Proof. Let $G = \langle x \rangle \times \langle y \rangle $ with x of order $p^a$ and y of order $p^b$ . Let C be a cyclic subgroup of G. Either C is a subgroup of $\langle (1,y) \rangle $ or there exist integers n and c so that C is generated by $(x^{p^n},y^c)$ for $0 \leq n < a$ and $0 \leq c < p^b$ . We claim that C is maximal exactly when $n = 0$ , or c is relatively prime to p or $C = \langle (1,y) \rangle $ . To see this, observe that in each of these cases, a generator of C does not lie in $G^{\{p\}}$ , so C is maximal. Furthermore, we claim that the generator for a maximal cyclic subgroup is unique if we make the additional restriction that $0 \leq c < \min (p^b, p^{a-n})$ . First note for an integer l that $(x^{p^n}, y^c)^{1 + lp^{a-n}} = (x^{p^n}, y^{c + clp^{a-n}})$ . Since c is coprime to p, as l runs through the integers modulo $p^b$ , then also $cl$ will run through all of ${\mathbb Z}_{p^b}$ . So we are getting as the exponents for y, all elements in the coset $c + p^{a-n} {\mathbb Z}_{p^b}$ in ${\mathbb Z}_{p^b}$ . It is not difficult to see that $\{0,1,\ldots ,p^{a-n} -1\}$ is a transversal for $p^{a-n}{\mathbb Z}_{p^b}$ in ${\mathbb Z}_{p^b}$ . Hence, there exists an integer $c'$ with $0 \le c' \le p^{a-n}-1$ so that $c' + p^{a-n}{\mathbb Z}_{p^b} = c + p^{a-n}{\mathbb Z}_{p^b}$ . It follows that $(x^{p^n},y^{c'})$ will lie in $\langle (x^{p^n},y^c) \rangle $ and since it has the same order, it will be a generator.

So, we claim the following are distinct maximal cyclic subgroups of G:

$$ \begin{align*} &\hspace{3pt}\quad\qquad\qquad\langle (x, y^c) \rangle\quad \mbox{for } 0 \leq c \leq p^{b}-1\quad \mbox{and}\quad \langle (1, y) \rangle \\ &\langle (x^{p^n}, y^c) \rangle\quad\mbox{for } 1 \leq n < a,\, 0 \leq c < \min(p^b, p^{a-n})\mbox{ and } {\textrm{gcd}}(c,p) =1. \end{align*} $$

To count the number of subgroups of the form $\langle (x^{p^n}, y^c) \rangle $ , we consider two cases: when $1 \leq n \leq a-b$ and when $a-b+1 \leq n < a$ . For $1 \leq n \leq a-b$ , we have $\phi (p^b) = p^{b-1}(p-1)$ such subgroups, where $\phi $ is the Euler totient function. For $a-b+1 \leq n \leq a-1$ , we have $\phi (p^{a-n})$ maximal cyclic subgroups of the form $ \langle x^{p^n},y^c \rangle $ , where p does not divide c and $1 \leq c \leq p^{a-n}$ . With this range on b, we obtain $1 \leq a-n \leq b-1$ . Observe that $\phi (p^{a-n}) = p^{a-n-1} (p-1)$ . We can view this as $p^{i-1} (p-1)$ for i running from 1 to $b-1$ or $p^i (p-1)$ for i running from 0 to $b-2$ . So, in total, we have

$$ \begin{align*} p^b + (a-b)&p^{(b-1)} (p-1) + \sum_{i=0}^{b-2} p^i(p-1) + 1 \\ & = p^b + (a-b) p^{(b-1)} (p-1) + [(p^{(b-1)} - 1)/(p-1)] (p-1) + 1 \\ & = p^b + (a-b) p^{(b-1)} (p-1) + p^{(b-1)} \\ & = p^{(b-1)} ((a - b)(p-1) + p + 1) \end{align*} $$

maximal cyclic subgroups, as desired. A simple inductive argument shows that $p^{b-1}((a-b)(p-1) + p + 1) \geq a+b$ .

