Proof. We use $*$ to denote the operation of $Q_{a,a^q}$. If y − x is a square, then $x*y = x+(y-x)a = x+ (y-x)\circ a= x*_a y$. If y − x is a non-square, then $x*y = x+(y-x)a^q = x+ (y-x)\circ a = x*_a y$, for all $x,y \in \mathbb F$. It follows that $Q_{a,a^q} = (\mathbb F_{q^2},*_a)$, and then by substituting a^q for a, we find that $Q_{a^q,a} = (\mathbb F_{q^2},*_{a^q})$. The fact that $x\mapsto x^q$ is an isomorphism $Q_{a,a^q}\cong Q_{a^q,a}$ was shown in [Reference Wanless14] (and also follows from Proposition 2.1).

Proof. To prove (vii), consider $x,y \in \mathbb F$. If $\chi(y-x) = 1$, then $\zeta(x*y) = \zeta x + a\zeta(y-x)$, while $\zeta x * \zeta y = \zeta x + b\zeta(y-x)$. If $\chi(y-x) = -1$, then $\zeta(x*y) = \zeta x + b\zeta(y-x)$, while $\zeta x * \zeta y = \zeta x + a\zeta(y-x)$.

To prove (viii), note that $\chi(y-x) = \chi(\alpha(y-x)) =\chi(\alpha(y)-\alpha(x))$ for all $x,y \in Q$. Hence, $x*y = x+a(y-x)$ in $Q_{a,b}$ if and only if

\begin{equation*}\alpha(x)*\alpha(y) = \alpha(x)+\alpha(a)(\alpha(y)-\alpha(x)) =\alpha(x*y)\end{equation*}

in $Q_{\alpha(a),\alpha(b)}$. The argument remains true if a is replaced by b, so $\alpha(x)*\alpha(y) =\alpha(x*y)$ in all cases.

Proof. The statement is concerned with two special cases of a more general problem that asks if for $\varepsilon_i\in \{-1,1\}$, $-1\leqslant i \leqslant 1$, there exists $x \in \mathbb F$ such that $\chi(x+i) = \varepsilon_i$. A consequence of Weil’s bound (e.g., as stated in [Reference Drápal and Wanless6, Theorem 1.6]) implies that such an x exists if

\begin{equation*} 2^{-3}q -(3/2)(\sqrt q + 1) + \sqrt q(1-2^{-3}) \gt 0, \end{equation*}

where $q= |\mathbb F|$. This is true for each prime power q > 43. For prime values q = 11, 13, 17, 19, 23, 29, 31, 37, 41 and 43, put u = 7, 6, 6, 13, 20, 11, 12, 14, 12 and 19, respectively. For q = 25, set $u = 2\sqrt 2$. In all these cases, set $v = u-1$. For q = 27, set u = x and $v=2x^2$ in $\mathbb F_3[x]/(x^3+2x+1)$.

Proof. First suppose that a = b, so that $(Q,*)$ is defined by $x*y=(1-a)x+ay$ for all $x,y\in\mathbb F$. So Q is isotopic to the additive group of $\mathbb F$.

For the remainder of this proof, suppose that $a\ne b$. We use the well known quadrangle criterion (see e.g. [Reference Evans8]) to show that Q is not isotopic to any group. This criterion states that if Q is isotopic to a group and $r_1,r_2,c_1,c_2,r_1^{\prime},r_2^{\prime},c_1^{\prime},c_2^{\prime}$ are any elements of Q such that $r_1*c_1=r_1^{\prime}*c_1^{\prime}$, $r_1*c_2=r_1^{\prime}*c_2^{\prime}$ and $r_2*c_1=r_2^{\prime}*c_1^{\prime}$, then it follows that $r_2*c_2=r_2'*c_2'$. For $|\mathbb F|=7$, we apply this criterion to the following quadrangles in $Q_{3,5}$:

\begin{equation*} \begin{array}{r|cc} *&0&1\\ \hline 0&0&3\\ 1&3&1 \end{array} \qquad \begin{array}{r|cc} *&6&5\\ \hline 2&0&3\\ 4&3&0 \end{array} \end{equation*}

It follows that $Q_{3,5}$ (and also, by Proposition 2.1(iv), its isomorph $Q_{5,3}$) is not isotopic to a group.

Similarly, for $|\mathbb F|=9$, the following quadrangles

\begin{equation*} \begin{array}{r|cc} *&1&2i\\ \hline 1&1&2\\ 1+i&2&1 \end{array} \qquad \begin{array}{r|cc} *&2+i&2+2i\\ \hline 1+2i&1&2\\ i&2&0 \end{array} \end{equation*}

show that $Q_{1+i,1+2i}$ is not isotopic to any group (where $i=\sqrt{-1}$). This property is necessarily inherited by the opposite quasigroup $Q_{2i,i}$ and translate $Q_{2+i,2+2i}$, as well as by $Q_{1+2i,1+i}$, $Q_{i,2i}$ and $Q_{2+2i,2+i}$. There are no other solutions to Condition (1.2) for $|\mathbb F|\leqslant9$.

If $|\mathbb F| \gt 9$, then let $u,v\in\mathbb F$ be as given by Lemma 2.2. For such $u,v$, we have the following violation of the quadrangle criterion:

\begin{equation*} \begin{array}{r|cc} *&u-bu&u-bu+1\\ \hline -bu&0&b\\ 1-bu&1-b&1 \end{array} \qquad \begin{array}{r|cc} *&v-bv&v-bv+1\\ \hline -bv&0&b\\ 1-bv&1-a&1 \end{array} \end{equation*}

Proof. We use Proposition 2.1(v). If $|\mathbb F|\equiv1\bmod4$, then $Q_{a,b}$ is commutative if and only if $a=1-a$ and $b=1-b$. If $|\mathbb F|\equiv3\bmod4$, then $Q_{a,b}$ is commutative if and only if $a=1-b$ and $b=1-a$. The result follows.

Proof. If a = b, then Q is entropic (this is a well-known fact that may be verified directly). Idempotent entropic quasigroups are left distributive since an idempotent entropic quasigroup fulfils $x\cdot yz = xx \cdot yz = xy\cdot xz$. By Proposition 2.1(v), it thus suffices to assume that Q is left distributive and show that then a = b.

