The Sym(3) Conjecture and Alt(8)

TL;DR

This paper provides a new, computer-free proof that in the context of the Sym(3) conjecture, certain minimal counterexamples cannot have Soc(G)' isomorphic to Alt(8).

Contribution

It offers an alternative proof avoiding computational methods for a key aspect of the Sym(3) conjecture.

Findings

Minimal counterexamples' Soc(G)' cannot be isomorphic to Alt(8).

Provides a computer-free proof of a known result.

Strengthens understanding of the structure of potential counterexamples.

Abstract

We give an alternate computer-free proof of a result of Z. Arad, M. Muzychuk, and A. Oliver: if G is a minimal counterexample to the Sym(3) conjecture, then Soc(G)' cannot be isomorphic to Alt(8).

Full text

1. The $Sym(3)$ Conjecture

The $Sym(3)$ conjecture states that if $G$ is a non-trivial finite group whose conjugacy classes have distinct cardinalities, then $G$ is isomorphic to $Sym(3)$, the symmetric group of order $6$. With additional assumptions on $G$, the conclusion is known to be true: for example when $G$ is supersolvable [Markel, 1973] or when $G$ is solvable [Zhang, 1994]. Let $\mathcal{H}$ be the hypothesis of the $Sym(3)$ conjecture:

($\mathcal{H}$) The group $G$ is a finite, non-trivial group whose conjugacy classes have distinct cardinalities.

The authors Z. Arad, M. Muzychuk, and A. Oliver [2004] proved that if $G$ satisfies $\mathcal{H}$, then either $Soc(G)^{\prime}=1$ or $Soc(G)^{\prime}$ is isomorphic to one of the following:

1. $Alt(5)^{a}$, for $1\leq a\leq 5,a\neq 2$

2. $Alt(8)$

3. $PSL(3,4)^{e}$, for $1\leq e\leq 10$

4. $Alt(5)\times PSL(3,4)^{e}$, for $1\leq e\leq 10$

The authors go on to show that if $G$ is a minimal counterexample to the $Sym(3)$ conjecture, then the list of possibilities for $Soc(G)^{\prime}$ can be shortened; either $Soc(G)^{\prime}=1$ or $Soc(G)^{\prime}$ is isomorphic to one of the following:

1. $Alt(5)^{a}$, for $3\leq a\leq 5$

2. $PSL(3,4)^{e}$, for $1\leq e\leq 10$

So in particular, in a minimal counterexample to the $Sym(3)$ conjecture, $Soc(G)^{\prime}$ cannot be isomorphic to $Alt(8)$. Their proof makes use of computer programs in the algebra system GAP. The purpose of this note is to give an alternate computer-free proof of their result for $Alt(8)$:

Theorem: Let $G$ be a minimal counterexample to the $Sym(3)$ conjecture. Then $Soc(G)^{\prime}$ is not isomorphic to $Alt(8)$.

In what follows, we will let $Sym(n)$ be the symmetric group on $n$ letters, and $Alt(n)$ the alternating group on $n$ letters. For any group $G$, we will let $Soc(G)$ be the socle of $G$ (the subgroup generated by the minimal normal subgroups of $G$), $Z(G)$ the center of $G$, $G^{\prime}$ the derived subgroup of $G$, $Aut(G)$ the automorphism group of $G$, $Inn(G)$ inner automorphism group of $G$, and $Out(G)$ the outer automorphism group of $G$. If $X$ is a subset of $G$, then $C_{G}(X)$ will be the centralizer of $X$ in $G$, and $N_{G}(X)$ will be the normalizer of $X$ in $G$. If $g\in G$, we will let $g^{G}$ denote the set of all elements of $G$ that are conjugate in $G$ to $g$. In other words, $g^{G}$ is the conjugacy class in $G$ containing $g$. Euler’s totient function will be denoted by $\phi$.

A finite group is said to be a rational group if every ordinary (i.e. complex) character is rational-valued. That is, for every ordinary character $\chi$ of $G$ and for every $g\in G$, we have $\chi(g)\in\mathbb{Q}$. This is equivalent to only requiring that the irreducible ordinary characters of $G$ are rational-valued. We have the following characterization of rational groups (see [Isaacs], [Serre], or [Ellard, 2013]):

Lemma 1.1**.**

*Let $G$ be a finite group. Then the following are equivalent:

(i) $G$ is a rational group

(ii) For all $g\in G$, all generators of $\langle g\rangle$ are conjugate in $G$.

