 = 1$ and $g^h = g^{-1}$. Note that $h^g = hg^2$. As Q is elementary abelian and maximal non-normal, we get $\langle Q, h \rangle \unlhd G$. As also $Q \times\langle z \rangle \unlhd G$, we have that $[h,q]$ has order at most 2 and so hq has order at most 4 for every $q \in Q$. Note for this that $h^2 = z$ is central in G. But as $h^g = hg^2 \in \langle Q, h\rangle$, we have $g^2 \in \langle Q, h \rangle$, a contradiction, since g ² has order at least 8.

In particular Claim 2 implies that G is metabelian. Moreover, there exists a cyclic subgroup C of G containing z such that $A = C \times Q$ is a maximal abelian subgroup of G. We are now finally ready to prove that G has at most one matrix component by applying lemma 2.5. So assume $\mathbb{Q}e(G,H,K)$ is a non-commutative component of $\mathbb{Q}G$. As K lies in the kernel of a representation corresponding to this component, we have $G' \not\leq K$, i.e. $z \notin K$ by Claim 2. Hence, again by Claim 2, the unique maximal element of $\{B \leq G \ | \ A\leq B \ \ \text{and} \ \ B' \leq K \leq B \}$ is A and $e(G,H,K) = e(G,A,K)$ with $A/K$ a cyclic group. By lemma 2.5, it is thus sufficient to prove that all subgroups of A which do not contain z and have cyclic quotients are conjugate. We call such subgroups ‘good’. Note that good subgroups are elementary abelian, so contained in $\langle z \rangle \times Q$. A good subgroup U is determined by $(\langle z \rangle \times Q)/U$ and these quotients are exactly the images of the groups $\langle z q\rangle$, where q runs through the elements of Q. Hence there are $|Q|$ good subgroups. It is clear that Q itself is a good subgroup. So we need to prove that Q has $|Q|$ conjugates in G, i.e. $[G:N_G(Q)] = |Q|$. By the maximality of Q, and since $z \in N_G(Q)$, we have $N_G(Q) \unlhd G$. As Gʹ is a central subgroup of order p, the group $G/N_G(Q)$ is elementary abelian. So we can view $V = Q \times G/N_G(Q)$ as an $\mathbb{F}_p$-vector space of dimension $|Q| + |G/N_G(Q)|$. We define a non-degenerate symplectic bilinear form

\begin{equation*}V \times V \rightarrow \langle z \rangle, \ \ (v,w) \mapsto [v,w]. \end{equation*}

As no element of $G/N_G(Q)$ leaves all elements of Q fixed under conjugation, Q and $N_G(Q)$ are maximal isotropic subspaces of V, so that each of them has dimension $\frac{1}{2}\cdot(|Q| + |G/N_G(Q)|)$ by [Reference Huppert20, II, Satz 9.11], i.e. $|G/N_G(Q)| = |Q|$.

Proof. Assume that every involution in G is central. We first note that G is not a generalized quaternion group. Indeed Q ₈ and Q ₁₆ have SSN [Reference Liu30, Theorem 2.3, BJ6] and if $G = \langle g, h \ | \ g^{2^n} = h^4 = 1, \ g^{2^{n-1}} = h^2, \ g^h = g^{-1} \rangle$ for some $n \geq 4$, then G does not have SN. This can be observed by taking $N = \langle g^4 \rangle$, $Y = \langle h \rangle$, so that $N \not\subseteq Y$ and $NY \ntrianglelefteq G$ as $h^g = hg^2 \notin NY$. In particular, the centre of G is not cyclic.

By lemmas 2.10 and 2.11, we can assume G contains a non-normal subgroup Q isomorphic to Q ₈. We fix $a,b \in Q$ as generators of Q and $c = a^2$. We will prove several small facts on G which will lead to the proof of the lemma.

1. (i) $\mathcal{Z}(G)$ has rank 2:

   $\mathcal{Z}(G)$ is not cyclic, as G is not generalized quaternion. So assume the rank of $\mathcal{Z}(G)$ is bigger than 2. Say $z_1, z_2 \in \mathcal{Z}(G)$ are independent elements of order 2 such that $(\langle z_1 \rangle \times \langle z_2 \rangle) \cap Q = 1$. Then $Q \times \langle z_1 \rangle$ and $Q \times \langle z_2 \rangle$ are both normal subgroups of G. Hence $[G,Q] \leq Q\langle z_1 \rangle \cap Q\langle z_2 \rangle = Q$ which would imply $Q\unlhd G$.

2. Convention: We let $z \in G$ be an involution not lying in Q, so $\langle c \rangle \times \langle z \rangle$ is the unique maximal elementary abelian subgroup of G.

3. (ii) $|G/\Phi(G)| \leq 16$, i.e. G is at most 4-generated:

   This follows from (i) using [Reference MacWilliams35, Four Generator Theorem].

4. (iii) The groups $\langle a \rangle$, $\langle b \rangle$, and $\langle ab \rangle$ are not normal in G:

   Say $\langle a \rangle \unlhd G$. As $\langle a \rangle \not\subseteq \langle b \rangle$ and G has SN this implies $\langle a \rangle \langle b \rangle = Q \unlhd G$, a contradiction. Similarly $\langle b \rangle$ and $\langle ab \rangle$ are not normal in G.

5. (iv) $a^G = \{a, a^{-1}, az, a^{-1}z \}$, $b^G = \{b, b^{-1}, bz, b^{-1}z \}$, and $(ab)^G = \{ab, (ab)^{-1}, abz, (ab)^{-1}z \}$:

   As G has SN we have $\langle a \rangle \times \langle z \rangle \unlhd G$. So by (iii) there is $g \in G$ such that $a^g = az$ or $a^g = a^{-1}z$. As $a^b = a^{-1}$ and $(az)^b = a^{-1}z$ the claim for a^G follows. Similarly the conjugacy classes of b and ab follow from (iii).

6. (v) For $g \in G$, we have $g^2 \in C_G(Q)$:

   By (iv) we have $[g,a] \in \{1,c, z, cz \}$, in any case a central element of order at most 2. So $[g^2, a] = [g,a]^g[g,a] = 1$. Of course for b and ab we similarly have $[g^2,b] = [g^2, ab] =1$.

7. (vi) If $g \in C_G(Q)$ and $g \notin \langle c \rangle$, then $c \notin \langle g \rangle$:

   This is clear if g has order 2. So assume the order of g is 2ⁿ for $n \geq 2$ and such that $g^{2^{n-1}} = c$. Then $(g^{2^{n-2}}a)^2 = c\cdot c = 1$, so $g^{2^{n-2}}a$ is an involution. Here we used that $g \in C_G(Q)$. As $g^b = g$ we have $(g^{2^{n-2}}a)^b = g^{2^{n-2}}a^{-1}$, so this involution is not central, contradicting the assumptions on G.

8. (vii) If $g \notin N_G(Q)$, but g is centralizing a, b, or ab, then $c \notin \langle g \rangle$:

   Say $g \in C_G(a)$, $\circ(g) = 2^n$ and assume $c \in \langle g \rangle$. As $g\notin N_G(Q)$, we must have $g \notin N_G(\langle b \rangle)$, so $b^g = b^{\pm 1}z$ by (iv). Note that $g^2 \in C_G(Q)$ by (v). So $g^{2^{n-2}}a$ is an involution with $(g^{2^{n-2}}a)^b = g^{2^{n-2}}ac$, if n > 2, or $(g^{2^{n-2}}a)^b \in \{gaz, gacz \}$, if n = 2. In any case we would have a non-central involution.

9. (viii) If $g \notin N_G(\langle a \rangle)$ and $g \notin N_G(\langle b \rangle)$, then $g \in N_G(\langle ab \rangle)$. The same holds for every permutation of a, b, and ab:

   If $g \notin N_G(\langle a \rangle)$ and $g \notin N_G(\langle b \rangle)$, then $a^g = a^{\pm 1}z$ and $b^g = b^{\pm 1}z$, so that

   \begin{equation*}(ab)^g = a^{\pm 1}b^{\pm 1} z z \in \{ab, a^{-1}b, ab^{-1}, a^{-1}b^{-1} \}. \end{equation*}

   Noting that $a^{-1}b = ab^{-1} = (ab)^{-1}$ and $a^{-1}b^{-1} = ab$, the claim follows.

10. (ix) $N_G(Q) \unlhd G$ and $G/N_G(Q) \cong C_2 \times C_2$:

    $N_G(Q)\unlhd G$ follows, as $N_G(Q)$ contains z and is thus bigger than Q. By (v) the group $G/N_G(Q)$ is elementary abelian. If $G/N_G(Q) \cong C_2$ would hold, then by (viii) one of $\langle a \rangle$, $\langle b \rangle$, or $\langle ab \rangle$ would be normal in G, which would contradict (iii). On the other hand, if $g,h \notin N_G(\langle a \rangle)$, then $gh \in N_G(\langle a \rangle)$ by (iv). This implies that G has only three non-trivial ways to act on the cyclic subgroups of the normal subgroup $Q \times \langle z \rangle$, implying $|G/N_G(Q)| \leq 4$.

11. Convention: By (viii) and (ix), we can choose $x,y \in G$ such that $\circ(x) \geq \circ(y)$ and $x \notin N_G(\langle a \rangle)$, $x \in N_G(\langle b \rangle)$ as well as $y \in N_G(\langle a \rangle)$, $y \notin N_G(\langle b \rangle)$. To assure the condition $\circ(x) \geq \circ(y)$ we might have to rename the elements a and b.

12. (x) $C_G(Q) = \Phi(G)$ and $\{a,b,x,y \}$ is a minimal generating set of G:

    First note that if $g \in G$, then $Q^g \leq Q \langle z \rangle$ by (iv). As z is central, this implies that for every $h \in C_G(Q)$, the element h is also centralizing Q^g. Hence $h^g \in C_G(Q)$, implying that $C_G(Q)$ is normal in G. Now, by the action of a, b, x, and y on $Q \times \langle z \rangle$, we see that no element of the form $a^\alpha b^\beta x^\gamma y^\delta$ with at least one of the α, β, γ, and δ odd is centralizing Q. Hence the images of a, b, x, and y in $G/C_G(Q)$ generate an elementary abelian subgroup of order 16. By (ii), this is a maximal elementary abelian quotient of G, so the well-known properties of Frattini subgroups of p-groups, as recorded for instance in [Reference Huppert20, III, Section 3], imply the claim.

13. (xi) $G^2 = \Phi(G)$. Moreover for any $g,h \in G$ we have $[g,h]^g = [g,h]$ and $(gh)^2 = g^2h^2[h,g]$:

    The equation $G^2 = \Phi(G)$ holds in every 2-group [Reference Huppert20, III, Satz 3.14(b)]. Let $i \in G$ be an involution so that $i \notin \langle g \rangle$. Hence $\langle g \rangle \times \langle i \rangle \unlhd G$ and so $[g,h] \in \langle g^2, i \rangle$, which implies $[g,h]^g = [g,h]$. Moreover, this gives $(gh)^2 = g^2h[h,g]h = g^2h^2[h,g]$.