Since the number of squares in $\mathbb F$ exceeds $|\mathbb F|/2$, the squares cannot form a subquasigroup of Q. Hence, there exist squares $x,y\in \mathbb F$ such that $x*y$ is a non-square. By Condition (1.1), $0*(x*y) = b(x*y)$. If a is a square, then $(0*x) * (0*y) =ax * ay=a(x*y)$, by Proposition 2.1(iii). Thus, in this case, the left distributivity clearly implies a = b. If a is a non-square, then b is a non-square too. If $a\ne b$, then $0*(x*y)=b(x*y)\ne bx * by = (0*x)*(0*y)$ by Proposition 2.1(vii).

Proof. We use Proposition 2.1(vi) and the fact that $Q_{a,b}$ is semisymmetric if and only if it equals its translate. First suppose that $\chi(a)\ne\chi(-1)$. Then $Q_{a,b}$ equals its translate if and only if $a-1=ab=b-1$, which is equivalent to $a=b=a^2+1$. Next suppose that $\chi(a)=\chi(-1)$. Then $Q_{a,b}$ equals its translate if and only if $a,b$ are both solutions to $x=(x-1)/x$. There are two possibilities. The first is that $a=b=a^2+1$. The second is that $a,b$ are the two distinct roots of $x^2-x+1=0$, in which case $a+b=1$. The result now follows from the observation that if $a^2-a+1=0$ and $a+b=1$, then $\chi(a)=\chi(b)=\chi(-a^2)=\chi(-1)$.

Proof. Since total symmetry is the combination of semi-symmetry with commutativity, we combine Lemmas 2.6 and 2.4. Together they imply that the necessary and sufficient conditions for total symmetry are that at least one of

(2.2)\begin{align} a^2-a+1=0\text{ and }a=b=1/2,\text{ or} \end{align}

(2.3)\begin{align} a^2-a+1=0\text{ and }a+b=1\text{ and }\chi(-1)=-1 \end{align}

$3/4=0$

$\operatorname{char}(\mathbb F) = 3$

$\operatorname{char}(\mathbb F) = 3$

$a=b=-1$

Condition (2.3) implies that $a(1-b)-a=-1$ and hence ab = 1. Moreover, if ab = 1 and $a+b=1$, then $a^2=a(1-b)=a-1$. The characterization of quadratic Steiner quasigroups follows.

Assume Condition (2.3) holds. Then $b-b^2=b(1-b)=(1-a)a=1$. So $a,b$ are both roots of $x^2-x+1=0$ and hence also of $x^3+1=(x+1)(x^2-x+1)=0$. From $a^3=b^3=-1$ and $a+b=1$, we get $\chi(a)=\chi(b)=\chi(-1)=\chi(1-b)=\chi(1-a)$.

The polynomial $x^2-x+1$ has at most two roots (these roots coincide if and only if $\operatorname{char}(\mathbb F) = 3$). That explains why $\{a,b\} = \{c,d\}$ if $Q_{a,b}$ and $Q_{c,d}$ are Steiner quadratic quasigroups on $\mathbb F$, with $a\ne b$ and $c\ne d$. If $\{a,b\} = \{c,d\}$, then $Q_{a,b}\cong Q_{c,d}$ by Proposition 2.1(iv).

Proof. First note that $x*(x*x)=x*x=(x*x)*x$ for all $x\in\mathbb F$, by idempotence. Thus, consider distinct $x,y\in \mathbb F$ and let $z=x-y$. Then

\begin{equation*} x*(y*x)=x*(y+\theta_{z}z)=x+\theta_{z(a-1)}z(\theta_{z}-1), \end{equation*}

and

\begin{equation*} (x*y)*x=(x-\theta_{-z}z)*x=x-\theta_{-z}z+\theta_{az}\theta_{-z}z =x+\theta_{-z}z(\theta_{az}-1). \end{equation*}

It follows that Q is flexible if and only if

(2.4)\begin{equation} \theta_{z(a-1)}(\theta_{z}-1)=\theta_{-z}(\theta_{az}-1) \end{equation}

for all z. Both sides of Condition (2.4) are members of $\Phi=\{a(a-1),a(b-1),b(a-1),b(b-1)\}$. If a = b, then $|\Phi|=1$, so Condition (2.4) is automatically satisfied. Henceforth, we assume $a\ne b$. In this case, $|\Phi|=4$ unless $a(a-1)=b(b-1)$, which requires $a=1-b$.

Suppose for the moment that $\chi(z)=1$, meaning that Condition (2.4) is equivalent to $\theta_{a-1}(a-1)=\theta_{-1}(\theta_{a}-1)$. From the above observations, this condition can only be satisfied if

(2.5)\begin{align} \theta_{a-1}=\theta_{-1}\quad \text{and}\quad \theta_{a}=a,\ \text{or} \end{align}

(2.6)\begin{align} a=1-b\quad \text{and}\quad \theta_{a-1}=a\quad \text{and}\quad \theta_{-1}=\theta_{a}=b. \end{align}

Now Condition (2.5) is equivalent to condition (ii), whereas Condition (2.6) implies (iii). It follows that for $Q_{a,b}$ to be flexible, it is necessary that (i), (ii) or (iii) holds.

To check sufficiency, we first note that if (iii) is true, then $\chi(a)=\chi(1-a)$, so either (ii) or Condition (2.6) holds. Hence, if (ii) or (iii) holds, then at least one of Conditions (2.5) or (2.6) holds. Moreover, Condition (2.5) implies that $\theta_{z(a-1)}=\theta_{-z}$ and $\theta_{z}=\theta_{az}$, whereas Condition (2.6) implies that $\theta_{-z}=1-\theta_{z}$ and $\theta_{az}=\theta_{(1-a)z}=1-\theta_{z(a-1)}$. In either case, Condition (2.4) holds for all z.