(iii) For all $g\in G$, $[N_{G}(\langle g\rangle){:}C_{G}(\langle g\rangle)]=\phi(ord(g))$.

The finite symmetric groups $Sym(n)$ for $n\geq 1$ are rational groups [Isaacs]. Also, any group satisfying hypothesis $\mathcal{H}$ is a rational group:

Lemma 1.2**.**

Let $G$ be a group which satisfies hypothesis $\mathcal{H}$. Then $G$ is a rational group.

Proof.

Let $G$ be a group which satisfies hypothesis $\mathcal{H}$, and let $g\in G$. By the above lemma, it suffices to show that all generators of $\langle g\rangle$ are conjugate in $G$. Assume that $g_{1}$ and $g_{2}$ are elements of $G$ which generate $\langle g\rangle$. We wish to show that $g_{1}$ and $g_{2}$ are conjugate. Note that $C_{G}(\langle g_{1}\rangle)=C_{G}(\langle g_{2}\rangle)$. Therefore

$|g_{1}^{G}|=[G{:}C_{G}(g_{1})]=[G{:}C_{G}(\langle g_{1}\rangle)]=[G{:}C_{G}(\langle g_{2}\rangle)]=[G{:}C_{G}(g_{2})]=|g_{2}^{G}|$

and therefore the conjugacy classes $g_{1}^{G}$ and $g_{2}^{G}$ have the same cardinalities. So by hypothesis $\mathcal{H}$, $g_{1}^{G}$ and $g_{2}^{G}$ are the same conjugacy class, so $g_{1}$ and $g_{2}$ are conjugate in $G$. ∎

We also will make use of the fact that if $H$ and $K$ are finite groups, then $H\times K$ is rational if and only if both $H$ and $K$ are rational. This is true because the irreducible ordinary characters of $H\times K$ are precisely the functions of the form $(h,k)\mapsto\chi_{1}(h)\cdot\chi_{2}(k)$ where $\chi_{1}$ and $\chi_{2}$ are irreducible ordinary characters of $H$ and $K$, respectively. [Isaacs]

The term Frattini’s Argument refers to the theorem stating that if G is a finite group and if $N$ is a normal subgroup of $G$, and if $P$ is a Sylow $p$-subgroup of $N$, then $G=N_{G}(P)N$. This theorem is often generalized as follows: if $G$ is a finite group acting transitively on a set $\Omega$, and if $N$ is a normal subgroup of $G$ which is also transitive on $\Omega$, then for any $\omega\in\Omega$, $G=Stab_{G}(\omega)N$. One can go a little further with the following:

Lemma 1.3**.**

*(Extended Frattini Argument) Let $G$ be a finite group acting (not necessarily transitively) on a set $\Omega$, let $N$ be a normal subgroup of $G$ and let $\omega\in\Omega$. Then:

(i) $|\omega^{G}|/|\omega^{N}|=[G{:}Stab_{G}(\omega)N]$

(ii) $|\omega^{G}|/|\omega^{N}|=1$ iff $G=Stab_{G}(\omega)N$

(iii) $|\omega^{G}|/|\omega^{N}|=[G{:}N]$ iff $Stab_{G}(\omega)\leq N$

(iv) $|\omega^{G}|/|\omega^{N}|<[G{:}N]$ iff $Stab_{G}(\omega)$ is not a subgroup of $N$

Proof.

(i) $|Stab_{G}(\omega)N|=|Stab_{G}(\omega)||N|/|Stab_{G}(\omega)\cap N|=|Stab_{G}(\omega)||N|/|Stab_{N}(\omega)|$

So $|Stab_{G}(\omega)N|/|Stab_{G}(\omega)|=|N|/|Stab_{N}(\omega)|$

Therefore, $[Stab_{G}(\omega)N{:}Stab_{G}(\omega)]=[N{:}Stab_{N}(\omega)]$

So $|\omega^{G}|=[G{:}Stab_{G}(\omega)]=[G{:}Stab_{G}(\omega)N][Stab_{G}(\omega)N{:}Stab_{G}(\omega)]=$

$[G{:}Stab_{G}(\omega)N][N{:}Stab_{N}(\omega)]=[G{:}Stab_{G}(\omega)N]|\omega^{N}|$.