14. (xii) For $g,h \in G$, we have $[g^2,h] = [g,h]^2$ and $[g^2,h] \in \langle g^4 \rangle \cap \langle h^4 \rangle$. Furthermore, $\langle g^2 \rangle \unlhd G$:

    In general $[g^2,h] = [g,h]^g[g,h]$, so $[g^2,h] = [g,h]^2$ holds by (xi). Moreover, if $i,j \in G$ are involutions such that $i \notin \langle g \rangle$ and $j \notin \langle h \rangle$, then $\langle g \rangle \times \langle i \rangle$ and $\langle h \rangle \times \langle j \rangle$ are normal subgroups of G, so that $[g,h] \in \langle g^2, i \rangle$ and $[g,h] \in \langle h^2, j \rangle$. So $[g^2,h] = [g,h]^2 \in \langle g^4 \rangle \cap \langle h^4 \rangle$. Finally, as $\langle g \rangle \times \langle i \rangle \unlhd G$ we have $[g,G] \subseteq \langle g^2, i\rangle$, so that $[g^2,G] \subseteq \langle g^4 \rangle$ by the previous, implying that $\langle g^2 \rangle \unlhd G$.

15. (xiii) If $g,h \in C_G(Q)$ and both have order at least 4, then $\langle g \rangle \cap \langle h \rangle \neq 1$:

    Say $\langle g \rangle \cap \langle h \rangle =1$. By (vi), we know that c is not contained in $\langle g \rangle$ or $\langle h \rangle$. So we can assume $z \in \langle g \rangle$ and $cz \in \langle h \rangle$. Assume first that $\circ(g) = 2^n \geq 8$ and say $\circ(h) = 2^m$. By (xii) and the assumption $\langle g \rangle \cap \langle h \rangle = 1$, we have $[g^2,h] = 1$, so that $(g^{2^{n-2}}h^{2^{m-2}})^2 = zzc = c$. Hence $g^{2^{n-2}}h^{2^{m-2}}$ is an element of order 4 in $C_G(Q)$ squaring to c, contradicting (vi). Now assume g and h are both of order 4. As $g \in C_G(Q)$ and $G^2 = \Phi(G) = C_G(Q)$ by (x) and (xi), there are $g_1,\ldots,g_k \in G$ such that $g = g_1^2g_2^2 \cdots g_k^2$. By the general commutator formulas and (xii), we get

    \begin{equation*}[g,h] = [g_1^2 \cdots g_k^2,h] = ([g_1,h]^2)^{g_2^2 \cdots g_k^2}([g_2,h]^2)^{g_3^2 \cdots g_k^2} \cdots [g_k,h]^2. \end{equation*}

    As $[g_i,h]^2 \in \langle h^4 \rangle = 1$ for all i by (xii) we get $[g,h] = 1$ and so $(gh)^2 = g^2h^2 = zcz = c$, again contradicting (vi).

16. Convention: In case $C_G(Q)$ contains an element g of order 4, we set $z = g^2$. By (vi) and (xiii), this is well defined. If there is no such element, we just keep the z from before.

17. (xiv) There is $\tilde{z} \in C_G(Q)$ such that $z \in \langle \tilde{z} \rangle$ and $C_G(Q) = \langle c \rangle \times \langle \tilde{z} \rangle$:

    Let $g,h\in C_G(Q)$. As $C_G(Q) = G^2$ by (x) and (xi) as in the proof of (xiii) we have $[g,h] \in \langle h^4 \rangle$, where we also need that $\langle h^2 \rangle \unlhd G$ by (xii). Hence $[C_G(Q), C_G(Q)] \leq C_G(Q)^4$, so that $C_G(Q)$ is a powerful 2-group (cf. the definition in [Reference Dixon, du Sautoy, Mann and Segal13, I, Definition 2.1]). Hence $C_G(Q)^2 = \{g^2 \ | \ g \in C_G(Q) \}$ [Reference Dixon, du Sautoy, Mann and Segal13, I, Proposition 2.6]. By (vi), this implies $c \notin \Phi(C_G(Q)) = C_G(Q)^2$, hence $\langle c \rangle$ is a direct factor of $C_G(Q)$ and we have $C_G(Q) = \langle c \rangle \times H$ for a subgroup H containing only the involution z. Then H is cyclic or generalized quaternion by lemma 2.12, but as quaternion groups are not powerful, H must be cyclic and the claim follows.

18. (xv) For $g,h \in G$, we have $[g^2, h^2] = 1$:

    By (v), we know $g^2, h^2 \in C_G(Q)$ which is an abelian group by (xiv).

19. (xvi) Without breaking the conventions, we can assume that y has order 4, $a^y = a$ and $z \in \langle y \rangle$:

    We first aim to replace y by an element of order 4. By (xiv) we know $x^4, y^4 \in \langle \tilde{z} \rangle$, so, as by convention $\circ(x) \geq \circ(y)$, there is $\ell \in \mathbb{Z}$ such that $x^{4\ell}y^4 = 1$. If x has order 4, then also y. So assume x has order at least 8. Note that as $x^2, y^2 \in \langle c \rangle \times \langle \tilde{z} \rangle$ by (v) and (xiv) we have $y^2 \in \langle x^2, c \rangle$, so that $[x,y^2] = 1$. Hence by (xi) and (xii)

    \begin{equation*}(x^\ell y)^4 = (x^{2\ell}y^2[y,x^\ell])^2 = x^{4 \ell}y^4[y^2,x^\ell] = x^{4\ell}y^4 = 1. \end{equation*}

    Note that by the defining properties of x and y we have $x^\ell y \notin N_G(\langle b \rangle)$, so $x^\ell y$ cannot be an involution. So, $x^\ell y$ has order 4 and we replace y by $x^\ell y$, where we replace a by ab if $\ell$ is odd, so that the convention is kept. In case with the new y, we have $a^y \neq a$, we replace y by by. Finally, if $z \notin \langle y \rangle$, then $cz \in \langle y \rangle$ as $c \in \langle y \rangle$ is impossible by (vii). Then ay satisfies all the conventions and moreover $(ay)^2 = a^2y^2 = ccz = z$, so that we choose ay as the new y.

20. Convention: We choose y as described in (xvi).

21. (xvii) $\circ(x) = 4$:

    Assume $\circ(x) = 2^n \gt 4$. Then by (v) and (xiv) we have $x^4 \in \langle \tilde{z} \rangle$ and so $z \in \langle x^4 \rangle$. As $[x^2,y] \in \langle y^4 \rangle = 1$ by (xii) we then get $(x^{2^{n-2}}y)^2 = zz=1$ and $x^{2^{n-2}}y$ is an involution. But as it is not centralizing b this gives a contradiction.

22. (xviii) $\tilde{z} = z$:

    Assume $\circ(\tilde{z}) \gt 2$ holds. First note that as x and y have order 4 by (xvi) and (xvii), the defining properties of x and y then imply that $\tilde{z}$ is not a power of x or y. Moreover, again as x has order 4, we have $x^2 \in Z(G)$ and so $1 = [x^2,y] = [x,y]^2$ by (xii). In particular, $[x,y]$ has order at most 2, implying that it is an element of the maximal elementary abelian subgroup $\langle c \rangle \times \langle z \rangle$ which is contained in $\langle c \rangle \times \langle \tilde{z}^2 \rangle$. So, $G/(\langle c \rangle \times \langle \tilde{z}^2 \rangle)$ is an elementary abelian group where the images of a, b, x, y, and $\tilde{z}$ are independent elements. To see the this use again the fact that $\circ(x) = \circ(y) = 4$. But this cannot be true, as G is 4-generated by (x).

23. (xix) We can assume $b^x = b$ and $x^2 = z$ without breaking the conventions:

    As $x \in N_G(\langle b \rangle)$, we have $b^x = b^{\pm 1}$. If $b^x = b^{-1}$, we can replace x by ax. Next, $x^2 = c$ is not possible by (vii). So if $x^2 \neq z$, we must have $x^2 = cz$ (note that $\circ(x) = 4$ by (xvii)). If this is the case, we replace x by bx noting that $(bx)^2 = ccz = z$.

24. Convention: We choose x as described in (xix).

25. (xx) $a^x = az$ and $b^y = bz$:

    Assume $a^x \neq az$. Then $a^x = a^{-1}z$ by the properties of x and (iv). Then using (xi) we get $(ax)^2 = a^2x^2[x,a] = czcz = 1$, so that ax would be an involution not centralizing b. Similarly, if $b^y \neq bz$, then by would be an involution not centralizing a.

26. Relations: We summarize the obtained relations:

    \begin{align*} x^4 &= y^4 = 1, \ \ x^2 = y^2 = z, \\ a^x &= az, \ \ b^x = b, \ \ a^y = a, \ \ b^y = bz. \end{align*}

    We also note $(ab)^x = abz$ and $(ab)^y = abz$.

27. (xxi) $\langle x, y \rangle$ and $\langle bx, ay \rangle$ are normal subgroups of G isomorphic to Q ₈:

    By the relations already obtained for x and y to show that $\langle x,y \rangle \cong Q_8$ it suffices to show $[x,y] = z$. As x and y have order 4 the commutator $[x,y]$ has order at most 2 by (xii) and the fact that involutions are central. We consider the other possible values for $[x,y]$. If $[x,y]=1$, then xy is an involution no centralizing b. If $[x,y] = cz$, then by (xi) we get $(axy)^2 = a^2 (xy)^2 [xy,a] = c (x^2y^2cz)z = 1$, so that axy would be an involution not centralizing a. Similarly, if $[x,y] = c$, then abxy is an involution not centralizing a. Hence $[x,y] = z$ and $\langle x,y \rangle \cong Q_8$. We next observe $\langle bx, ay \rangle \cong Q_8$. This follows from calculating $(bx)^2 = (ay)^2 = cz$ as well as $(bx)^{ay} = bxcz = (bx)^{-1}$. It remains to show that both these subgroups are normal, but as $\{a,b,x,y \}$ is a generating set of G by (x), it is sufficient to consider their conjugates under these four elements. A direct calculation using the relations above then gives the claim.

28. (xxii) $G = \langle x,y \rangle \times \langle bx, ay \rangle \cong Q_8 \times Q_8$:

    By (xxi), both groups $\langle x,y \rangle$ and $\langle bx, ay \rangle$ are normal subgroups of G and isomorphic to Q ₈. As they have trivial intersection, $\langle x,y \rangle \times \langle bx, ay \rangle$ is a subgroup of G. This subgroup contains a, b, x, and y which is a generating set of G by (x).

Finally, we show that if $G \cong Q_8 \times Q_8$, then G has SN but not SSN using the notation for the elements of G as in the $Q_8 \times Q_8$ we just found. First, as $a^x = az$ the subgroup $\langle a, b \rangle$ is not normal in G and G does not have SSN by lemma 2.10. Next, note that every subgroup containing the three non-trivial involutions is normal in G, as $G' = \langle c \rangle \times \langle z \rangle$. Moreover, it is easy to see that a cyclic subgroup Y of order 4 is non-normal in G if and only if $Y^2 = \langle c \rangle$. Hence, a general subgroup Y is non-normal if and only if $\Phi(Y) = \langle c \rangle$ and Y is either cyclic of order 4 or a quaternion group of order 8. Hence, for every normal subgroup N and non-normal subgroup Y, the relation $N \not\leq Y$ implies that N contains an involution different from c, giving $NY \unlhd G$. Overall, G has SN.