Proof. Put $n=|\Omega|$. By the assumptions there exists an orbit Γ of G_ω such that $|\Gamma|+1 \gt n/2$. This implies that G is primitive. Now, elements of Ω may be identified with elements of a group N (which is isomorphic to the Frobenius kernel), and G_ω may be identified with a subgroup of $\operatorname{Aut}(N)$. Since there are at most two nontrivial orbits of G_ω, there are at most two integers that occur as an order of a non-trivial element of N. One of these integers has to be a prime, and the other (if it exists) is either a square of this prime or another prime. This means that N is solvable. A finite solvable group is either elementary abelian or it contains a non-trivial proper characteristic subgroup. However, such a subgroup yields a block $B\subseteq \Omega$. That is not possible since G is primitive.

Proof. Denote by N the Frobenius kernel of G and recall that non-identity elements of N act without fixed points. Since $G/N$ is abelian, we must have $[g,h]\in N$ for any $g,h\in G$. To prove that $[g,h]$ is fixed point free, it therefore suffices to find any $\omega\in \Omega$ such that $gh(\omega) \ne hg(\omega)$. We take $\omega=\psi^{-1}(\alpha)$ and observe that

\begin{equation*}\varphi\varphi^\psi(\omega) = \varphi\omega \quad\text{and }\quad \varphi^\psi\varphi(\omega) = \psi^{-1}\varphi\psi(\varphi\omega).\end{equation*}

Suppose that $\psi(\varphi\omega)=\varphi\psi(\varphi\omega)$. By definition, α is the only fixed point of φ and β is the only fixed point of ψ, so we can deduce in turn that $\psi(\varphi\omega)=\alpha$, $\varphi\omega =\omega$, $\omega = \alpha$, $\psi(\alpha)=\alpha$ and thus $\alpha = \beta$. This contradiction proves that $\varphi\varphi^\psi(\omega)\ne\varphi^\psi\varphi(\omega)$, from which the result follows.

Proof. If s = t, then u = v, so we may use $x\mapsto x+u-s$ for α. So assume $t\ne s$, meaning that $u\ne v$ as well. Put $\lambda = (u-v)/(s-t)$ and $\mu = (vs-ut)/(s-t)$. The mapping $x \mapsto \lambda x + \mu$ is an automorphism of $Q_{a,b}$ that sends s to u and t to v.

Proof. Let U be a minimal subquasigroup of Q. Set $m = |U|$. For $\varepsilon\in \{-1,1\}$, denote by $\sigma(\varepsilon)$ the number of $(u,v)\in U\times U$ such that $\chi(v-u) = \varepsilon$. Since $\sigma(1)+\sigma(-1)=m(m-1)$, there exists $\varepsilon\in \{-1,1\}$ such that $\sigma(\varepsilon)\geqslant \binom m2$. Fix such ɛ, and fix also $s,t \in U$ such that $\chi(t-s)=\varepsilon$.

Denote by A the permutation group on U that is induced by affine automorphisms acting on U. Since s and t generate U, we see that $\psi(U) = U$ whenever $\psi \in \operatorname{Aut}(Q)$ is such that $\psi(s),\psi(t)\in U$. By Proposition 3.3, for each $(u,v) \in U\times U$ with $\chi(v-u)=\varepsilon$, there exists $\psi \in A$ such that $\psi(s) = u$ and $\psi(t) = v$. Therefore, $|A|\geqslant \binom m2$.

Denote by f the number of fixed point free elements of A. The aggregate number of fixed points of elements of A is equal to $m+|A| - 1 -f$. By Burnside’s lemma, this is equal to $|A|k$, where k is the number of orbits of A. If $k\geqslant 2$, then

\begin{equation*}m-1-f = |A|(k - 1) \geqslant|A|\geqslant \frac{1}{2} m(m-1).\end{equation*}

It follows that $2(-1-f) \geqslant m(m-3)$, and hence m < 3. However, this is impossible since there is no idempotent quasigroup of order two and $|U| \gt 1$. Hence, k = 1; in other words, A is transitive. As each non-trivial element of A fixes at most one element, it follows that A is either a regular group or a Frobenius group. Since $|A| \geqslant m(m-1)/2$, the former alternative may take place if and only if m = 3. We distinguish two cases.

First, suppose that $|A|=3$. This means that A is cyclic, and $|U|=m=3$ as well. Let A be generated by an affine permutation $x\mapsto \lambda x +\mu$. If λ = 1, then U is a coset of a subspace and $\operatorname{char}(\mathbb F)=3$. Suppose $\lambda \ne 1$. Then $\lambda^3 = 1$, so $\lambda^2 + \lambda + 1=0$. Our intention is to show that $ab = 1 = a+b$. By Proposition 2.1(ii), it may be assumed that $0 \in U$ since automorphisms map minimal subquasigroups to minimal subquasigroups. Suppose that $u \in U\setminus\{0\}$. Then $u=\mu$ and $U=\{0,u,(\lambda+1)u\}$. Since U is idempotent, $u*0=0*u = (\lambda+1)u = -\lambda^2 u$. Suppose that −1 is a square. Then $-\lambda^2u=u*0=u-0*u=(1+\lambda^2)u=-\lambda u$, resulting in λ = 1. Hence, −1 is a non-square. If u is a square, then $0*u=au$, so $a=-\lambda^2$ and $u=0*(-\lambda^2 u) = -\lambda^2 bu = abu$. Alternatively, if u is a non-square, then $0*u=bu$, so $b=-\lambda^2$ and $u=0*(-\lambda^2 u) = -\lambda^2 au = abu$. Therefore, ab = 1 in every case. Since at least one of a ³ and b ³ is equal to $(-\lambda^2)^3 = - 1$, it must be that $a^3 = -1 =b^3$.

Thus, $\chi(a) = \chi(b) = \chi(-1) = -1$ and $0 = a^3+1=(a+1)(a^2-a+1)$. If $a = -1$, then $b=-1$ and $-u = 0*u=u * 0 = u -(-u) = 2u$. In such a case, $0 = 3$, $U = \{0,u,-u\}$ is a subspace of $\mathbb F$, and $a+b = - 2=1$. Assume $a\ne -1$. Then $b\ne -1$, $a^2-a=-1=b^2-b$ and $a+b = a+a^{-1} = (a^2+1)/a=1$. Hence, Condition (2.1) holds, by Lemma 2.7.