Therefore, $|\omega^{G}|/|\omega^{N}|=[G{:}Stab_{G}(\omega)N]$. This proves (i).

(ii) $|\omega^{G}|/|\omega^{N}|=1$ iff $[G{:}Stab_{G}(\omega)N]=1$ iff $G=Stab_{G}(\omega)N$. This proves (ii).

(iii) $|\omega^{G}|/|\omega^{N}|=[G{:}N]$ iff $[G{:}Stab_{G}(\omega)N]=[G{:}N]$ iff $Stab_{G}(\omega)N=N$ iff $Stab_{G}(\omega)\leq N$. This proves (iii).

(iv) $|\omega^{G}|/|\omega^{N}|<[G{:}N]$ iff $[G{:}Stab_{G}(\omega)N]<[G{:}N]$ iff $|N|<|Stab_{G}(\omega)N|$ iff $Stab_{G}(\omega)$ is not a subgroup of $N$. This proves (iv). ∎

A particular instance of this Extended Frattini Argument is the following:

Lemma 1.4**.**

*Let $G$ be a finite group, let $N$ be a normal subgroup of $G$, and let $n\in N$. Then:

(i) $|n^{G}|/|n^{N}|=[G{:}C_{G}(n)N]$

(ii) $|n^{G}|/|n^{N}|=1$ iff $G=C_{G}(n)N$

(iii) $|n^{G}|/|n^{N}|=[G{:}N]$ iff $C_{G}(n)\leq N$

(iv) $|n^{G}|/|n^{N}|<[G{:}N]$ iff $C_{G}(n)$ is not a subgroup of $N$

Proof.

Let $\omega=n$, let $\Omega=N$, and apply Lemma 1.3 ∎

The group $Alt(8)$ is a simple subgroup of index $2$ in $Sym(8)$, and so we have $|Alt(8)|=\frac{1}{2}\cdot 8!=20{,}160$. $Alt(8)$ has two classes of involutions: the first with cycle structure $(ab)(cd)$ and cardinality $210$ and the second with cycle structure $(ab)(cd)(ef)(gh)$ and cardinality $105$. $Alt(8)$ also has two classes of elements of order $3$; cycle structures $(abc)$ and $(abc)(def)$ with cardinalities $112$ and $1{,}120$ respectively. $Alt(8)$ has two classes of elements of order $4$ with cycle structures $(abcd)(efgh)$ and $(abcd)(ef)$ and cardinalities $1{,}260$ and $2{,}520$ respectively. And finally, $Alt(8)$ has two classes of elements of order $6$ with cycle structures $(abcdef)(gh)$ and $(abc)(de)(fg)$ and cardinalities $3{,}360$ and $1{,}680$ respectively.

Note that $Alt(8)$ has only one class of cardinality $1{,}680$, since an element $a$ in such a class would satisfy $|C_{Alt(8)}(a)|=12$, and therefore the order of $a$ would be a divisor of $12$. The order of $a$ cannot be $1$, and it cannot be $12$ since $Alt(8)$ has no element of order $12$. Therefore the order of $a$ is $2,3,4,$ or $6$. But among these orders, from the previous paragraph, only one has a class of cardinality $1{,}680$, the class of an element with cycle structure $(abc)(de)(fg)$ and order $6$. Similarly, $Alt(8)$ has only one class of cardinality $2{,}520$, since an element $a$ in such a class would satisfy $|C_{Alt(8)}(a)|=8$, and therefore the order of $a$ would be a divisor of $8$. The order of $a$ cannot be $1$, and it cannot be $8$ since $Alt(8)$ has no element of order $8$. So the order of $a$ is $2$ or $4$. But among these orders, only one has a class of cardinality $2{,}520$, the class of an element with structure $(abcd)(ef)$ and order $4$.