Proof of Proposition 2.8

Let G be a nilpotent group which has SN but not SSN. By lemma 2.9, G is a p-group. If p is odd, or p = 2 and G contains a non-central involution, then the result is contained in Proposition 2.13. Finally consider the case p = 2 and that all involutions of G are central which only applies to the group $Q_8 \times Q_8$ by lemma 2.14. As $\mathcal{Z}(G)$ is not cyclic, G has no faithful irreducible representations. It hence suffices to consider the components of the maximal quotients of G. Say c and z are the involutions of the direct factors. Then $G/\langle c \rangle$, $G/\langle z \rangle$, and $G/\langle cz \rangle$ are the maximal quotients. The first two of those are isomorphic to $Q_8 \times C_2 \times C_2$ which is a Hamiltonian group not contributing matrix components to $\mathbb{Q}G$. Finally, we show that the group $G/\langle cz \rangle$ has SN, but not SSN, and contains non-central involutions, so that the result then follows from Proposition 2.13. To see this say a is an element of order 4 in the first direct factor and x an element of order 4 in the second. Then $(ax)^2 = cz$, so that ax is mapped to a non-central involution when mapping to $G/\langle cz \rangle$. To see that this quotient does not have SSN consider the subgroup generated by the first direct factor and x: then $\langle x \rangle$ is mapped to a normal subgroup not contained in the image of $\langle a \rangle$, but $\langle a,x \rangle$ is not mapped to a normal subgroup.

Proof. We first show that G being a solvable and non-nilpotent group with SN implies the described properties. Assume first that G contains a normal elementary abelian subgroup of rank at least 2 for some prime p. Then by lemma 2.4, we know that $P \unlhd G$ for $P \in \text{Syl}_p(G)$ and G contains a nilpotent pʹ-Hall subgroup H. By lemma 2.3, the action of H on P is irreducible and faithful and P is elementary abelian. It remains to show that the Sylow subgroups of H all have rank 1. The action of H on P corresponds to a faithful and irreducible representation of H over $\mathbb{F}_p$. Let χ be the character of this representation, M the $\mathbb{F}_pG$-module and F a field extension of $\mathbb{F}_p$ which is a splitting field for H. Then by [Reference Isaacs21, Theorem 9.21] the character of the module $F \otimes_{\mathbb{F}_p} M$ is a sum of certain Galois-conjugate characters of an irreducible F-character η of H. If H contains an elementary-abelian subgroup Q of rank at least 2, then by the structure of Dedekind groups, Q is central in H and η has a non-trivial kernel on Q. But this kernel is then also contained in the kernel of χ, contradicting the fact that the action of H on P is faithful. Note also that if D is a representation corresponding to η and $h \in H \setminus \{1 \}$, then D(h) has no eigenvalue 1. This follows again from the structure of Dedekind groups, as this is true for faithful characters of the quaternion group of order 8 and cyclic groups. Hence there is no $g \in P \setminus \{1\}$ such that $g^h = g$.

We can hence assume that every elementary abelian normal subgroup of G has rank 1. Let P ₀ be such a normal p-subgroup, so P ₀ is a cyclic group of order p, and let H be a nilpotent pʹ-Hall subgroup of G which exists by lemma 2.4. We first consider the case that some Sylow subgroup of H acts trivially on P ₀. Let $Q \in \text{Syl}_q(G)$ be a such a Sylow subgroup, i.e. $[P_0, Q] = 1$. Then $P_0 \times Q \unlhd G$, as G has SN, and so $Q \unlhd G$, as it is characteristic in $P_0 \times Q$. So by lemma 2.4, we have G = QR for R a qʹ-Hall subgroup of G. Moreover, the action of R on Q is not faithful, as P ₀ acts trivially on Q. We conclude by lemma 2.3 that G is an SSN group of unfaithful type.

So we can assume that every Sylow subgroup of H acts non-trivially on P ₀. Let $Q \in \text{Syl}_q(H)$. We first show that the rank of Q is 1. Assume it is not and that $\langle x \rangle \times \langle y \rangle$ is an elementary abelian group of rank 2 contained in Q. As $\text{Aut}(P_0)$ is cyclic, some element of $\langle x \rangle \times \langle y \rangle$ must act trivially on P ₀, say this is x. Then $P_0 \times \langle x \rangle \unlhd G$, as G has SN and so $\langle x \rangle \unlhd G$, as $\langle x \rangle$ is a characteristic subgroup of $P_0 \times \langle x \rangle$. Hence, again using the SN property of G, also $\langle x \rangle \times \langle y \rangle \unlhd G$, but this contradicts our assumption that G contains no normal elementary abelian subgroup of rank at least 2. Hence Q has rank 1. So, every Sylow subgroup of H is cyclic or generalized quaternion by lemma 2.12.

We show that H contains no quaternion group. Indeed, assume $Q = \langle g,h \ | \ g^{2^n} = g^4 = 1, g^{2^{n-1}} = h^2, g^h = g^{-1} \rangle$ is a subgroup of H for some $n \geq 2$. As $\text{Aut}(P_0)$ is cyclic, some element of order 4 in Q must act trivially on P ₀. As in the previous paragraph, this element must generate a normal subgroup of G. When $n \geq 3$ the only normal subgroup of Q of order 4 is $\langle g^{2^{n-2}} \rangle$, so it must act trivially. When n = 2 we can assume this without loss of generality. Now $G/\langle g^{2^{n-2}}, h\rangle$ is a Dedekind group, so that the image of Q in this quotient acts trivially on the image of P ₀. Hence g acts trivially on P ₀ and h non-trivially. As before we get then $\langle g \rangle \unlhd G$ and the SN property implies $\langle g \rangle \langle h \rangle = Q \unlhd G$. But this cannot be as h acts non-trivially on P ₀. We conclude that all the Sylow subgroups of H are cyclic.

We now show that under all the assumptions G has a normal Sylow p-subgroup. Let $\{q_1,q_2,\ldots,q_k \}$ be the prime divisors of $|H|$ with $q_1 \lt q_2 \lt \cdots \lt q_k$. Note that as a Sylow q_i-subgroup of H acts non-trivially on P ₀ we have $q_i \mid (p-1)$ and so $q_i \lt p$ for each i. By successively applying the famous corollary of Burnside’s p-complement Theorem on cyclic Sylow subgroups for minimal primes [Reference Huppert20, IV, Satz 2.8], we obtain that G contains a normal q ₁-complement H ₁, which contains a normal q ₂-complement H ₂,…, which contains a normal q_k-complement P which must be a Sylow p-subgroup of G. Note that in each step the normal complement found is characteristic, so that all these groups are also normal in G, in particular P. So $G = P \rtimes H$. It follows from lemma 2.3 that the structure of G is as claimed. Moreover, as all the Sylow subgroups of G are cyclic and the action of each Sylow subgroup of H on P is now faithful we get $[g,h] \neq 1$ for all $h \in H \setminus \{1\}$ and $g \in P \setminus \{1\}$.

Finally, we also show that the described groups are groups with SN. This is clear for the SSN groups of unfaithful type, as these have even SSN by [Reference Liu and Passman33, Theorem 2.7]. So assume G is as described in (ii). As the action of H on P is irreducible and faithful, P is the unique minimal normal subgroup of G. So if $N \unlhd G$ and N ≠ 1, then $P \leq N$. Let moreover $Y \leq G$. If N is not a subgroup of Y, then $NY/N \unlhd G/N$, as $G/N$ is a quotient of H and hence a Dedekind group. This implies $NY \unlhd G$ and so G indeed has SN.

Proof. Let G be a non-solvable group with SN. Recall that minimal normal subgroups are direct products of isomorphic simple groups, see [Reference Aschbacher2, (8.3)]. By lemma 2.4, we cannot have an abelian minimal normal subgroup. Hence every minimal normal subgroup is the direct product of non-abelian simple groups, the socle S of G is non-abelian and the Fitting subgroup trivial. In particular, S equals the generalized Fitting subgroup which is the unique largest normal semisimple subgroup.

Next, write S as the internal direct product $\prod_{i=1}^n A_i$ with A_i a minimal normal subgroup of G. Using lemma 2.2, we see that $G/A_1$ is Dedekind which is only possible if all $A_i=1$ for i ≠ 1. Thus S is the unique minimal normal subgroup. Moreover, S is the direct product of isomorphic non-abelian simple groups. Furthermore, as S is non-abelian, lemma 2.2 also yields that $G /S$ is Dedekind.

Conversely, suppose $S \leq G $ is the unique minimal normal subgroup of G, that S is non-abelian and $G/S$ is Dedekind. Now, every $N \unlhd G$ contains the unique minimal normal subgroup S. Hence if $Y \leq G$, then $S \leq NY$ and so $NY / S \leq G/S$ is normal. Therefore, $NY \unlhd G$ as needed.

Finally, suppose that G has SN and $S = \operatorname{Soc}(G)$ is simple. As S is also the generalized Fitting subgroup, it contains its own centralizer. In other words, $C_G(S) = \mathcal{Z}(S)$ is trivial and hence G acts faithfully on S. Thus one may identify G with a subgroup of ${\rm Aut} (S)$ with S simple, i.e. G is almost simple.

Proof. Following [Reference Jespers and del Rio24, Theorem 3.5.5], ${\mathbb Q} G e(G,H,K) \cong M_{[G : N_G(K)]}({\mathbb Q}(\zeta_{[H:K]}) \star N_G(K)/H)$ for some crossing that can be made explicit (see [Reference Jespers and del Rio24, Remark 3.5.6]). Therefore, one has that

\begin{align*} dim_{{\mathbb Q}} {\mathbb Q} G e(G,H,K) & = [G : N_G(K)] ^2 \phi([H:K]) [N_G(K) : H] \\ & = [G:H] [G:N_G(K)] \phi([H:K]). \end{align*}

Proof. As A is nilpotent, i.e. $A^n=0$ for some n, the multiplicativity of the determinant gives $\det(A)=0$, implying xw = yz. Hence,

\begin{equation*}A^2 = \begin{pmatrix} x^2 + yz & xy + yw \\ xz + zw & w^2 + yz \end{pmatrix} = \begin{pmatrix} x^2 + xw & y(x + w) \\ z(x + w) & w^2 + xw \end{pmatrix} = \text{tr}(A)A. \end{equation*}

So, $0 = \text{tr}(A)^{n-1}A$, which gives $\text{tr}(A) = 0$. Hence, $x=-w$ and from xw = yz we also get $x^2=-yz$.