Let us now turn to the case $|A| \gt 3$. The group A is a Frobenius group on an m-element set U with complements of order $\geqslant (m-1)/2$. The kernel of A is elementary abelian, by Lemma 3.1. To finish it suffices to prove that all elements of the kernel of A are induced by translations $x\mapsto x+\mu$. Assume the contrary. This means that the kernel of A contains a permutation that is a restriction of an affine automorphism $\psi\colon x\mapsto \lambda x + \mu$, where $\lambda\ne1$. Choose an affine automorphism φ that induces a non-trivial permutation belonging to a complement of A. Both φ and ψ are elements of the Frobenius group formed by affine automorphisms of Q, and they belong to different Frobenius complements of the latter group. Hence, $[\psi,\psi^\varphi]$ is a fixed point free permutation of Q, by Lemma 3.2. Denote by $\tilde \psi$ the restriction of ψ to A. Both $\tilde \psi$ and $\tilde \psi^\varphi$ belong to the kernel of A. This kernel is abelian. Hence, $[\psi,\psi^\varphi]$ fixes each point of U, a contradiction.

Proof. Since $0*1 = a$, we have $0*f=\varphi(0)*\varphi(1) = \varphi(a) = bf$. Therefore, f is a non-square.

Denote by U_a the subfield of $\mathbb F$ generated by a and by U_b the subfield of $\mathbb F$ generated by b. The field U_a coincides with the set of all sums $a^{i_1}+\dots + a^{i_k}$, $k\geqslant 0$, by Lemma 4.1. The assumptions of the statement imply that

(4.1)\begin{equation} \varphi (a^{i_1}+\dots +a^{i_k} )= (b^{i_1}+\dots +b^{i_k}) f. \end{equation}

This means that the bijection φ maps U_a to U_bf. Therefore, there exists a bijection $\alpha\colon U_a\to U_b$ such that φ sends $u\in U_a$ to $\alpha(u)f$. The form of α follows from Condition (4.1), and this form stipulates that α is an isomorphism of fields $U_a \cong U_b$. Both of them are subfields of $\mathbb F$. Since they are of the same order, they have to coincide.

Note that any subfield that contains both a and b forms a subquasigroup of Q, by Condition (1.1). Hence, U_a is a subquasigroup that contains both 0 and 1. Therefore, $U_a \supseteq U$. Since $0*1 =a \in U$, we have $U\supseteq U_a$ as well, so $U=U_a$.

The field automorphism $\alpha \in \operatorname{Aut}(U)$ extends to an automorphism of $\mathbb F$ that sends each $x\in \mathbb F$ to x ^γ, where γ divides $|U|$, which in turn divides $|\mathbb F|$.

Every element of a subfield $\mathbb K$ of $\mathbb F$ is a square in $\mathbb F$ if and only if $|\mathbb F:\mathbb K|$ is an even number. If $|\mathbb F:U|$ is even, then γ ² divides $|\mathbb F|$.

Suppose that $|\mathbb F:U|$ is odd. Then there exists $u\in U$ such that u is a non-square in $\mathbb F$. Since $0*u = bu=a^\gamma u$ and φ is an automorphism, $\varphi(a^\gamma u)=\varphi(0) *\varphi(u) = 0*u^\gamma f = au^\gamma f$. Since $a^\gamma u \in U$, we also have $\varphi(a^\gamma u) = a^{\gamma^2} u^\gamma f$. Therefore, $a^{\gamma^2} = a$. Since $U=U_a$ is the least subfield containing a, we see that $x^{\gamma^2} = x$ for every $x\in U$, by Lemma 4.1. Also $a\ne b=a^\gamma$, so $\gamma\ne1$ and $U = \mathbb F_{\gamma^2}$. The rest follows from Proposition 1.4.

Proof. The first step is to prove for each $i\geqslant 0$ that

(4.2)\begin{align} \varphi(a^ib^i) = a^ib^i f,\ \text{and} \end{align}

(4.3)\begin{align} \varphi(a^{i+1}b^{i}) =a^{i}b^{i+1} f. \end{align}

Now $\varphi(a)=\varphi(0*1)=\varphi(0)*\varphi(1)=0*f=bf$, so Condition (4.3) holds for i = 0. Also Condition (4.2) holds for i = 0 by the definition of f. Note that $a^ib^i$ is a square and $a^ib^{i-1}$ is a non-square. So by induction on i, we find that

\begin{gather*} \varphi(a^{i}b^{i})=\varphi(0*a^ib^{i-1})=\varphi(0)*\varphi(a^ib^{i-1})=0*a^{i-1}b^if=a^{i}b^{i}f\ \text{and} \\ \varphi(a^{i+1}b^{i})=\varphi(0*a^ib^i)=\varphi(0)*\varphi(a^ib^i)=0*a^ib^if=a^{i}b^{i+1}f, \end{gather*}

completing the proof of Conditions (4.2) and (4.3).

Denote by C the subfield of $\mathbb F$ generated by ab. By Lemma 4.1, each element of C may be expressed as a sum $(ab)^{i_1}+\dots + (ab)^{i_k}$. Hence, $\varphi(c) = cf$ for each $c\in C$, by Condition (4.2).