Lemma 1.5**.**

$Alt(8)$* is not a rational group.*

Proof.

Let $a=(1234567)$, an element of order $7$ in $Alt(8)$. By the previous characterization of rational groups, it suffices to show that

$[N_{Alt(8)}(\langle a\rangle){:}C_{Alt(8)}(\langle a\rangle)]\neq\phi(ord(a))$.

We have $C_{Alt(8)}(\langle a\rangle)=\langle a\rangle$, and so $|C_{Alt(8)}(\langle a\rangle)|=7$. Therefore,

$[Alt(8){:}C_{Alt(8)}(\langle a\rangle)]=\frac{1}{7}(\frac{1}{2}8!)\equiv 3\hskip 2.84544pt(mod\hskip 2.84544pt7)$.

But $\langle a\rangle$ is a Sylow $7$-subgroup of $Alt(8)$ and so by a Sylow theorem we have

$[Alt(8){:}N_{Alt(8)}(\langle a\rangle)]\equiv 1(mod\hskip 2.84544pt7)$. Therefore,

$[N_{Alt(8)}(\langle a\rangle){:}C_{Alt(8)}(\langle a\rangle)]\equiv 3(mod\hskip 2.84544pt7)$, and so

$[N_{Alt(8)}(\langle a\rangle){:}C_{Alt(8)}(\langle a\rangle)]\neq 6=\phi(7)=\phi(ord(a))$. ∎

Note that since $[N_{Alt(8)}(\langle a\rangle){:}C_{Alt(8)}(\langle a\rangle)]$ is a divisor of $\phi(ord(a))=6$ and congruent to $3\hskip 2.84544pt(mod\hskip 2.84544pt7)$, it must be $3$. Therefore $Alt(8)$ does not induce an automorphism of order $2$ by conjugation on $\langle a\rangle$. Therefore, $a$ is not conjugate to $a^{-1}$ in $Alt(8)$.

Since $Alt(8)$ is not a rational group, it follows that $Alt(8)$ cannot satisfy hypothesis $\mathcal{H}$. Also, $Sym(8)$ does not satisfy hypothesis $\mathcal{H}$; in fact the permutations $(123456)$ and $(123456)(78)$ have the same centralizer in $Sym(8)$ (and therefore their conjugacy classes have the same cardinality) but since they have different cycle structures, they are not conjugate in $Sym(8)$. Finally, we note that $Aut(Alt(8))\cong Sym(8)$, and therefore $|Out(Alt(8))|=2$.

We can now prove the main theorem:

Theorem 1.6**.**

Let $G$ be a minimal counterexample to the $Sym(3)$ conjecture. Then $Soc(G)^{\prime}$ is not isomorphic to $Alt(8)$.

Proof.

Let G be a minimal counterexample to the $Sym(3)$ conjecture; so $G$ satisfies the hypothesis $\mathcal{H}$, but $G$ is not isomorphic to $Sym(3)$ and $G$ is a group of least cardinality satisfying hypothesis $\mathcal{H}$ but not isomorphic to $Sym(3)$. We wish to prove that $Soc(G)^{\prime}$ is not isomorphic to $Alt(8)$. Assume to the contrary that $Soc(G)^{\prime}$ is isomorphic to $Alt(8)$. We wish to derive a contradiction.

Let $A=Soc(G)^{\prime}$ and let $B=C_{G}(A)$. $A=Soc(G)^{\prime}$ char $Soc(G)\trianglelefteq G$, and so $A\trianglelefteq G$. Also, $B=C_{G}(A)\trianglelefteq N_{G}(A)=G$, and so $B\trianglelefteq G$. Therefore $A\cap B\trianglelefteq A$, so since $A$ is simple, $A\cap B=1$ or $A\cap B=A$. But $A\cap B=A$ would imply $A\leq B=C_{G}(A)$ which would imply $A$ was abelian. So $A\cap B=1$. Therefore $B$ is a proper subgroup of $G$ and $AB\cong A\times B$.

Claim: $B$ is finite, and $B\neq 1$.