Proof of Theorem 3.6

Let

\begin{equation*}G(2,2,3) = G = \langle a, b \ | \ a^4 = b^8 = 1, \ a^b = a^{-1} \rangle. \end{equation*}

We note that $G' = \langle a^2 \rangle$ and $\mathcal{Z}(G) = \langle a^2, b^2 \rangle$. Then $G/G' \cong C_2 \times C_8$, so the algebra $\mathbb{Q}G$ has a direct summand

\begin{equation*}\mathbb{Q}[C_2 \times C_8] \cong 4\mathbb{Q} \oplus 2\mathbb{Q}(i) \oplus \mathbb{Q}(\zeta_8).\end{equation*}

Moreover, $G/\langle a^2b^2 \rangle \cong Q_8$ and $G/\langle b^2 \rangle \cong D_8$, so that $\mathbb{Q}G$ has also direct summands $\mathbb{H}_{\mathbb{Q}}$, the standard rational quaternions, and $M_2(\mathbb{Q})$. Moreover, $G/\langle a^2b^4 \rangle$ is a group of order 16 sometimes denoted by $D_{16}^+$. The rational group algebra of this group has one matrix component isomorphic to $M_2(\mathbb{Q}(i))$. We mention that it has been used in [Reference Bächle, Maheshwary and Margolis5] to solve another problem on integral group rings. Overall

\begin{equation*} \mathbb{Q}G \cong 4\mathbb{Q} \oplus 2\mathbb{Q}(i) \oplus \mathbb{Q}(\zeta_8) \oplus \mathbb{H}_{\mathbb{Q}} \oplus M_2(\mathbb{Q}) \oplus M_2(\mathbb{Q}(i)). \end{equation*}

This can also be easily checked using GAP [45] and the wedderga package therein [Reference Bakshi, Broche Cristo, Herman, Konovalov, Maheshwary, Olivieri, Olteanu, del Rio and Van Gelder7].

Furthermore, the following representations correspond to the non-commutative components (in the order as above):

(7)\begin{align} &G \rightarrow \mathbb{H}_{\mathbb{Q}}, \ \ a \mapsto i, \ \ b \mapsto j, \nonumber \\ &G \rightarrow M_2(\mathbb{Q}), \ \ a \mapsto \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \ \ b \mapsto \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix},\\ &G \rightarrow M_2(\mathbb{Q}(i)), \ \ a \mapsto \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix}, \ \ b \mapsto \begin{pmatrix} 0 & 1 \\ i & 0 \end{pmatrix}. \nonumber \end{align}

Here we denote by i and j the standard generators of $\mathbb{H}_{\mathbb{Q}}$.

We first derive the properties which are equivalent to having a nilpotent element in $\mathbb{Z}G$. Let n be a generic nilpotent element in $\mathbb{Z}G$ and write

\begin{equation*}n = \sum_{g \in G} \alpha(g) g. \end{equation*}

In the quotient $G/G'$, the element n must map to 0 which is equivalent to the fact that for each $h \in G$ one has $\sum_{g \in G'} \alpha(hg) = 0$. This is, in turn, equivalent to

(8)\begin{equation} \alpha(g) = - \alpha(ga^2) \ \forall g \in G, \end{equation}

since $G' = \langle a^2 \rangle$. As in the three non-commutative components the element a ² is always send to −1, this will always give a factor 2 in the considerations below and reduces the number of indeterminants by half. So we will always replace an expression of shape $\alpha(g) - \alpha(ga^2)$ by $2\alpha(g)$.

Next we consider the $\mathbb{H}_{\mathbb{Q}}$-component where the projection of n must equal 0. With the representation above n is sent to

\begin{align*} 2&(\alpha(1) - \alpha(b^2) + \alpha(b^4) - \alpha(b^6) \\ +&i(\alpha(a) - \alpha(ab^2) + \alpha(ab^4) - \alpha(ab^6)) \\ +&j(\alpha(b) - \alpha(b^3) + \alpha(b^5) - \alpha(b^7)) \\ +&ij(\alpha(ab) - \alpha(ab^3) + \alpha(ab^5) - \alpha(ab^7))). \end{align*}

Setting this equal to 0 gives the equations

(9)\begin{align} \alpha(1) &= \alpha(b^2) + \alpha(b^6) - \alpha(b^4), \nonumber \\ \alpha(a) &= \alpha(ab^2) + \alpha(ab^6) - \alpha(ab^4), \nonumber \\ \alpha(b) &= \alpha(b^3) + \alpha(b^7) - \alpha(b^5),\\ \alpha(ab) &= \alpha(ab^3) + \alpha(ab^7) - \alpha(ab^5). \nonumber \end{align}

Together with (8), this can be written compactly as

(10)\begin{align} \alpha(g) + \alpha(gb^4) = \alpha(gb^2) + \alpha(gb^6) \ \forall g \in G. \end{align}

We denote the representation of n in the $M_2(\mathbb{Q})$-component by $\begin{pmatrix} x_1 & y_1 \\ z_1 & w_1 \end{pmatrix}$ and in the $M_2(\mathbb{Q}(i))$-component by $\begin{pmatrix} x_2 & y_2 \\ z_2 & w_2 \end{pmatrix}$.

From the representation given above, we get

\begin{equation*}x_1 = 2(\alpha(1) + \alpha(b^2) + \alpha(b^4) + \alpha(b^6) - \alpha(b) - \alpha(b^3) - \alpha(b^5) - \alpha(b^7)) \end{equation*}

which using (9) transforms to

(11)\begin{equation} x_1 = 4(\alpha(b^2) + \alpha(b^6) - \alpha(b^3) - \alpha(b^7)). \end{equation}

Similarly, we get

(12)\begin{equation} w_1 = 4(\alpha(b^2) + \alpha(b^6) + \alpha(b^3) + \alpha(b^7)). \end{equation}

As one of our conditions on n is $x_1 = -w_1$ by lemma 3.7 this gives, using also (9),

(13)\begin{equation} \alpha(b^2) = -\alpha(b^6), \ \alpha(1) = -\alpha(b^4). \end{equation}

Furthermore, we compute

\begin{equation*}y_1 = 2(-\alpha(a) - \alpha(ab^2) - \alpha(ab^4) - \alpha(ab^6) -\alpha(ab) - \alpha(ab^3) - \alpha(ab^5) - \alpha(ab^7)) \end{equation*}

which we convert using (9) to

(14)\begin{equation} y_1 = 4(-\alpha(ab^2) - \alpha(ab^6) - \alpha(ab^3) - \alpha(ab^7)). \end{equation}

Similarly

(15)\begin{equation} z_1 = 4(\alpha(ab^2) + \alpha(ab^6) - \alpha(ab^3) - \alpha(ab^7)). \end{equation}

We now calculate the $\mathbb{Q}(i)$-representation. We get

\begin{align*} x_2 &= 2(\alpha(1) - \alpha(b^4) + \alpha(ab^2) - \alpha(ab^6) \\ & \quad + i(-\alpha(a) + \alpha(ab^4) + \alpha(b^2) - \alpha(b^6))). \end{align*}

Using (9) and (13), this becomes

\begin{align*} x_2 &= 2(-2\alpha(b^4) + \alpha(ab^2) - \alpha(ab^6) \\ &\quad + i(2\alpha(ab^4) - 2\alpha(b^6) - \alpha(ab^2) - \alpha(ab^6))). \end{align*}

Similarly,

\begin{align*} w_2 &= 2(-2\alpha(b^4) - \alpha(ab^2) + \alpha(ab^6) \\ &\quad + i(-2\alpha(ab^4) - 2\alpha(b^6) + \alpha(ab^2) + \alpha(ab^6))). \end{align*}

With the condition $x_2 = -w_2$ from lemma 3.7, this gives $\alpha(b^4) = 0 = \alpha(b^6)$ and together with (13) we conclude

(16)\begin{equation} \alpha(1) = \alpha(b^2) = \alpha(b^4) = \alpha(b^6) = 0 \end{equation}

and

(17)\begin{align} x_2 &= 2(\alpha(ab^2) - \alpha(ab^6)\\ &\quad + i(2\alpha(ab^4)- \alpha(ab^2) - \alpha(ab^6))) \nonumber \end{align}

as well as

(18)\begin{align} w_2 &= 2( - \alpha(ab^2) + \alpha(ab^6)\\ &\quad + i(-2\alpha(ab^4) + \alpha(ab^2) + \alpha(ab^6))). \nonumber \end{align}

We compute the other coefficients as

\begin{align*} y_2 &= 2(\alpha(b) - \alpha(b^5) + \alpha(ab^3) - \alpha(ab^7) \\ & \quad + i(\alpha(b^3) - \alpha(b^7) - \alpha(ab) + \alpha(ab^5)) \end{align*}

which by (9) transforms to

(19)\begin{align} y_2 = & 2(-2\alpha(b^5) + \alpha(b^3) + \alpha(b^7) + \alpha(ab^3) - \alpha(ab^7)\\ & + i(2\alpha(ab^5) + \alpha(b^3) - \alpha(b^7) -\alpha(ab^3) -\alpha(ab^7) )). \nonumber \end{align}

Similarly, also by (9),

(20)\begin{align} z_2 = & 2(2\alpha(ab^5) - \alpha(b^3) + \alpha(b^7) - \alpha(ab^3) - \alpha(ab^7)\\ & + i(-2\alpha(b^5) + \alpha(b^3) + \alpha(b^7) - \alpha(ab^3) + \alpha(ab^7) )). \nonumber \end{align}

Moreover, from (11) and (16), we have

(21)\begin{equation} x_1 = -4(\alpha(b^3)+\alpha(b^7)). \end{equation}

We now compute the quadratic equations from lemma 3.7. Then

(22)\begin{equation} x_1^2 = 16(\alpha(b^3)^2 + 2\alpha(b^3)\alpha(b^7) + \alpha(b^7)^2) \end{equation}

and from (14) and (15) we get

(23)\begin{align} -y_1z_1 &= -16(-\alpha(ab^2)^2 - 2\alpha(ab^2)\alpha(ab^6) - \alpha(ab^6)^2 + \alpha(ab^3)^2 \nonumber\\ &\quad + 2\alpha(ab^3)\alpha(ab^7) + \alpha(ab^7)^2). \end{align}

The analogues equations for the $M_2(\mathbb{Q}(i))$-component give by (17)

(24)\begin{align} x_2^2 &= 8(-2\alpha(ab^4)^2 + 2\alpha(ab^2)\alpha(ab^4) + 2\alpha(ab^6)\alpha(ab^4) - 2\alpha(ab^2)\alpha(ab^6)\\ & \quad + i(2\alpha(ab^2)\alpha(ab^4) - 2\alpha(ab^6)\alpha(ab^4) - \alpha(ab^2)^2 + \alpha(ab^6)^2 )) \nonumber \end{align}

and by (19) and (20)

(25)\begin{align} -y_2z_2 &= -8(2\alpha(b^3)\alpha(b^5) - 2\alpha(b^7)\alpha(b^5) - \alpha(b^3)^2 + \alpha(b^7)^2 \nonumber \\ & \quad +2\alpha(ab^3)\alpha(ab^5) -2 \alpha(ab^7)\alpha(ab^5) - \alpha(ab^3)^2 + \alpha(ab^7)^2\\ & \quad + 2i(\alpha(ab^5)^2 - \alpha(ab^3)\alpha(ab^5) - \alpha(ab^7)\alpha(ab^5) \nonumber \\ & \quad + \alpha(b^5)^2 - \alpha(b^3)\alpha(b^5) - \alpha(b^7)\alpha(b^5) \nonumber \\ & \quad + \alpha(b^3)\alpha(b^7) + \alpha(ab^3)\alpha(ab^7) )). \nonumber \end{align}