Let us now assume that there exists $\zeta\in C$ such that $\chi(\zeta) = -1$. In such a case, the non-squares of C coincide with the non-squares of $\mathbb F$ contained in C, and, as we shall prove, whenever $x,y \in \mathbb F$ such that $\chi(x) = \chi(y)$ and that $\varphi(x) = yf$, then

(4.4)\begin{equation} (\forall c\in C\colon\ \varphi(xc) = ycf) \quad \Rightarrow \quad (\forall c\in C\colon\ \varphi(axc) = bycf). \end{equation}

Let us assume that the hypothesis of Condition (4.4) is true. Our aim is to show that then $\varphi(axc) = bycf$ for each $c\in C$. The first step is a choice of $x^{\prime},y^{\prime}\in \mathbb F$. If x is a square, put $(x^{\prime},y^{\prime})=(x,y)$. If x is a non-square, put $(x^{\prime},y^{\prime}) = (x\zeta,y\zeta)$, where ζ is as above. By the hypothesis of Condition (4.4), $\varphi(x^{\prime}s) = y^{\prime}sf$ for each square $s\in C$. Since $x^{\prime}s$ and $y^{\prime}s$ are squares, $\varphi(ax^{\prime}s)=\varphi(0*x^{\prime}s)=\varphi(0)*\varphi(x^{\prime}s)=0*y^{\prime}sf=by^{\prime}sf$ for each square $s\in C$. If $t\in C$ is another square, then $\varphi(ax^{\prime}(s+t))=by^{\prime}(s+t)f$ since φ is additive. Therefore, $\varphi(ax^{\prime}c) = by^{\prime}cf$ for each $c\in C$, by Lemma 4.2. That finishes the proof of Condition (4.4) since $\zeta \in C$.

Assuming the existence of ζ, Condition (4.4) implies $\varphi(a^ic) = b^i c f$, for each $i \geqslant 0$ and $c\in C$. Indeed, since we have proved that $\varphi(c) = cf$ for each $c\in C$, the equality holds for i = 0. The induction step follows from Condition (4.4), by setting $(x,y) = (a^i,b^i)$.

We have shown that if C carries a non-square in $\mathbb F$, then $\varphi(a^i) =b^if$ for every $i \geqslant 0$. That allows us to draw the needed conclusions from Lemma 4.3. So, for the rest of the proof, we may assume that each element of C is a square in $\mathbb F$. In particular, $\chi(-1) = 1$.

Now $\varphi(a)=\varphi(0*1) =\varphi(0)*\varphi(1) = 0*f =bf$ and $\chi(a-1)=\chi(b-1)$, by Condition (1.2). Thus, the claim

(4.5)\begin{equation} \varphi(a^i)=b^if\quad \text{and } \chi(a^i-1)=\chi(b^i-1) \end{equation}

holds for i = 1. Let us now show that if $\varphi(a^j)=b^jf$ holds for each positive $j\leqslant i$, then $\chi(a^i-1)=\chi(b^i-1)$. Since $\chi(a-1)=\chi(b-1)$, it suffices to show that $\chi(A)=\chi(B)$, where $A=\sum_{j=0}^{i-1}a^j$ and $B=\sum_{j=0}^{i-1}b^j$. Note that $\varphi(A) = Bf$. It follows that A = 0 if and only if B = 0, so we may assume that $A\ne0$ and $B\ne0$. Suppose that A is a non-zero square, so that $0*A=aA$. Since φ is an additive automorphism, we must have $0*fB=\varphi(aA)=fbB$. However, that is possible if and only if B is a square. Conversely, if B is a square, then $\varphi(0*A)=0*fB=fbB=\varphi(aA)$, implying that $0*A = aA$. Hence, A is a square in $\mathbb F$ if and only if B is a square in $\mathbb F$, and the same holds for $a^i-1$ and $b^i-1$.

Thus, Condition (4.5) holds for all $i\geqslant 1$ if we can prove that its validity for a given $i\geqslant 1$ implies $\varphi(a^{i+1}) = b^{i+1}f$. For even i, we need only observe that $\varphi(a^{i+1})=\varphi(0*a^i)=\varphi(0)*\varphi(a^i)=0*b^if=b^{i+1}f$. So we may suppose that i is odd.

Assume first that $a^i-1$ is a square. Then

\begin{equation*}\varphi(a^{i+1}-a)=\varphi(0*(a^i-1))=\varphi(0)*\varphi(a^i-1)=0*(b^i-1)f = b^{i+1}f - bf.\end{equation*}

Since $\varphi(-a) =-bf$ and φ is additive, we have $\varphi(a^{i+1})=b^{i+1}f$.

Suppose now that $a^i-1$ is a non-square. Then $b^i-1$ is a non-square too, and $(b^i-1)(-a^i) = a^i - (ab)^i$ is a square. By the inductive assumption and Condition (4.2), we see that

\begin{equation*}\varphi(a^i - (ab)^i) = (b^i -(ab)^i)f=(a^i-1)(-b^i)f\end{equation*}

is a non-square. Hence, by Condition (4.3),

\begin{align*} \varphi(a^{i+1})-a^ib^{i+1}f&=\varphi(a^{i+1} - a^{i+1}b^i)=\varphi(0*(a^i-(ab)^i)) = 0*(b^i -(ab)^i)f\\ &=b^{i+1}f - a^ib^{i+1}f. \end{align*}

Therefore, $\varphi(a^{i+1})=b^{i+1}f$, completing the proof of Condition (4.5). We see that Lemma 4.3 may be applied in this case too.

Proof. If a and b are non-squares, then the result follows from Lemma 4.4, so we assume that a and b are squares. It is then easy to show by induction that $\varphi(a^i) = b^i f$ for every $i \geqslant 1$, where $f=\varphi(1)$. Indeed $\varphi(a) = bf$ since $0*1 =a $, while $\varphi(a^{i+1}) = \varphi(0*a^i) = 0 * b^if = b^{i+1}f$ yields the induction step. By Lemma 4.3, U is a subquasigroup, $b =a^\gamma$ for γ > 1 such that $\gamma^2 \bigm| |\mathbb F|$, and the statement is true if U contains an element that is a non-square in $\mathbb F$. Thus, for the rest of the proof, it will be assumed that each element of U is a square in $\mathbb F$. Let us also stipulate that γ > 1 is the least possible.

The next step is to show that if $\chi(\varphi(f))=1$, then $\varphi(u^\gamma f) = u \varphi(f)$, while for $\chi(\varphi(f))=-1$ we have $\varphi(uf) = u \varphi(f)$, for each $u \in U$.