Since $G$ is finite and $B$ is a subgroup of $G$, $B$ is finite, too. The homomorphism $\theta:G\rightarrow Aut(A)$ induced by conjugation has kernel $B$ and maps $AB$ onto $Inn(A)$, and so $G/B\cong Im(\theta)$ is a subgroup of $Aut(A)$ which contains $Inn(A)$. Since $Aut(A)\cong Sym(8)$ and $Inn(A)\cong Alt(8)$, it follows that either $G/B\cong Alt(8)$ or $G/B\cong Sym(8)$. Since neither $Alt(8)$ nor $Sym(8)$ satisfy hypothesis $\mathcal{H}$, it follows that we cannot have $B=1$.

Claim: $[G:AB]=2$.

Since $\theta$ maps $AB$ onto $Inn(A)$, then by the correspondence theorem we have $[G{:}AB]=[Im(\theta){:}Inn(A)]$, which divides $[Aut(A){:}Inn(A)]=[Sym(8){:}Alt(8)]=2$, and so $[G{:}AB]$ divides $2$. If $[G{:}AB]$ were to equal 1, we would have $G\cong A\times B$; but since $G$ is a rational group, this would imply that $A\times B$, and therefore both $A$ and $B$, were rational groups. But $A$ is not a rational group, and so we must have $[G{:}AB]=2$.

Claim: $C_{G}(B)\leq AB$.

Suppose not. We wish to derive a contradiction. Then since $[G{:}AB]=2$, we would have $G=C_{G}(B)AB$. But since $B=C_{G}(A)$, we have $A\leq C_{G}(B)$, and so $G=C_{G}(B)AB=C_{G}(B)B$.

We will consider two cases: $Z(B)=1$ and $Z(B)\neq 1$.

Case 1: Assume that $Z(B)=1$. We have just shown that $G=C_{G}(B)B$. Also, $C_{G}(B)\cap B=Z(B)=1$. Note that $C_{G}(B)\trianglelefteq N_{G}(B)=G$ and so both $C_{G}(B)$ and $B$ are normal in $G$. Thus $G\cong C_{G}(B)\times B$. In the proof of our first claim above, we have shown that either $G/B\cong Alt(8)$ or $G/B\cong Sym(8)$. But since $|G/B|=|G|/|B|=2|A||B|/|B|=2|A|=2|Alt(8)|$, we must have $G/B\cong Sym(8)$. So $C_{G}(B)\cong Sym(8)$ and so $G\cong Sym(8)\times B$. $Sym(8)$ does not satisfy hypothesis $\mathcal{H}$; in fact the permutations $s_{1}=(123456)$ and $s_{2}=(123456)(78)$ have the same centralizer in $Sym(8)$ but since they have different cycle structures they are not conjugate in $Sym(8)$, and therefore, since $G\cong Sym(8)\times B$, they are not conjugate in $G$. Therefore,

$C_{G}(s_{1})=C_{Sym(8)}(s_{1})\times B=C_{Sym(8)}(s_{2})\times B=C_{G}(s_{1})$

so

$|s_{1}^{G}|=[G{:}C_{G}(s_{1})]=[G{:}C_{G}(s_{2})]=|s_{2}^{G}|$.

But since $G$ satisfies hypothesis $\mathcal{H}$, and since the conjugacy classes $s_{1}^{G}$ and $s_{2}^{G}$ have the same cardinality, they must be the same conjugacy class, and so $s_{1}$ and $s_{2}$ must be conjugate in $G$, a contradiction.

Case 2: Assume that $Z(B)\neq 1$. Let $z\in Z(B)$ with $z\neq 1$. So $C_{G}(B)\leq C_{G}(z)$ and $B\leq C_{G}(z)$, so $G=C_{G}(B)B\leq C_{G}(z)$, so $z\in Z(G)$. But this would give us two distinct conjugacy classes $z^{G}=\{z\}$ and $1^{G}=\{1\}$ with the same cardinality, 1. But $G$ satisfies hypothesis $\mathcal{H}$, so this is a contradiction.

This proves the claim that $C_{G}(B)\leq AB$.

Claim: If $b_{1}$, $b_{2}\in B$ and if $|b_{1}^{B}|=|b_{2}^{B}|$, and if neither $C_{G}(b_{1})\leq AB$ nor $C_{G}(b_{2})\leq AB$, then $b_{1}$ and $b_{2}$ are conjugate in $B$.