We now show certain congruences modulo 2 which will provide the key for the final argument. First note that the imaginary part of $-y_2z_2$ is divisible by 16. So this is also true for the imaginary part of $x_2^2$ implying $-\alpha(ab^2)^2 + \alpha(ab^6)^2 \equiv 0 \bmod 2$ which means

(26)\begin{equation} \alpha(ab^2) \equiv \alpha(ab^6) \mod 2. \end{equation}

We next show that $\alpha(b^3) \equiv \alpha(b^7) \bmod 2$ and also $\alpha(ab^3) \equiv \alpha(ab^7) \bmod 2$. Assume that $\alpha(b^3) \not\equiv \alpha(b^7) \bmod 2$. Then one of them is even and the other is odd which implies by (22) that $\frac{x_1^2}{16} \equiv 1 \bmod 4$. Note that (26) implies that

\begin{equation*}\alpha(ab^2) + 2\alpha(ab^2)\alpha(ab^6) + \alpha(ab^6)^2 = (\alpha(ab^2) + \alpha(ab^6))^2 \equiv 0 \mod 4. \end{equation*}

So from (23)

\begin{equation*}\frac{-y_1z_1}{16} \equiv -(\alpha(ab^3)^2 + 2\alpha(ab^3)\alpha(ab^7) + \alpha(ab^7)^2) = -(\alpha(ab^3) + \alpha(ab^7))^2 \mod 4 \end{equation*}

which can only be congruent to 0 or −1 modulo 4, contradicting $x_1^2 = -y_1z_1$. Hence

(27)\begin{equation} \alpha(b^3) \equiv \alpha(b^7) \mod 2. \end{equation}

We now consider the real parts of $x_2^2$ and $-y_2z_2$. The real part of $x_2^2$ is divisible by 16. So by (25) and (27)

\begin{equation*}0 \equiv Re(-y_2z_2) \equiv -8(-\alpha(ab^3)^2 + \alpha(ab^7)^2) \mod 16\end{equation*}

which implies

(28)\begin{equation} \alpha(ab^3) \equiv \alpha(ab^7) \mod 2. \end{equation}

Together with (8), (10), and (16), the congruences (26), (27), and (28) can be compactly written as

(29)\begin{align} \alpha(g) + \alpha(gb^4) \equiv 0 \mod 2 \ \ \forall g \in G. \end{align}

These are all the equations and congruences we need.

Let now $e\in \operatorname{PCI}({\mathbb Q} G) $. If e corresponds to a component which is not a matrix component, then ne = 0 which is clearly an element in $\mathbb{Z}G$. To analyse the other elements of $\operatorname{PCI}({\mathbb Q} G)$, we will deploy lemma 2.7. Let first $e \in \operatorname{PCI}({\mathbb Q} G)$ be the element corresponding to the $M_2(\mathbb{Q})$-representation and let χ be its character. Then from the representation given above we get

\begin{equation*} \chi(g) = \left\{\begin{array}{lll} 2, & g \in \langle b^2 \rangle, \\ -2, & g \in a^2\langle b^2 \rangle,\\ 0, & \text{else}. \end{array}\right.\end{equation*}

So by lemma 2.7 we can compute the coefficient of ne at a generic element $g \in G$ in the following way, where we use first the values of χ, then (8) and then (10):

\begin{align*} \frac{k}{|G|}\sum_{h \in G}&\alpha(gh^{-1})\chi(h^{-1}) \\ &= \frac{2}{32}\left(2(\alpha(g) + \alpha(gb^2) + \alpha(gb^4) + \alpha(gb^6) \right.\\ &\quad \left.- \alpha(ga^2) - \alpha(ga^2b^2) -\alpha(ga^2b^4) - \alpha(ga^2b^6))\right)\\ &= \frac{1}{8}\left(2(\alpha(g) + \alpha(gb^2) + \alpha(gb^4) + \alpha(gb^6))\right) \\ &= \frac{1}{4}\left(2(\alpha(g)+\alpha(gb^4))\right) = \frac{1}{2}\left(\alpha(g) + \alpha(gb^4)\right). \end{align*}

By (29), all these numbers are integers and hence $ne \in \mathbb{Z}G$.

Finally let $e \in \operatorname{PCI}({\mathbb Q} G)$ be the element corresponding to the component $M_2({\mathbb Q} (i))$ and χ its character. The argument will be similar to the previous case. Note that we consider χ as a character of a $\mathbb{Q}$-representation, so that each of the entries in the representation given in (7) corresponds to a $2\times 2$-matrix and i corresponds to a matrix with trace 0, e.g. to its rational canonical form $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. Hence

\begin{equation*} \chi(g) = \left\{\begin{array}{lll} 4, & g \in \langle a^2b^4 \rangle, \\ -4, & g \in \{a^2, b^4 \}, \\ 0, & \text{else}. \end{array}\right.\end{equation*}

We again use lemma 2.7 to compute the coefficient of ne at g, where we use first the values of χ and then (8):

\begin{align*} \frac{k}{|G|}\sum_{h \in G}&\alpha(gh^{-1})\chi(h^{-1}) = \frac{2}{32}\left(4(\alpha(g) + \alpha(ga^2b^4) - \alpha(ga^2) - \alpha(gb^4))\right) \\ &= \frac{1}{4}\left(2(\alpha(g) + \alpha(gb^4))\right) = \frac{1}{2}\left(2\alpha(g)+\alpha(gb^4)\right). \end{align*}

Hence again by (29), all the coefficients of ne are integers. Overall we conclude that G has the (ND) property.

Proof. By (i), we have rs = sr, so that by (i) $(y(r+s)/p)^2 = y^2(r+s)^2/p^2 = 0$. So by (iii) $y(r+s)/p$ is a nilpotent element in $\mathbb{Z}G$ and moreover $ey(r+s)/p = yr/p$ by (ii). So by (iv) $(y(r+s)/p$ is non-zero and G does not have ND.

Proof. In this situation, it was already shown by Liu that G does not have ND when either m = 2 and $n \geq 4$ or m = 3 [Reference Liu30, Lemma 2.8]. While Liu’s statement of the lemma is different, the proof shows exactly that these groups do not have (ND). As our proof is uniform for all cases, this will include a repetition of Liu’s result. The goal is to use lemma 3.8 and the items (i)–(iv) refer to this lemma. We set p = 2.

Let

\begin{align*} r &= a(a^{p^{m-1}}+b)(1-a^{p^{m-1}})(1+b^p)\widetilde{b^{p^2}}, \\ s &= a(a^{p^{m-2}}+b)(1-a^{p^{m-1}})(1-b^p)\widetilde{b^{p^2}}, \\ y &= 1 + a^{p^{m-2}}. \end{align*}

Note that

\begin{equation*}(1+b^p)(1-b^p)\widetilde{b^{p^2}} = (1-b^{p^2})\widetilde{b^{p^2}} = 0, \end{equation*}

so that $rs = sr = 0$ follows using that b^p is central in G. Furthermore,

\begin{align*} a&(a^{p^{m-1}} + b)a(a^{p^{m-1}}+b) = a^p(a^{p^{m-1}} + ba^{p^{m-1}})(a^{p^{m-1}}+b) \\ &= a^p(1 + ba^{p^{m-1}} + b + b^pa^{p^{m-1}}) = a^p((1-b^p) + (1+a^{p^{m-1}})(b+b^p)). \end{align*}

As

(30)\begin{equation} (1-b^p)(1+b^p)\widetilde{b^{p^2}} = 0 = (1+a^{p^{m-1}})(1-a^{p^{m-1}}), \end{equation}

this implies $r^2 = 0$. Moreover,

\begin{align*} a&(a^{p^{m-2}}+b)a(a^{p^{m-2}}+b) = a^p (a^{p^{m-2}} + ba^{p^{m-1}})(a^{p^{m-2}} + b) \\ &= a^p (a^{p^{m-1}} + ba^{p^{m-2}} + ba^{p^{m-1} + p^{m-2}} + b^pa^{p^{m-1}}) \\ &= a^p(a^{p^{m-1}}(1+b^p) + ba^{p^{m-2}}(1 + a^{p^{m-1}})) \end{align*}

which by (30) also implies $s^2 = 0$. So (i) follows.

To construct the idempotent for (ii), note again that b^p is central in G so that if $f \in \operatorname{PCI}(\mathbb{Q}G)$, then $b^pf = \zeta I$ for a certain root of unity ζ, where I denotes the identity matrix of some size. The same argument applies to $a^{p^{m-1}}$. Now let $e \in \operatorname{PCI}(\mathbb{Q}G)$ which is the sum of all primitive central idempotents f such that $fa^{p^{m-1}}$ has order p and fb^p is the identity. Note that as $G' = \langle a^{p^{m-1}} \rangle$, this implies that fb is not central. So for each such f we have rf ≠ 0. It is clear that se = 0 as $(1-b^p)e=0$. Furthermore, if fʹ is any primitive central idempotent such that $f'b^p$ is not the identity, then $f'(1+b^p) \widetilde{b^{p^2}} = 0$. Similarly, if $f'a^{p^{m-1}}$ does not have order p, it must be the identity, so $f'(1-a^{p^{m-1}}) = 0$. We conclude $r(1-e) = 0$ and so re = r and (ii) holds.

Next,

\begin{align*} y(r+s) &= ya(1-a^{p^{m-1}})\widetilde{b^{p^2}}((a^{p^{m-1}} + b)(1+b^p) + (a^{p^{m-2}} + b)(1-b^p)) \\ &= ya(1-a^{p^{m-1}})\widetilde{b^{p^2}}(a^{p^{m-1}} + b^pa^{p^{m-1}} + a^{p^{m-2}} - b^pa^{p^{m-2}} + 2b) \\ &= ya(1-a^{p^{m-1}})\widetilde{b^{p^2}}((a^{p^{m-2}}(1 + a^{p^{m-2}}) + b^pa^{p^{m-2}}(-1 + a^{p^{m-2}}) + 2b). \end{align*}

Using that $1 + a^{p^{m-2}} \equiv -1 + a^{p^{m-2}} \equiv 1 - a^{p^{m-2}} \bmod 2$ this means

\begin{align*} y(r+s) &\equiv a(1+a^{p^{m-2}})(1-a^{p^{m-2}})(1-a^{p^{m-1}})\widetilde{b^{p^2}}(a^{p^{m-2}} + b^pa^{p^{m-2}}) \\ &= a(1 - a^{p^{m-1}})(1-a^{p^{m-1}})\widetilde{b^{p^2}}(a^{p^{m-2}} + b^pa^{p^{m-2}}) \\ &\equiv a(1 + a^{p^{m-1}})(1-a^{p^{m-1}})\widetilde{b^{p^2}}(a^{p^{m-2}} + b^pa^{p^{m-2}}) = 0 \mod 2. \end{align*}

Hence $y(r+s)/p \in \mathbb{Z}G$ and (iii) holds.