Proof. Without loss of generality, we may assume $a \ge b$ . Set $n = a+b$ , so that $a = n - b$ . We work to show $g_p (a,b) \ge (p-1)(n-2) + p + 1 \geq n + 1$ . We observe that $1 \le b \le n/2$ . We see that $g_p (n-b,b) = p^{b-1} ((n-2b)(p-1) + p+1)$ . We can view n as fixed, and this becomes a function in one variable:

$$ \begin{align*}f(x) = g_p (n-x,x) = p^{x-1}((n-2x)(p-1) + p+1).\end{align*} $$

We want to find the minimal value for $f(x)$ on $[1,n/2]$ . We use calculus to see that

$$ \begin{align*}f'(x) = p^{x-1}(((\ln p)(n-2x) - 2)(p-1) + (\ln p)(p+1)).\end{align*} $$

Setting this equal to $0$ , we obtain the critical value of

$$ \begin{align*}x = n/2 + (p+1)/(2(p-1)) - 1/(\ln p).\end{align*} $$

We claim that $(p+1)/(2(p-1)) - 1/(\ln p)$ is positive for all primes p. (This can be shown using calculus or graphing using a computer.) Hence, this critical point is not in our interval.

Now, we remark that $f(1) = ((n-2)(p-1)) + p +1 = p(n-1) -n +3$ and $f(n/2) = p^{n/2 - 1}(p+1)$ . Recall that Fermat’s theorem tells us that any local extreme values would occur when the derivative was zero. Since the derivative of this function is never zero on this interval, we know that our minimum is the smallest of these two values. Set $f_1(x) = ((x-2)(p-1)) + p + 1$ and $f_{n/2}(x) = p^{x/2 -1}(p+1)$ for $x \geq 2$ . Then $f_1(2) = f_{n/2}(2)$ , but $f^{\prime }_{n/2} (x)> f^{\prime }_{1} (x)$ ; so $f_{n/2}(x)> f_{1}(x)$ for $x> 2$ . Thus, $g_p (n-b,b) \geq (p-1)(n-2) + p + 1$ . Finally, note that $(p-1)(n-2) + p + 1 \geq n-2 + 3= n+1.$

Proof. We write $Y = \langle y \rangle \cong C_{p^a}$ , so that $G = H \times Y$ . Let $\{ h_i \}$ , for $1 \leq i \leq \eta (H)$ , be representatives of generators for the maximal cyclic subgroups of H. We now prove that $\langle (h_i, y^{p^c})\rangle $ , for $0 \le c \le a$ and $1 \leq i \leq \eta (H)$ , along with $\langle (1,y)\rangle $ are distinct maximal cyclic subgroups of G. This gives the required count for the lower bound.

First note that $h_i \in H \setminus H^{\{p\} }$ , since $\langle h_i \rangle $ is maximal in H, and thus, $(h_i, 1)$ , $(h_i, y^{p^c})$ and $(1, y)$ all lie in $G \setminus G^{\{p\} }$ for $c \in \{ 0, \ldots , a \}$ . (It is not difficult to see that $G^{\{p\} } = \{ (\alpha ,\beta ) \mid \alpha \in H^{\{ p\} }, \beta \in \langle y^p \rangle \}$ .) Thus, each of these elements generate a maximal cyclic subgroup of G. We need to show that they generate different cyclic subgroups of G. Suppose $\langle (h_i, y^{p^c})\rangle = \langle (h_j, y^{p^d})\rangle $ for $0 \le c,d \le a$ . In both cases, by projecting into H, it is not difficult to see that $\langle h_i\rangle = \langle h_j\rangle $ , and thus, $i = j$ . We are left to consider $\langle (h_i, y^{p^c})\rangle = \langle (h_i, y^{p^d})\rangle $ . Projecting into Y, this forces $\langle y^{p^c}\rangle = \langle y^{p^d} \rangle $ , and thus, $c=d$ as required.

Proof. We work by induction on $|G|$ . Suppose first that $G = C_{p^a} \times C_{p^b}$ for positive integers a and b. By Lemma 2.1, we have $\eta (G) = g_p (a,b)$ and applying Lemma 2.2, $g_p (a,b) \ge (p-1)(a+b-2) + p + 1 \ge a+b + 1$ .

Now suppose H is not cyclic and $|H| = p^h$ and $G = H \times C_{p^a}.$ By induction, $\eta (H) \ge (p-1)(h-2) + p + 1$ . By Lemma 2.3,

$$ \begin{align*} \eta (G) = \eta (H \times C_{p^a}) & \ge (a + 1) \eta (H)+ 1 \\ & \ge (a + 1)((h-2)(p-1) + p + 1) + 1 \\ & = (a+1)(h-2)(p-1) + a (p+1) + p + 1 + 1 \\ & = (a+1)(h-2)(p-1) + a (p-1) + 2a + p + 2 \\ & = ((a+1)(h-2) + a)(p-1) + 2a + p + 2 \\ & \ge (h - 2 + a)(p-1) + p + 1. \end{align*} $$

Since $n = h + a$ , this gives the result.