Assume that $b_{1}$, $b_{2}\in B$ and that $|b_{1}^{B}|=|b_{2}^{B}|$, and that neither $C_{G}(b_{1})\leq AB$ nor $C_{G}(b_{2})\leq AB$. We wish to show that $b_{1}$ and $b_{2}$ are conjugate in $B$. Since $[G{:}AB]=2$, $AB\trianglelefteq G$ and we have $C_{G}(b_{1})(AB)=G=C_{G}(b_{2})(AB)$. So by the Extended Frattini Argument, (using $AB$ for $N$), we get

$|b_{1}^{G}|/|b_{1}^{AB}|=1=|b_{2}^{G}|/|b_{2}^{AB}|$

and therefore $b_{1}^{AB}=b_{1}^{G}$ and $b_{2}^{AB}=b_{2}^{G}$. But

$|b_{1}^{(AB)}|=[AB{:}C_{(AB)}(b_{1})]=[AB{:}AC_{B}(b_{1})]$ and

$|b_{2}^{(AB)}|=[AB{:}C_{(AB)}(b_{2})]=[AB{:}AC_{B}(b_{2})]$

and so since $|C_{B}(b_{1})|=|C_{B}(b_{2})|$, it follows that $|b_{1}^{(AB)}|=|b_{2}^{(AB)}|$. So $|b_{1}^{G}|=|b_{2}^{G}|$. Since $G$ satisfies hypothesis $\mathcal{H}$, we get $b_{1}^{G}=b_{2}^{G}$. So

$b_{1}^{B}=b_{1}^{(AB)}=b_{1}^{G}=b_{2}^{G}=b_{2}^{(AB)}=b_{2}^{B}$

and so $b_{1}^{B}=b_{2}^{B}$, so $b_{1}$ and $b_{2}$ are conjugate in $B$. This proves the claim.

Claim: $B$ satisfies hypothesis $\mathcal{H}$.

We have already shown that $B$ is finite and non-trivial. So we now only need to show that distinct conjugacy classes of $B$ have distinct cardinalities. Assume to the contrary that there are elements $b_{1}$ and $b_{2}$ in $B$ such that $|b_{1}^{B}|=|b_{2}^{B}|$ but $b_{1}$ and $b_{2}$ are not conjugate in $B$. We wish to derive a contradiction.

Note that $|b_{1}^{B}|=|b_{2}^{B}|$ implies that $[B{:}C_{B}(b_{1})]=[B{:}C_{B}(b_{2})]$ and therefore implies that $|C_{B}(b_{1})|=|C_{B}(b_{2})|$.

Since $b_{1}$ and $b_{2}$ are not conjugate in $B$, then by the previous claim, we must have either $C_{G}(b_{1})\leq AB$ or $C_{G}(b_{2})\leq AB$. Without loss of generality, assume that $C_{G}(b_{1})\leq AB$. Let $a_{1}$ be a 7-cycle of $Alt(8)$ in $A$ and let $a_{2}=a_{1}^{-1}$. Then $a_{1}$ and $a_{2}$ are not conjugate in $Alt(8)$, but they generate the same cyclic subgroup of $G$ of order $7$; since $G$ is a rational group, $a_{1}$ and $a_{2}$ are therefore conjugate in $G$. Therefore, $a_{1}^{A}$ is a proper subset of $a_{1}^{G}$. Also, $|C_{A}(a_{1})|=|C_{A}(a_{2})|$. Since $B=C_{G}(A)$, $a_{1}^{A}=a_{1}^{AB}$. So by the Extended Frattini Argument, $1<|a_{1}^{G}|/|a_{1}^{AB}|=[G{:}C_{G}(a_{1})(AB)]\leq 2$. So $[G{:}C_{G}(a_{1})(AB)]=2$, so $G_{G}(a_{1})\leq AB$. Similarly, $G_{G}(a_{2})\leq AB$. Therefore (using $A\trianglelefteq G$, $B\trianglelefteq G$, and $AB\cong A\times B$), we get:

$C_{G}(a_{1}b_{1})=C_{G}(a_{1})\cap C_{G}(b_{1})=C_{AB}(a_{1})\cap C_{AB}(b_{1})=C_{A}(a_{1})\times C_{B}(b_{1})$.