Finally, it is easy to see that the coefficient of ab in the element

\begin{equation*}yr = (1+a^{p^{m-2}})a(a^{p^{m-1}} + b)(1-a^{p^{m-1}})(1+b^p)\widetilde{b^{p^2}}\end{equation*}

is 1, so that $yr/p \notin \mathbb{Z}G$. This proves (iv) and the fact that G does not have (ND) hence follows.

Proof. To simplify notation, we write for a positive integer $\ell$:

\begin{equation*}\gamma(x,\ell) = 1 + x + x^{2} + \cdots + x^{\ell}.\end{equation*}

We will several times use the equation

(31)\begin{equation} (1-x^{\ell}) = (1-x)\gamma(x, \ell-1). \end{equation}

We will also use that

(32)\begin{equation} (1-x^p) \equiv (1-x)^p \mod p. \end{equation}

We proceed to show the first congruence using first $1-x^{-p} = -x^{-p}(1-x^p)$ and later (31) and finally (32). We also use $x^{-p} = x^{p^2-p}$ and $x^{-3p} = x^{p^2-3p}$.

\begin{align*} (&1-x^p)(1-x^p z) + (1-x^{-p})(1-x^{-p} z) \\ =& (1-x^p)(1-x^p z) - x^{-p}(1-x^p)(1-x^{-p} z) \\ =& (1-x^p)(1-x^p z - x^{-p}(1-x^{-p} z)) \\ =& (1-x^p)((1-x^{-p}) - x^p z (1-x^{-3p})) \\ =& (1-x^p)((1-x)\gamma(x, p^2-p-1) - x^p z(1-x)\gamma(x, p^2-3p-1)) \\ =& (1-x^p)(1-x) (\gamma(x, p^2-p-1) - x^p z\gamma(x, p^2-3p-1)) \\ \equiv & (1-x)^{p+1}(\gamma(x, p^2-p-1) - x^p z \gamma(x, p^2-3p-1)) \mod p. \end{align*}

Note that when p = 3, then in the fourth line $1-x^{-3p} = 0$, so that $p^2-3p-1$ does not appear later in this case. This shows the first congruence.

Next, we show the second congruence, also using $1-x^{-p} = -x^{-p}(1-x^p)$, (31) and (32) and also that $x^{-3p-2} = x^{p^2-3p-2}$ for p ≠ 3 and $x^{-3p-2} = x^{p^2-2}$ for p = 3:

\begin{align*} & x(1-x^p)(1-x^p z) + x^{-1}(1-x^{-p})(1-x^{-p} z) \\ =& x(1-x^p)(1-x^p z) - x^{-1}x^{-p}(1-x^p)(1-x^{-p} z) \\ =& (1-x^p)(x(1-x^p z) - x^{-p-1} (1-x^{-p} z)) \\ =& (1-x^p)(x - x^{-p-1} - x^{p + 1}z + x^{-2p-1} z) \\ =& (1-x^p)(x(1-x^{-p-2}) - x^{p + 1} z(1-x^{-3p-2})) \\ =& (1-x^p)(1-x)(x\gamma(x, p^2-p-3) - x^{p + 1} z\gamma(x, p^2-3p-3)) \\ \equiv& (1-x)^{p+1}(x\gamma(x, p^2-p-3) - x^{p + 1}\gamma(x, p^2-3p-3)) \mod p, \end{align*}

where in the last two lines in case p = 3 the expression $p^2-3p-3$ has to be replaced by $p^2-3$. This shows the second congruence.

Proof. Set

\begin{align*} r &= (b - a^{p^{m-2}})a(1-a^{p^{m-1}}) \widetilde{b^{p^2}} \prod_{i=0}^{p-2}(1-b^pa^{ip^{m-1}}), \\ s &= (b- a^{-p^{m-2}})a(1-a^{-p^{m-1}}) \widetilde{b^{p^2}} \prod_{i=2}^p (1- b^pa^{ip^{m-1}}),\\ y &= (1-a^{p^{m-2}})^{p(p-1)-1}. \end{align*}

We will again show (i)–(iv) from lemma 3.8 which will imply that G does not have (ND). We analyse the irreducible representations of G in which r and s are not mapped to 0. Note that as a^p and b^p are central, they are mapped to a central matrix under every irreducible representation. Let R be an irreducible $\mathbb{Q}$-representation and e the corresponding primitive central idempotent of $\mathbb{Q}G$. Denote by I the identity matrix. If $R(b^{p^2}) \neq I$, then $e\widetilde{b^{p^2}} = 0$, as the sum over all the powers of a primitive $p^\ell$th root of unity equals 0 when $\ell \geq 1$. Moreover $R(b^p) = I$ implies $e(1-b^p) = 0$. Hence, as both r and s contain the factor $(1-b^p)\widetilde{b^{p^2}}$ the inequality er ≠ 0 implies that $R(b^p) = \zeta I$ for ζ a primitive pth root of unity while es ≠ 0 implies $R(b^p) = \zeta'I$ for $\zeta'$ a primitive pth root of unity. Moreover, if $R(a^{p^{m-1}}) = I$, then $e(1-a^{p^{m-1}}) = 0$. As $a^{p^{m-1}}$ has order p, we conclude that er ≠ 0 implies $R(a^{p^{m-1}}) = \xi I$ while es ≠ 0 implies $R(a^{p^{m-1}}) = \xi'I$ for certain primitive pth roots of unity ξ and $\xi'$. The factor $\prod_{i=1}^{p-2} (1-b^p a^{ip^{m-1}})$ in r means that er ≠ 0 implies $R(b^p) \neq R(a^{ip^{m-1}})^{-1}$ for every $1 \leq i \leq p-2$. From the fact that both b^p and $a^{p^{m-1}}$ are mapped to elements of order p, we conclude that er ≠ 0 means $R(b^p) = R(a^{p^{m-1}})$, i.e. $eb^p = ea^{p^{m-1}}$. Similarly $es \ne 0$ implies $eb^p = ea^{-p^{m-1}}$. It follows that r and s live in different components of $\mathbb{Q}G$, so that $rs = sr = 0$ and also (ii) holds.

We next show that $r^p = 0$. The non-central factors of r give

\begin{align*}((b-a^{p^{m-2}})a)^p &= a^p(b - a^{p^{m-2}})(ba^{(p-1)p^{m-1}} - a^{p^{m-2}})(ba^{(p-2)p^{m-1}} - a^{p^{m-2}})\\ &\quad \cdots(ba^{p^{m-1}} - a^{p^{m-2}}). \end{align*}

From the paragraph before we know that when er ≠ 0, then $R(a^{p^{m-1}}) = \zeta I$ for some primitive pth root of unity ζ. So then

\begin{equation*}e((b-a^{p^{m-2}})a)^p = ea^p(b - a^{p^{m-2}})(b\zeta^{-1} - a^{p^{m-2}})(b\zeta^{-2} - a^{p^{m-2}})\cdots(b\zeta - a^{p^{m-2}}). \end{equation*}

We claim that the coefficient of $a^{ip^{m-2}}$ in the expression

\begin{equation*}(b - a^{p^{m-2}})(b\zeta^{-1} - a^{p^{m-2}})(b\zeta^{-2} - a^{p^{m-2}})\cdots(b\zeta - a^{p^{m-2}})\end{equation*}

is 0 for every $1 \leq i \leq p-1$. Indeed, using $\prod_{j=0}^{p-1}(X-\zeta^j) = X^p - 1$, as an equation in the polynomial ring $\mathbb{Z}[X]$, up to the factor bⁱ and possibly a sign this coefficient is the same as in $\prod_{j=0}^{p-1} (a^{p^{m-2}} - \zeta^j) = (a^{p^{m-2}})^p - 1$. So,

\begin{equation*}e((b-a^{p^{m-2}})a)^p = ea^p(b^p - a^{p^{m-1}}) = 0, \end{equation*}

where the last equality follows from the previous paragraph. Hence, $r^p = 0$. A similar calculation shows also s^p, so that (i) follows.

We proceed to show (iii). We first calculate

\begin{align*} y(r+s) &= y\widetilde{b^{p^2}}(1-b^p)\prod_{i=2}^{p-2}(1-a^{ip^{m-1}}b^p) \\ & \quad \cdot ((b-a^{p^{m-2}})a(1-a^{p^{m-1}})(1-a^{p^{m-1}}b^p) \\ &\quad + (b-a^{-p^{m-2}})a(1-a^{-p^{m-1}})(1-a^{-p^{m-1}}b^p)) \\ &= y\widetilde{b^{p^2}}(1-b^p)\prod_{i=2}^{p-2}(1-a^{ip^{m-1}}b^p) \\ &\quad \cdot (ba((1-a^{p^{m-1}})(1-a^{p^{m-1}}b^p) + (1-a^{-p^{m-1}})(1-a^{-p^{m-1}}b^p)) \\ & \quad -a(a^{p^{m-2}}(1-a^{p^{m-1}})(1-a^{p^{m-1}}b^p)\\ &\quad +a^{-p^{m-2}}(1-a^{-p^{m-1}})(1-a^{-p^{m-1}}b^p))). \end{align*}

So the last two lines mean that, taking into account the factor y, it is enough to show that

(33)\begin{equation} y ((1-a^{p^{m-1}})(1-a^{p^{m-1}}b^p) + (1-a^{-p^{m-1}})(1-a^{-p^{m-1}}b^p)) \equiv 0 \mod p \end{equation}

and

(34)\begin{align} &y(a^{p^{m-2}}(1-a^{p^{m-1}})(1-a^{p^{m-1}}b^p) \nonumber\\ &\quad + a^{-p^{m-2}}(1-a^{-p^{m-1}})(1-a^{-p^{m-1}}b^p)) \equiv 0 \mod p. \end{align}

Note that

\begin{align*}y(1-a^{p^{m-2}})^{p+1} &= (1-a^{p^{m-2}})^{p(p-1)-1+p+1} = (1-a^{p^{m-2}})^{p^2} \\ &\equiv (1 - a^{p^m}) = 0 \mod p\end{align*}

by (32). So to prove (33) and (34), it is enough to show that the factors to the right of y contain a factor $(1-a^{p^{m-2}})^{p+1}$ modulo p. This follows by applying lemma 3.11 with $x = a^{p^{m-2}}$ and $z = b^p$. This shows (iii).

Finally, to get (iv), note that the coefficient of ba in yr is 1. This follows as none of the products one can get by factoring out the element yr gives ba except the trivial one which in turns follows as the powers appearing for a and b are otherwise not big enough to sum up to p^m or pⁿ, respectively, when a and b are both taken in as a factor.

Proof. Again we will show (i)–(iv) from lemma 3.11, this time using the elements:

\begin{align*} r &= (a-b^{p^{n-2}})b(1-b^{p^{n-1}})\prod_{i=0}^{p-2}(1-a^pb^{ip^{n-1}}), \\ s &= (a-b^{-p^{n-2}})b(1-b^{-p^{n-1}})\prod_{i=2}^p(1-a^pb^{ip^{n-1}}), \\ y & = (1-b^{p^{n-2}})^{p(p-1)-1}. \end{align*}

We again first analyse the properties of primitive central idempotents which do not map r or s to 0. Note that a^p and $b^{p^{n-1}}$ both have order p. Let $e \in \operatorname{PCI}(\mathbb{Q}G)$. The factor $(1-a^p)$ in both r and s means that er ≠ 0 implies that ea^p has order p and also es ≠ 0 means that ea^p has order p. From the factor $(1-b^{p^{n-1}})$ in r and the factor $(1-b^{-p^{n-1}})$ in s, we also get that er ≠ 0 implies that $eb^{p^{n-1}}$ has order p and es ≠ 0 implies the same. Finally, the rest of the factors appearing on the right from b then mean that er ≠ 0 implies $ea^p = eb^{p^{n-1}}$ and es ≠ 0 implies $ea^p = eb^{-p^{n-1}}$. This, in particular, gives that $rs = sr = 0$ and (ii).