We first assume $\chi(\varphi(f)) = 1$ and employ induction to prove that $\varphi(b^if) = a^i\varphi(f)$ for each $i\geqslant 0$. The case i = 0 is trivial, and $\varphi(b^{i+1}f) = \varphi(0*b^if) = 0*a^i\varphi(f) = a^{i+1}\varphi(f)$ completes the induction. Each $u\in U$ may be expressed as $\sum a^{i_j}$, by Lemma 4.1. In such a case, $u^\gamma = \sum b^{i_j}$ and $\varphi(u^\gamma f) = u\varphi(f)$, by the additivity of φ.

Assume now that $\chi(\varphi(f)) = -1$. Since b is the image of a under the field automorphism $x\mapsto x^\gamma$, we have $a = b^{\gamma'}$, where $x\mapsto x^{\gamma'}$ is the inverse automorphism. Thus, each $u\in U$ may be expressed as $\sum b^{i_j}$, by Lemma 4.1. Also, $\varphi(b^if) = b^i\varphi(f)$ for each $i\geqslant 0$ by induction, since $\varphi(b^{i+1}f) = \varphi(0*b^if) =0 * b^i\varphi(f) = b^{i+1}\varphi(f)$. Therefore, $\varphi(uf) = u\varphi(f)$ for each $u \in U$, as claimed.

The automorphism φ may be replaced by its composition with the affine isomorphism $x\mapsto cx$, where c is a square. Hence, f may be equal to any non-square in $\mathbb F$. By Lemma 4.2, we may choose f in such a way that $1+f$ is a square. Then

(4.6)\begin{align} a^\gamma f + \varphi(af)& =\varphi(a+af)=\varphi(0*(1+f)) \\ & =0*\varphi(1+f) \in \{af + a\varphi(f), \, a^\gamma f + a^\gamma \varphi(f)\}, \end{align}

by Condition (1.1). If $\varphi(f)$ is a non-square, then $\varphi(af) = a\varphi(f)$, by the results above. This contradicts Condition (4.6) since $a^\gamma \ne a$. Hence, $\varphi(f)$ is a square and $\varphi(af) = a^{1/\gamma} \varphi(f)$. Suppose first that

\begin{equation*} a^\gamma f + a^{1/\gamma}\varphi(f) = af + a\varphi(f)\end{equation*}

and choose $d\in \mathbb F$ such that

\begin{equation*} d^\gamma = \frac{a^\gamma - a}{a-a^{1/\gamma}}.\end{equation*}

Then $\varphi(f) = d^\gamma f$. Because $d\in U$, we also have $\varphi(d) = d^\gamma f$, by Lemma 4.3. Thus, d = f. This is a contradiction since d is a square and f is a non-square. Therefore, $a^{1/\gamma}\varphi(f) =a^\gamma \varphi(f)$, and $a^{\gamma^2} =a$. Hence, $|U|$ divides γ ² and admits a non-trivial involutory automorphism $x\mapsto x^\gamma$. Since γ has been chosen to be the least possible, γ is a proper divisor of $|U|$. Thus, $|U| = \gamma^2$.

Proof. Put $Q=Q_{a,b}$. Since $\operatorname{Aut}(Q)$ is transitive, there exists $u\in Q$ such that $\{0,u\}$ generates Q. If u is a square, then Q is generated by $\{0,1\}$ since $x\mapsto ux$ belongs to $\operatorname{Aut}(Q)$. Assume that u is a non-square. Then $x\mapsto u^{-1} x$ is an isomorphism $Q_{a,b}\cong Q_{b,a}$, by Proposition 2.1(iv), and this isomorphism sends $\{0,u\}$ to $\{0,1\}$.

Proof. By Proposition 2.1, G contains all mappings $x\mapsto \lambda x + \mu$, where $\lambda \in \mathbb F^*$ is a square. Suppose first that G is 2-transitive. Then G is sharply 2-transitive since an automorphism that fixes generators pointwise has to be the identity mapping. The mappings $x\mapsto x+\mu$ thus form a normal subgroup of G, and that makes each $\varphi \in G_0$ additive (where G ₀ is the stabilizer of 0 in G). Proposition 4.5 hence confirms that G may be 2-transitive only in the cases described above. All mappings mentioned so far are automorphisms of Q, and they form a 2-transitive group. No other automorphism of Q may thus exist.

Let us now turn to the case when G is not 2-transitive. Let Q be generated by $\{0,u\}$. Then $\chi(u) = \chi(\varphi(u))$ for each $\varphi \in G_0$ since otherwise G is 2-transitive. For each $\varphi \in G_0$, there thus exists a square $\lambda \in \mathbb F^*$ such that $\varphi(u) = \lambda u$. Since φ and $x\mapsto \lambda x$ agree on a set of generators, they agree everywhere. Nothing else is needed.

Proof. By Proposition 2.1, only the direct implication needs to be proved. Fix an isomorphism $\psi\colon Q\mapsto Q_{c,d}$ and put $G=\operatorname{Aut}(Q)$. The group G is 2-transitive if and only if $\operatorname{Aut}(Q_{c,d})$ is 2-transitive. If a = b, then c = d since this is the only 2-transitive case in which G ₀ is abelian, by Theorem 4.7. Hence, Theorem 4.7 implies that G is equal to $\operatorname{Aut}(Q_{c,d})$ in all cases. Therefore, ψ normalizes G, and hence ψ also normalizes the group of translations $x\mapsto x+\mu$. Since G is transitive, $\psi(0) = 0$ may be assumed. The normalizing property means that ψ is additive and normalizes G ₀.

By Lemma 4.6, it may be assumed that Q is generated by 0 and 1. Then 0 and $\psi(1)$ generate $Q_{c,d}$. After a possible switch of c and d, we may thus assume that $\psi(1) = 1$ as well, by Proposition 2.1(iii), (iv).

Denote by σ_s the multiplication $x\mapsto sx$, where $s\in \mathbb F^*$. If $\sigma_s\in G_0$, then $\psi \sigma_s \psi^{-1} = \sigma_t$ for some $t\in \mathbb F^*$ since ψ normalizes G ₀. Since $\psi(1) = 1$, we must have $t=\psi(s)$. Hence, $\psi(sy)=\psi\sigma_s(y) = \sigma_{t}\psi(y)=\psi(s)\psi(y)$ for all $y\in\mathbb F$, whenever $\sigma_s \in G_0$. This shows that $\psi\in \operatorname{Aut}(\mathbb F)$ if a = b.