Finally, note that $(p-1)(n-2) + p + 1 \geq n-2 + 3= n+1.$

Proof. We apply the result from Theorem 2.4:

$$ \begin{align*} (p-1)(n-2) + p + 1 & = (p-1)n -2p + 2 + p + 1 \\ & = (p+1)n/2 + (p-3)n/2 - p + 3\\ & = (p+1)n/2 + (p-3)(n-2)/2. \end{align*} $$

Since $p \ge 3$ and $n \ge 2$ , this is at least $(p+1)n/2$ .

Proof. We work by induction on l. We begin with the case $l = 2$ . If $|G^2| \le p^{n/3}$ , then we have the result. Thus, we may assume that $|G^2|> p^{ n/3}$ and so, $|G/G^2| < p^{2n/3}$ . We can find $a_1, \ldots , a_k \in G$ so that $G/G^2 = \langle a_1 G^2 \rangle \times \langle a_2 G^2 \rangle \times \cdots \times \langle a_k G^2 \rangle $ . We know that $G^2$ is central and so it is abelian. Also, it is generated by $\{ [a_i,a_j] \mid 1 \le i < j \le k \}$ . Since G is a p-group, it suffices to show that $o([a_i,a_j])$ is less than or equal to $p^{n/3}$ for all $1 \le i < j \le k$ . Observe that $o(a_iG^2)o(a_jG^2)$ divides $|G/G^2| < p^{2n/3}$ and so, without loss of generality, $o(a_iG^2)$ is less than or equal to $p^{n/3}$ . Now, $[a_i,a_j]^{\lfloor n/3 \rfloor } = [a_i^{\lfloor n/3\rfloor },a_j] = 1$ . This completes the proof when $l = 2$ .

We now assume that $l \ge 2$ . If $|G^l| \le p^{n/(l+1)}$ , then the result holds. Thus, we may assume that $|G^l|> p^{n/(l+1)}$ , and so, $|G/G^l| < p^{nl/(l+1)}$ . We see that $G^l$ is generated by the set $\{[a,b] \mid a \in G^{l-1}, b \in G \}$ . As above, since $G^l$ is central (thus abelian) and a p-group, it suffices to show that $o([a,b]) \le p^{n/{(l+1)}}$ when $a \in G^{l-1}$ and $b \in G$ . By induction, we know since $G/G^l$ has nilpotence class $l-1$ that $o(a G^l)$ is less than or equal to $p^{(nl/(l+1))/l} = p^{n/(l+1)}$ . We then have $[a,b]^{\lfloor n/(l+1) \rfloor } = [a^{\lfloor n/(l+1) \rfloor },b] = 1$ . This proves the desired result.

Proof of Theorem 1.1

We work by induction on l. If $l = 1$ , then this is Theorem 2.4.

We now suppose that $l \ge 2$ . Note that $G/G' = G/G^2$ is not cyclic, so the induction hypothesis is valid even if $l =2$ . If $|G/G^l| \ge p^{n(l-1)/l}$ , then by the induction hypothesis,

$$ \begin{align*}\eta \bigg(\frac G{G^l}\bigg) \ge (p-1) \bigg(\frac {n(l-1)/l}{l-1} - 2\bigg) + p + 1 = (p-1)\bigg(\frac nl - 2\bigg) + p + 1.\end{align*} $$

As we noted before the start of the proof, $\eta (G) \ge \eta (G/G^l)$ , and so we have the conclusion in this case.

Thus, we may assume that $|G/G^l| < p^{n(l-1)/l}$ and so $|G^l|> p^{(n/l)} > p^{n/(l+1)}$ . By Lemma 3.1, this implies that $G^l$ is not cyclic. It follows that $|Z(G)|> p^{n/l}$ and $Z(G)$ is not cyclic. By Theorem 2.4, $\eta (Z(G)) \ge (p-1)(n/l - 2) + p + 1$ , and as we noted before the start of the proof, we have $\eta (G) \ge \eta (Z(G))$ . This yields the result when $n = l$ and proves the theorem.