Computing r^p and s^p is also very similar to the previous case. Namely the non-central part of r gives

\begin{equation*}((a-b^{p^{n-2}})b)^p = b^p(a-b^{p^{n-2}})(aa^p-b^{p^{n-2}})(aa^{2p}-b^{p^{n-2}})\cdots(aa^{(p-1)p} - b^{p^{n-2}}).\end{equation*}

As ea^p has order p when er ≠ 0, the coefficient of $(b^{p^{n-2}})^i$ in the expression

\begin{equation*}e(a-b^{p^{n-2}})(aa^p-b^{p^{n-2}})(aa^{2p}-b^{p^{n-2}})\cdots(aa^{(p-1)p} - b^{p^{n-2}})\end{equation*}

is 0 for all $1 \leq i \leq p-1$, so that $e((a-b^{p^{n-2}})b)^p = eb^p(a^p - b^{p^{n-1}}) = 0$, implying $r^p = 0$. Similarly $s^p = 0$. Overall, we obtain (i).

Now,

\begin{align*} y(r+s) &= y(1-a^p)\prod_{i=2}^{p-2}(1-a^pb^{ip^{n-1}}) \\ &\quad \cdot ((a-b^{p^{n-2}})b(1-b^{p^{n-1}})(1-a^pb^{p^{n-1}})\\ &\quad + (a-b^{-p^{n-2}})b(1-b^{-p^{n-1}})(1-a^pb^{-p^{n-1}})) \\ &= \ y(1-a^p)\prod_{i=2}^{p-2}(1-a^pb^{ip^{n-1}}) \\ &\quad \cdot (ab((1-b^{p^{n-1}})(1-a^pb^{p^{n-1}}) + (1-b^{-p^{n-1}})(1-a^pb^{-p^{n-1}})) \\ &\quad - b(b^{p^{n-2}}(1-b^{p^{n-1}})(1-a^pb^{p^{n-1}}) \\ &\quad + b^{-p^{n-2}}(1-b^{-p^{n-1}})(1-a^pb^{-p^{n-1}}))). \end{align*}

As $y(1-b^{p^{n-2}})^{p+1} = (1-b^{p^{n-2}})^{p(p-1)-1+p+1} = (1-b^{p^{n-2}})^{p^2} \equiv 0 \mod p$, it is hence enough to show that the expressions

(35)\begin{equation} (1-b^{p^{n-1}})(1-a^pb^{p^{n-1}}) + (1-b^{-p^{n-1}})(1-a^pb^{-p^{n-1}}) \end{equation}

and

(36)\begin{equation} b^{p^{n-2}}(1-b^{p^{n-1}})(1-a^pb^{p^{n-1}}) + b^{-p^{n-2}}(1-b^{-p^{n-1}})(1-a^pb^{-p^{n-1}}) \end{equation}

when considered modulo p both contain a factor $(1-b^{p^{n-2}})^{p+1}$. This follows by applying lemma 3.11 for $x = b^{p^{n-2}}$ and $z = a^p$. So we have (iii). Moreover analysing yr, we see that the coefficient of ab equals 1, so also (iv) follows and G does not have (ND) in this case.

Proof. We note that the case p = 3 was considered in [Reference Liu and Passman32, Lemma 2.2], but we will use different arguments. We will again show (i)–(iv) from lemma 3.11 using

\begin{align*} r =& (b-a)(1-a^p)\prod_{i=0}^{p-2}(1-a^{ip}b^p), \\ s =& (b-a^{-1})(1-a^{-p})\prod_{i=2}^p(1-a^{ip}b^p), \\ y =& (1-a)^{p(p-1)-1}. \end{align*}

The proof of the fact that $r^p = s^p = 0$ is different from the cases before, so that we postpone this to the end. We start again by analysing the properties of primitive central idempotents not annihilating r and s. Similarly as in the other cases, we get that er ≠ 0 implies that ea^p and eb^p have order p and $ea^p = eb^p$ holds. Also, es ≠ 0 implies that ea^p and eb^p have order p and $ea^p = eb^{-p}$. So $rs = sr = 0$ and (ii) follow.

Similarly as we had to show (33) and (34) before we now need to obtain that

\begin{equation*}(1-a^p)(1-a^pb^p) + (1-a^{-p})(1-a^{-p}b^p) \end{equation*}

and

\begin{equation*}a(1-a^p)(1-a^pb^p) + a^{-1}(1-a^{-p})(1-a^{-p}b^p) \end{equation*}

both contain a factor $(1-a)^{p+1}$ when considered modulo p. This follows by applying lemma 3.11 for x = a and $z = b^p$ and so we obtain (iii). It is also easy to see that the coefficient of b in yr is 1, so that (iv) holds.

It remains to prove $r^p = s^p = 1$. We first claim that there is exactly one $e \in \operatorname{PCI}(\mathbb{Q}G)$ such that er ≠ 0. To see this, note that, since the centre of G is $\langle a^p \rangle \times \langle b^p \rangle$ and isomorphic to an elementary abelian group of rank 2, for each possible kernel different from the derived subgroup $\langle a^p \rangle$ there can be only one component with centre $\mathbb{Q}(\zeta)$ for dimension reasons. Here ζ denotes a primitive pth root of unity. Also, there is exactly one $ \in \operatorname{PCI}({\mathbb Q} G)e$ such that es ≠ 0. We will work in these unique components using explicit representations to see that $r^p = s^p = 0$. Set

\begin{equation*}A = \begin{pmatrix} 0 & 0 & \cdots & 0 & 1 \\ \zeta & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & & \ddots & & \vdots \\ 0 & 0 & \cdots & 1 & 0 \end{pmatrix}\end{equation*}

and

\begin{equation*}C = \begin{pmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & \zeta^{-1} & 0 & \cdots & 0 \\ 0 & 0 & \zeta^{-2} & \cdots & 0 \\ \vdots & & & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \zeta \end{pmatrix}.\end{equation*}

Then $A^p = \zeta I$, where I denotes the identity matrix, and $C^{-1}AC = A^{p+1}$. Hence each map $R_i: G \rightarrow M_p(\mathbb{Q}(\zeta))$ sending a to A and b to A ⁱC for $0 \leq i \leq p-1$ is a representation of G, since also $(A^iC)^p = A^{ip}$ is an element of order p. In fact, all these representations are irreducible: the degree of the representations is the smallest non-trivial divisor of the order of G, so any non-trivial decomposition would involve only linear representations. But the linear representations contain the derived subgroup $\langle a^p \rangle$ in their kernel, hence the character values of a^p under the sum of p linear representations is p, while the character value of a^p under every of the representations R_i is $p\zeta$. We note also that when D is a diagonal matrix with one of the diagonal entries being 0, then $(A^iD)^p=0$. This follows, since the characteristic polynomial of A ⁱD equals $-X^p + \zeta^id_1d_2\cdots d_p$, where $d_1,\ldots,d_p$ are the diagonal elements of D, so that the only eigenvalue of A ⁱD is 0 when one of the d_j’s equals 0.

The $e \in \operatorname{PCI}({\mathbb Q} G)$ which satisfies er ≠ 0 corresponds to the representation R ₁, as in general $R_i(a^{ip}) = R_i(b^p)$ and as we saw before $ea^p = eb^p$. Now $R_1(a-b) = AC-A = A(C-I)$ and C − I is a diagonal matrix containing 0 on the diagonal. So, $r^p = 0$ follows. Similarly $R_{p-1}$ is the representation corresponding to the primitive central idempotent not annihilating s and $D_{p-1}(b-a^{-1}) = A^{p-1}C - A^{-1} = A^{-1}(A^pC-I)$ and as also $A^pC-I$ is a diagonal matrix containing 0 on the diagonal, we get $s^p= 0$ by the paragraph before. This finishes the proof of (i) in this case and the theorem follows.

Proof of Theorem 3.5

By Theorem C, it remains to consider nilpotent group which have SSN. As observed in [Reference Liu and Passman33, Section 4], these were in fact classified in [Reference Božikov and Janko12]. They fall into nine categories. For eight of these categories, it is shown in [Reference Jespers and Sun27, Section 4] that if G lies in one of them, it has ND if and only if $\mathbb{Q}G$ has at most one matrix component. The last category which remains open in general are the groups $G(p,m,n)$.

We already know by Theorem 3.6 that $G(2,2,3)$ does have ND. So to exclude the cases of G having one matrix component by lemma 3.9, we can assume that $n \geq 2$ and $(m,n) \notin \{(2,2),(2,3) \}$ if p = 2. Then G does not have ND for any of the remaining cases by lemmas 3.10, 3.12, 3.13, and 3.14.

Proof of Theorem 3.15 for G solvable

For every finite group G whenever ${\mathbb Q} G$ has at most one matrix component, then G has ND. Conversely, assume that G has ND and hence property SN. As G is assumed to not be an SSN group of unfaithful type, Proposition 2.17 says that $G \cong P \rtimes H$ for an elementary abelian p-group P, where the action is faithful, irreducible and $[x,h] \neq 1$ for every non-trivial $x \in P$ and $h \in H$. Moreover H is Dedekind. In other words, by Theorem 2.1, either H is abelian or $H \cong Q_8 \times D$ with D an abelian group of odd order. In the latter case, we denote by $c \in Q_8$ the unique (central) element of order 2.

Claim 1: P is the unique maximal abelian subgroup. Moreover, it contains no non-trivial subgroup which is normal in G. Also, a subgroup N of G is normal in G iff $P \subseteq N$. If H is abelian, then G is metabelian with $G' = P$. If H is non-abelian, then $G' = \langle P, c \rangle$ and $G^{\prime\prime} = P.$

First notice that since P is elementary abelian and the action of H on P is irreducible it cannot contain a subgroup normal in G. Moreover, as $[x,h] \neq 1$ for all non-trivial $x \in P, h \in H$, P is indeed the unique maximal abelian subgroup. Now, if $G/P \cong H$ is abelian, then $G' \subseteq P$ and thus by the first part $G' = P$. If $H \cong Q_8 \times D, $ we directly see that $G' \subseteq \langle P, c \rangle$. Using that $G' \cap P \subseteq P$ is normal in G one has that $P \subset G'$ and hence $G' = \langle P, c \rangle$. Analogously we see that $G^{\prime\prime} \subseteq P$ and in fact $G^{\prime\prime} = P$ as Gʹʹ is normal. Finally, consider $N \unlhd G$. Then $N \cap P \subseteq P$ is normal in G, hence $N \subseteq P$ as H contains no normal subgroups. Conversely, if $P \subseteq N$ then $N/P \leq G/P$. As mentioned above $G/P$ is Dedekind and thus $N/P$ is normal as claimed.