Similarly,

$C_{G}(a_{1}b_{2})=C_{G}(a_{1})\cap C_{G}(b_{2})=C_{AB}(a_{1})\cap C_{AB}(b_{2})=C_{A}(a_{1})\times C_{B}(b_{2})$.

Thus $|C_{G}(a_{1}b_{1})|=|C_{A}(a_{1})|\times|C_{B}(b_{1})|=|C_{A}(a_{1})|\times|C_{B}(b_{2})|=|C_{G}(a_{1}b_{2})|$.

So $|(a_{1}b_{1})^{G}|=|(a_{1}b_{2})^{G}|$. Since $G$ satisfies hypothesis $\mathcal{H}$, it follows that $a_{1}b_{1}$ is conjugate to $a_{1}b_{2}$ in $G$. Choose $g\in G$ such that $(a_{1}b_{1})^{g}=a_{1}b_{2}$. Then $a_{1}^{g}b_{1}^{g}=a_{1}b_{2}$. Therefore (again using $A\trianglelefteq G$, $B\trianglelefteq G$, and $AB\cong A\times B$) we get $a_{1}^{g}=a_{1}$ and $b_{1}^{g}=b_{2}$. This implies that $g\in C_{G}(a_{1})$ and so $g\in AB$. But no element $ab$ of $AB$ can conjugate $b_{1}$ to $b_{2}$, because since $A$ centralizes $B$, this would imply that $b$ conjugates $b_{1}$ to $b_{2}$, a contradiction to our assumption that $b_{1}$ and $b_{2}$ are not conjugate in $B$. This proves the claim that $B$ satisfies hypothesis $\mathcal{H}$.

Claim: $B\cong Sym(3)$.

$B$ is a finite non-trivial proper subgroup of $G$ which satisfies the hypothesis $\mathcal{H}$ of the $Sym(3)$ conjecture. Since we are assuming that $G$ is a counterexample of least cardinality to the $Sym(3)$ conjecture, this implies that $B\cong Sym(3)$. This proves the claim.

Therefore $G$ has the normal subgroup $AB\cong Alt(8)\times Sym(3)$ of index $2$. Since $B\cong Sym(3)$, let $\sigma$ be an involution of $B$ and let $\delta$ be an element of order $3$ of $B$.

$Alt(8)$ has a unique conjugacy class of cardinality $1{,}680$ which is the class of an element $s$ of $Alt(8)$ of order $6$. Also, $Alt(8)$ has a unique conjugacy class of cardinality $2{,}520$ which is the class of an element $f$ of $Alt(8)$ of order $4$.

Since $Sym(3)\cong B\trianglelefteq G$, $\sigma^{G}\subseteq B$ and so $\sigma^{G}=\sigma^{B}$ and so by the Extended Frattini Argument, we have $G=C_{G}(\sigma)B$. So $C_{G}(\sigma)$ cannot be contained in $AB$. Similarly, we know that $C_{G}(\delta)$ cannot be contained in $AB$. Also, we have $s^{A}\subseteq s^{G}$ and $s^{G}$ is a union of $A$-conjugacy classes of elements of $A$ of order $6$ and thus has cardinality $1{,}680$ or $5{,}040$. Also, $B\leq C_{G}(s)$ and so $AB\leq C_{G}(s)A$. Then

$|s^{G}|/|s^{A}|=[G{:}C_{G}(s)A]\leq[G{:}AB]=2$, so

$|s^{G}|\leq 2|s^{A}|=3{,}360$. So $|s^{G}|=1{,}680=|s^{A}|$. So $s^{G}=s^{A}$. Similarly, we have $f^{A}\subseteq f^{G}$ and $f^{G}$ is a union of $A$-conjugacy classes of elements of $A$ of order $4$ and thus has cardinality $1{,}260$ or $3{,}780$. Also, $B\leq C_{G}(f)$ and so $AB\leq C_{G}(f)A$. Then

$|f^{G}|/|f^{A}|=[G{:}C_{G}(f)A]\leq[G{:}AB]=2$, so

$|f^{G}|\leq 2|f^{A}|=2{,}520$. So $|f^{G}|=1{,}260=|f^{A}|$. So $f^{G}=f^{A}$. Since $s^{G}=s^{A}$ and $f^{G}=f^{A}$, we know from the Extended Frattini Argument that neither $C_{G}(s)$ nor $C_{G}(f)$ can be contained in $AB$.