Suppose that $a\ne b$. We shall show that $\psi\in \operatorname{Aut}(\mathbb F)$ in this case too. Indeed, if $x\in \mathbb F$ is a non-square, then x may be expressed as u + v, where both u and v are squares, by Lemma 4.2. In such a case, $\psi(xy) = \psi(uy + vy) = (\psi(u)+\psi(v))\psi(y) = \psi(x)\psi(y)$ for all $y\in\mathbb F$.

To finish, note that in $Q_{c,d}$ both $c=0*1 = \psi(0)*\psi(1) =\psi(a)$ and $\psi(b)\psi(\zeta) = \psi(b\zeta) = \psi(0*\zeta) = 0*\psi(\zeta)= d\psi(\zeta)$ hold, where ζ is any non-square. Hence, $c = \psi(a)$ and $d=\psi(b)$.

Proof. If $\alpha(S) = S$, then $\alpha(X)\subseteq S$. Conversely, assume $\alpha(X)\subseteq S$, and denote by $S^{\prime}$ the subquasigroup generated by $\alpha(X)$. We have $\alpha(X)\subseteq S^{\prime}\cap S$, so $S^{\prime} \subseteq S$. However, α is an isomorphism from S to $S^{\prime}$ so $|S^{\prime}|=|S|$, which means that $\alpha(S)=S^{\prime} = S$.

Proof. The set is closed under sums and products, and hence a subfield. By Definition (1.1), it is closed under $*$ as well.

Proof. By Proposition 3.4 and Lemma 2.7, S is a subspace of $\mathbb F$. Moreover, if Q is affine Steiner, then S is equal to $\{0,1,-1\}$ and is a subfield that contains $a=b=-1$. For the rest of the proof, it thus may be assumed that Q is not a Steiner quasigroup.

Put $S_0 = \{\sum x_i:x_i\in S$ and $\chi(x_i) \geqslant 0 \}$ and note that $S_0\subseteq S$. If $s\in S$ is a non-zero square, then the automorphism $x\mapsto sx$ sends 0 to 0 and 1 to $s\in S$. Thus, 0 and s generate S, by Lemma 5.1. Hence, sS = S and $sS_0\subseteq S_0$. Since S ₀ is a subspace, $S_0 S_0\subseteq S_0$, and that implies that S ₀ is a subfield of $\mathbb F$. Furthermore, S is a vector space over S ₀, since $x_iS \subseteq S$ yields $(\sum x_i)S \subseteq S$.

If $\{a,b\}\subseteq S_0$, then $S=S_0$, by Lemma 5.2. Note that $a=0*1$ always belongs to S.

Let a be a square. Then $a^i\in S_0$ for each $i\geqslant 0$ since $0*a^i= a^{i+1}$. Hence, $S_0\supseteq S_a$ by Lemma 4.1, where S_a is the least subfield of $\mathbb F$ that contains a. If S ₀ consists only of squares, then S_a is a subquasigroup, and $S_a=S_0=S$. Suppose that S ₀ contains a non-square, say ζ. To show that $b\in S_0$, it suffices to show that $b\in S$, since b is a square. Now, $b\zeta = 0 * \zeta \in S$. Hence, $b=\zeta^{-1} (\zeta b)\in S$, since $\zeta^{-1} \in S_0$ and S ₀ acts on S.

For the remainder of the proof, let a and b be non-squares. If S ₀ contains a non-square, say ζ, then $b\zeta = 0*\zeta \in S_0$. Therefore, $b = \zeta^{-1}(\zeta b) \in S_0$. Since $a\in S$, $0*a = ab \in S_0$, and hence $a = b^{-1} (ab) \in S_0$. Thus, if S ₀ contains a non-square, then $\{a,b\}\subseteq S_0=S$, and S contains the subfield generated by a and b. In such a case, the subfield is equal to S, since the subfield is a subquasigroup containing a and b, by Lemma 5.2.

What remains is the case in which S ₀ contains only squares, i.e. the squares of S form a subfield. We shall show that this may be always brought to a contradiction. Consider distinct $s,t\in S_0$. If $s*t = s+a(t-s)$ is a square, i.e. $s*t\in S_0$, then $a(t-s) = s*t -s$ is a square too, a contradiction. Hence, $s*t \in N_1 = S\setminus S_0$. If $s\in S_0\setminus\{0\}$, then $0*s = as \in N_1$. If $n\in N_1$, then $0*n = bn \in S_0$. Therefore, $|S_0|=|N_1|+1$. Consider now the multiplication table of $(S,*)$. Note that S is a disjoint union of S ₀ and N ₁. The subtable $S_0\times S_0$ has elements of S ₀ on the diagonal, and the rest is occupied by elements of N ₁. Therefore, all entries in subtables $S_0\times N_1$ and $N_1\times S_0$ belong to S ₀. Therefore, all entries in $N_1\times N_1$ are from N ₁, and that makes N ₁ a subquasigroup. Since S is a minimal subquasigroup and $N_1\ne S$, the only possibility is that $N_1=\{a\}$ and $S_0 = \{0,1\}$. Since S is a subspace, $\operatorname{char}(\mathbb F) = 3$, $a=-1$ and $ab = 0*a = 0*-1 = 1$. That implies $b=-1$. By Lemma 2.7, this means that Q is a Steiner quasigroup, contrary to our assumptions.

Proof. Denote by K the minimal subquasigroup generated by 0 and 1. Lemma 5.3 implies that if K contains a non-square, then $K = \mathbb K= \mathbb K_0=\mathbb K_1$. The other possibility is that K consists of squares only. Then $K = \mathbb K_0$.

Suppose that $\lambda\in \mathbb F^*$ and $\mu \in \mathbb F$. If λ is a square, then $x\mapsto \lambda x + \mu$ is an automorphism of Q, by Proposition 2.1(iii). That makes $\lambda K + \mu$ a minimal subquasigroup of Q. If K contains a non-square ζ, then $\zeta K = K$ and $\lambda K +\mu =\lambda\zeta K +\mu$.