Next note that G is strongly monomial, being abelian-by-supersolvable, and hence by [Reference Jespers and del Rio24, Theorem 3.5.10] all primitive central idempotents of ${\mathbb Q} G$ are of the form $e(G,N,K)$ for some strong Shoda pairs (N, K). As recorded in [Reference Jespers and Sun27, Lemma 2.4.], ${\mathbb Q}[G]e(G,N,K)$ is commutative if and only if $G' \subseteq K$.

Claim 2: The tuples in $\{(P, K) \mid [P:K] = p \}$ are strong Shoda pair of G. Conversely, if (N, K) is a strong Shoda pair with $N \unlhd G$, then $P \subseteq K$ or N = P.

Let $K \unlhd N \unlhd G$. Then [Reference Jespers and del Rio24, Corollary 3.5.11] tells that (N, K) is a strong Shoda pair exactly when $N/K$ is cyclic and $N/K$ is a maximal abelian subgroup of $N_G(K)/K$. In particular, $N' \subseteq K$.

To start notice that $\langle P, h \rangle /K$ is non-abelian for every $K \lneq P$ and non-trivial $h \in H$. This follows from [Reference Aschbacher2, (24.6), p. 112] asserting that $P = [P, \langle h \rangle]$. Consequently, $P/K$ is maximal abelian in $N_G(K)/K$ and hence (P, K) is a strong Shoda pair when $[P:K]=p$. Next, by the first claim $P \leq N$ when $N \unlhd G$. Suppose $P \nsubseteq K$. If $N/P$ is abelian, then $N' \leq P \cap K \lneq P$. As $N' \unlhd G$, the first claim yields $N' = 1$ and so N = P. If $N/P$ is non-abelian, then H is non-abelian and $c \in N$. So in that case $G' = \langle P, c \rangle \leq N$, which entails that $G^{\prime\prime} = P \leq N' \leq K$, a contradiction. This proves the second claim.

By [Reference Jespers and del Rio24, Problem 3.4.4.], the number of simple components ${\mathbb Q} Ge(G,P,K)$ with $[P:K]=p$, denoted s, is equal to the number of orbits of H acting on $\mathcal{S} := \{K \mid [P:K] = p \}.$ To count the latter we decompose $\mathcal{S} = \bigcup_{d \mid |H|} \mathcal{S}_d$ with $\mathcal{S}_d := \{K \in \mathcal{S} \mid |N_H(K)|=d \}.$ Note that the action of H on $\mathcal{S}$ preserves each $\mathcal{S}_d$ and denote by s_d the number of H-orbits thereon. Thus $s = \sum_{d \mid |H|} s_d$ and

(37)\begin{equation} s_d = \frac{1}{|H|}\sum_{K \in \mathcal{S}_d} |N_H(K)| = \frac{d\, .\, |\mathcal{S}_d|}{|H|}. \end{equation}

Next let $\mathcal{T}_K$ be a left transversal for $N_H(K)$ in H. Then as every $K \in \mathcal{S}_d$ is a maximal subgroup, one has by Theorem 3.3 that

\begin{equation*}e_K := e(G,P,K) = \sum_{h \in \mathcal{T}_K} (\widehat{K}^h - \widehat{P}).\end{equation*}

Now consider any non-trivial nilpotent element of the form

\begin{equation*}x = (1-y) g \widetilde{H}\end{equation*}

with $1 \neq y \in H$ and $g \in G \setminus N_G(H)$ (which exists as H is non-normal).

Claim 3: If $x e_K \in {\mathbb Z} G$ for all $K \in \mathcal{S}$, then s = 1.

First write $y^g = t . v$ with $t \in P$ and $v \in H$. As x is non-trivial, also t ≠ 1 and $x = g (1 -y^{g}) \widetilde{H}= g (1- t) \widetilde{H}$. Therefore $ g^{-1}x e_{K} = (1-t) e_K \widetilde{H} \in {\mathbb Z} G.$ Because $K \subseteq P \unlhd G,$ one has that ${{\rm supp}}( (1-t)e_K )\subseteq P$ and so $(1-t)e_K \in {\mathbb Z} P$. Next note that $(1-t) \widehat{P} =0$ and $(1-t)\widehat{K}^h=0$ exactly when $t \in K^h$. Therefore $(1-t)e_K \in {\mathbb Z} P$ exactly means that

\begin{equation*}\frac{|\{h\in \mathcal{T}_K \mid t \notin K^{h} \}|}{|K|} \in {\mathbb Z}.\end{equation*}

Recall that $\operatorname{core}_G(K) = \ker \left( g \mapsto g e_K \right)$, by Theorem 3.3, which is trivial in this case. In particular $|\{h\in \mathcal{T}_K \mid t \notin K^{h} \}| \neq 0.$ Therefore, if $K \in \mathcal{S}_d$ then $|K| \leq |\mathcal{T}_K| = |H|/d$. Now (37) entails that $s_d |K| \leq |S_d|$ which sums up to $s |K| \leq |S| = \frac{|P|-1}{p-1}$, since $|S|$ is the number of maximal dimensional subspaces in the $\mathbb{F}_p$-vector space P. As $[P:K]=p$ the latter inequality simplifies to $s (p-1) \leq p - \frac{1}{|K|} \leq p-1$, hence $s \leq 1.$ In fact s = 1 by the second claim.

Next notice that by [Reference Jespers and Sun27, Lemma 3.3], ${\mathbb Q} H$ has no non-zero nilpotent elements. Therefore, when decomposing ${\mathbb Q} G$ as

\begin{equation*}{\mathbb Q} G \cong {\mathbb Q} G \widehat{P} \oplus {\mathbb Q} G (1-\widehat{P})\end{equation*}

the piece ${\mathbb Q} G \widehat{P} \cong {\mathbb Q} G/P \cong {\mathbb Q} H$ has no matrix components. So, it remains to prove that ${\mathbb Q} G (1-\widehat{P})$ is simple. By lemma 3.4

\begin{equation*}\dim_{{\mathbb Q} }({\mathbb Q} G e_K) = |H| (p-1) [G:N_G(K)].\end{equation*}

Furthermore, $[G:N_G(K)] = [H: N_H(K)] = |\mathcal{T}_d|$ if $K \in \mathcal{S}_d$. By the third claim, there is a unique $d \mid |H|$ such that $s = s_d$ and s = 1. So (37) translates to $|\mathcal{T}_d| = |S| = \frac{|P|-1}{p-1}.$ Altogether,

\begin{equation*}\dim_{{\mathbb Q}}({\mathbb Q} G\widehat{P}) + \dim_{{\mathbb Q} }({\mathbb Q} G e_K) = |H| + |H| (|P| -1) = |H| . |P| = \dim_{{\mathbb Q}}({\mathbb Q} G).\end{equation*}

Thus ${\mathbb Q} G (1 - \widehat{P}) = {\mathbb Q} G e_K$ is indeed simple, finishing the proof.

Proof of Theorem 3.15 for G non-solvable

If G has SN, then it does not have ND by [Reference Jespers and Sun27, Proposition 3.4]. So assume G has SN. By Proposition 2.18, we know G has a unique minimal normal subgroup S which is a direct product of isomorphic non-abelian simple groups and that $G/S$ is Dedekind.

Let $y \in G$ be an element of order 2 and $x \in G$ such that $y^x \notin \langle y \rangle$. Such x and y exist, as S has even order by the Feit–Thompson Theorem. Hence we can construct the non-trivial nilpotent element $n = (1-y)x(1+y)$. If we write n in the shape as in lemma 2.7, then

\begin{equation*}\alpha(g) = \left\{\begin{array}{lll} 1, & g = x \ \ \text{or} \ \ g = xy, \\ -1, & g = yx \ \ \text{or} \ \ g = yxy, \\ 0, & \text{else}. \end{array}\right. \end{equation*}

So by lemma 2.7 the coefficient of ne at x is

(38)\begin{align} \frac{k}{|G|}\sum_{h \in G} \alpha(h) \chi(x^{-1}h) &= \frac{k}{|G|} \chi\left(1+y - x^{-1}yx(1+y)\right) = \frac{k}{|G|} \chi\left(1-[x,y]\right), \end{align}

where in the last step we used $y=y^{-1}$ as well as the fact that y and $x^{-1}yx$ are conjugate and so have the same character value, which allows us to cancel them. If S is not contained in the kernel of χ, then the value appearing in (38) is not 0. Moreover, we have

\begin{equation*}\left|\frac{k}{|G|} \chi\left(1-[x,y]\right)\right| \leq \frac{2\chi(1)k}{|G|} = \frac{2\text{dim}_{\mathbb{Q}}(e\mathbb{Q}G)}{|G|}.\end{equation*}

So if the last is a rational number smaller than 1 for χ corresponding to a faithful representation, the product ne cannot lie in $\mathbb{Z}G$.

Now set $f = \widehat{S}$. Then $\mathbb{Q}G = f\mathbb{Q}G \oplus (1-f)\mathbb{Q}G$ and the direct summand $(1-f)\mathbb{Q}G$ corresponds to all the irreducible faithful representations of G. None of the indecomposable direct summands in $(1-f)\mathbb{Q}G$ is a division algebra, as $\operatorname{SL}(2,5)$ is the only non-solvable group which is a finite subgroup of a division algebra [Reference Shirvani and Wehrfritz44, 2.1.4]. If $(1-f)\mathbb{Q}G$ is decomposable, then one of its indecomposable direct summands must have dimension smaller than $\frac{|G|}{2}$, as $f\mathbb{Q}G$ has positive dimension and $|G| = \text{dim}(f\mathbb{Q}G) + \text{dim}((1-f)\mathbb{Q}G)$. Now the number of simple components of $\mathbb{Q}G$ equals the number of conjugacy classes of cyclic subgroups of G [Reference Jespers and del Rio24, Corollary 7.1.12]. The components in $f\mathbb{Q}G$ correspond to conjugacy classes in $G/S$, i.e. classes not lying in S except for the class of the trivial element. But as S certainly contains at least three conjugacy classes, we conclude that $(1-f)\mathbb{Q}G$ has at least two indecomposable summands.

Proof. For every $e \in \operatorname{PCI} ({\mathbb Q} G)_{\geq 2}$ one can choose an idempotent f_e in ${\mathbb Q} G$ such that ef_e is non-central in ${\mathbb Q} G e$. Consider the associated generalized bicyclic units $\text{GBic}^{\{f_e \}}({\mathbb Q} G)$, see [Reference Jespers and del Rio24, Section 11.2] for definition. Then following [Reference Janssens, Jespers and Schnabel22, Theorem 6.3] the group $\text{GBic}^{\{f_e \}}({\mathbb Q} G)$ contains a subgroup U_e of the form $1-e + E_{n_e}(J_e)$ for some non-zero ideal J_e of $\mathcal{O}_e$. As $\text{GBic}^{\{f_e \}}({\mathbb Q} G)$ is a subgroup of $\langle {\mathcal U}({\mathbb Z} G)_{un} \rangle$ the previous implies the first part of the statement.