Claim: $C_{G}(s\sigma)$ is not contained in $AB$.

Since $C_{G}(s)$ is not contained in $AB$, we can choose $g\in C_{G}(s)$ such that $g\notin A\times B$. Since $Sym(3)\cong B\trianglelefteq G$, $g$ permutes the three involutions of $B$. But $g$ cannot fix all three for otherwise we would have $g\in C_{G}(B)$ and by the previous claim, $C_{G}(B)\leq AB$. So either $g$ fixes one involution and transposes the other two, or $g$ permutes the three involutions in a $3$-cycle. But $g$ cannot permute the involutions in a $3$-cycle, because in this case, $g^{3}$ (which is also not in $AB$, since $[G{:}AB]=2$) would be in $C_{G}(B)$, which by a previous claim is a subgroup of $AB$. So $g$ must fix one involution and transpose the other two. Without loss of generality, we assume that $\sigma$ is the involution of $B$ fixed by $g$. Then $g\in C_{G}(s)\cap G_{G}(\sigma)$. Thus, $g\in C_{G}(s\sigma)$. So $g$ is an element of $C_{G}(s\sigma)$ which is not an element of $AB$. This proves the claim that $C_{G}(s\sigma)$ is not contained in $AB$.

Claim: $C_{G}(f\delta)$ is not contained in $AB$.

Since $C_{G}(f)$ is not contained in $AB$, we can choose $h\in C_{G}(f)$ such that $h\notin AB$. Since $Sym(3)\cong B\trianglelefteq G$, $h$ permutes the two elements of order $3$ of $B$. If $h$ fixes both of them, then $h\in C_{G}(\delta)$, and so we can take $h$ as our element in $C_{G}(f\delta)$ which is not in $AB$. Otherwise, assume that $h$ transposes the two elements of order $3$ of $B$. Since $C_{G}(\delta)$ cannot be contained in $AB$, there is an element $h^{\prime}\in G$ which is not in $AB$ which centralizes $\delta$. Every element in $G$ which is not $AB$ can be written as $hab$, so write $h^{\prime}=hab$. Then $(\delta)^{hab}=\delta$ so since $(\delta)^{h}\in B$, and $A$ centralizes $B$, we get $(\delta)^{ha}=(\delta)^{h}$ so $(\delta)^{hab}=(\delta)^{hb}$. So $hb$ centralizes $\delta$. But $hb$ also centralizes $f$ (since both $h$ and $b$ do), and so $hb$ is in in $C_{G}(f\delta)$. And $hb$ is not in $AB$ (since $b$ is, but $h$ isn’t). So $hb$ is an element of $C_{G}(f\delta)$ which is not an element of $AB$. This proves the claim that $C_{G}(f\delta)$ is not contained in $AB$.

Thus, $|C_{G}(s\sigma)|=2|C_{A}(s)\times C_{B}(\sigma)|=2|C_{A}(s)|\times|C_{B}(\sigma)|=2(12)(2)=48$.

Also, $|C_{G}(f\delta)|=2|C_{A}(f)\times C_{B}(\delta)|=2|C_{A}(f)|\times|C_{B}(\delta)|=2(8)(3)=48$.

Thus, $|C_{G}(s\sigma)|=48=|C_{G}(f\delta)|$. Therefore the conjugacy classes in $G$ of $s\sigma$ and $f\delta$ have the same cardinality. Since $G$ satisfies hypothesis $\mathcal{H}$, it follows that $s\sigma$ and $f\delta$ are conjugate in $G$. But $s\sigma$ has order $6$ while $f\delta$ has order $12$, and so they cannot be conjugate in $G$. So we have reached the desired contradiction. It follows that $Soc(G)^{\prime}$ is not isomorphic to $Alt(8)$. ∎