Let S be a minimal subquasigroup of Q that contains 0. By Proposition 2.1(ii), no other subquasigroups need to be considered.

Suppose there exists $s\in S$ and $\xi\in K$ such that $\chi(s)=\chi(\xi)\ne0$. Let $\lambda=s/\xi$ and note that $x\mapsto \lambda x$ is an automorphism of Q that maps $\{0,\xi\}$ to $\{0,s\}$. By minimality, the former set generates K, while the latter set generates S. Therefore $\lambda K=S$. Such a λ always exists if K contains a non-square.

Suppose that $K=\mathbb K_0$ consists of squares only. If S contains a non-zero square, then, as we have proved, $S= \lambda \mathbb K_0$, where λ is a square. In such a case, all elements of S are squares. What remains to be characterized are those minimal subquasigroups S where $0\in S$ and all non-zero elements are non-squares.

The mapping $x\mapsto x\zeta$ yields an isomorphism $Q_{a,b}\cong Q_{b,a}$ and sends S to $S\zeta$. Applying the earlier part of the proof to $Q_{b,a}$ yields $S\zeta = \lambda \mathbb K_1$, where λ is a square.

Proof. Let S be a 2-generated subquasigroup that is not minimal. We shall first investigate the situation when S is generated by 0 and 1. The treatment is divided into a sequence of claims. The case of general S is considered at the end of the proof.

In $Q_{b,a}$, the proper minimal subquasigroups of $\mathbb K$ containing 0 are exactly all of the sets $\mathbb K_0\xi$, where ξ is a non-square. Furthermore, the field $\mathbb K_0$ consists of squares only. The minimal subquasigroups of Q that include 0 are thus equal to $s\mathbb K_0$, where $\chi(s) = 1$.

Proof. It may be assumed that µ = 0. Let $x,y\in \mathbb F$ be distinct elements. Put i = 0 if $\chi(y-x)=1$, and i = 1 if $\chi(y-x)=-1$. Then $x*y = x + a^{\gamma^i}(y-x)$ and

\begin{equation*} \zeta(x*y)^\gamma = \zeta x^\gamma + \zeta b^{\gamma^i}(y-x)^\gamma = \zeta x^\gamma + b^{\gamma^i}(\zeta y^\gamma -\zeta x^\gamma)= \zeta x^\gamma * \zeta y^\gamma,\end{equation*}

since $\chi(y^\gamma-x^\gamma)=\chi((y-x)^\gamma)=\chi(y-x)$ because γ is odd.

Proof. Point (i) coincides with Proposition 4.5. Put $\lambda = \psi(1)$. Point (ii) will be first proved under the assumption that all minimal subquasigroups of Q are of the same order. In this case, $\psi(\mathbb K) = \lambda \mathbb K$ by Theorems 5.4 and 5.5 since a minimal subquasigroup is mapped to a minimal subquasigroup and a 2-generated subquasigroup is mapped to a 2-generated subquasigroup. The mapping $x\mapsto \lambda^{-1} \psi(x)$ hence is an automorphism of Q that sends the 2-generated subquasigroup $\mathbb K$ to itself. The restriction of the latter mapping to $\mathbb K$ is $\mathbb K$-linear by Theorem 4.7. Since the mapping is $\mathbb K$-semilinear, by Proposition 5.8, it has to be a $\mathbb K$-linear automorphism of Q. Hence, ψ is $\mathbb K$-linear as well.

Let us now assume that Q contains minimal subquasigroups of two distinct sizes. Let $\alpha\in \operatorname{Aut}(\mathbb K)$ be the automorphism such that $\psi(cx) = \alpha(c)\psi(x)$ for all $c\in \mathbb K$. In this case, $|\mathbb F|=q^2$ for some q > 2 and $\psi(\mathbb F_q) = \lambda\mathbb F_q$ by Corollary 5.7. Hence, $x\mapsto \lambda^{-1} \psi(x)$ is a $\mathbb K$-semilinear mapping with automorphism α, the restriction of which yields an automorphism of $(\mathbb F_q,*)$ such that $x*y = x+a(y-x)$ for all $x,y \in \mathbb F_q$. Therefore, $\alpha(a) = a$, by the earlier part of this proof. To finish it suffices to show that $\alpha(b) = b$ too. To that end, choose a non-square $\zeta \in \mathbb F$ and note that $\psi(\zeta \mathbb F_q) = \lambda^{\prime} \zeta \mathbb F_q$ for a square $\lambda^{\prime}$, by Corollary 5.7. The mapping $x\mapsto \zeta^{-1} (\lambda^{\prime})^{-1} \psi(\zeta x)$ is an automorphism of $Q_{b,a}$ that sends $\mathbb F_q$ to $\mathbb F_q$ and is associated, as a $\mathbb K$-semilinear mapping, with the automorphism $\alpha \in \operatorname{Aut}(\mathbb K)$. Since b is now in the position of a, we must have $\alpha(b) = b$.

Proof. The set of all additive permutations σ of $\mathbb F$ that satisfy Condition (6.1) forms a group. This group contains the mapping $x\mapsto\lambda x$ for each $\lambda \in \mathbb F^*$, with $\chi(\lambda) = 1$. Since $\chi(\sigma(1))=\chi(\sigma(1)-\sigma(0))=\chi(1)=1$, it suffices to prove the statement under the assumption that $\sigma(1) = 1$. However, if σ is a permutation of $\mathbb F$ such that $\sigma(0) = 0$, $\sigma(1) = 1$ and Condition (6.1) holds, then $\sigma\in \operatorname{Aut}(\mathbb F)$ as shown by Carlitz [Reference Carlitz4].

Proof. If $\chi(v-u) = 1$, then $\sigma(u*v)=\sigma(u+a(v-u)) = \sigma(u)+a(\sigma(v)-\sigma(u))$. Hence, $\sigma(u*v) = \sigma(u)*\sigma(v)$ implies that $\chi(\sigma(v)-\sigma(u)) = 1$. The argument when $\chi(v-u) = -1$ is similar.