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I'm dealing with finite groups $G$ in which the maximal order of an element is $6$.
With GAP I found out that for all groups with order $<1000$ the number of elements of order 6 $k := |\{x\in G : ord(x)=6\}|$ is $0$ or $2$ $\mod 6$.

I'm trying to understand, why $k \not\equiv 4 \mod 6$.

I'm collecting some facts about such a group $G$:

  • $|G|=2^\alpha 3^\beta 5^\gamma$, with $\alpha,\beta,\gamma$ being natural numbers.

  • $k$ must be even, as elements of order $6$ occur in pairs $x,x^{-1}$.

Let $n_a := |\{x\in G : x^a=e\}$ for a divisor $a$ of $|G|$. Frobenius theorem says $a$ divides $n_a$.

  • So $n_3 \equiv 3 \mod 6$, as elements of order $3$ occur pairwise.

Can anyone help me out or suggest some tools to tackle this problem.
Thanks!

Edit: I again computed with GAP in groups with small order: $n_4 \not\equiv 0 \mod 6$ if $4$ divides $|G|$. And $n_2 \not\equiv 0 \mod 6$ if $4$ does not divide $|G|$. If $5$ does not divide $|G|$ then $k=n_6 - n_4 -n_3 + 1\equiv n_4 + 4 \mod 6$. This means it suffices to prove $n_4 \not\equiv 0 \mod 6$ if $4$ divides $|G|$ or analogous for $n_2$.

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    $\begingroup$ Surely, I meant that I did not find a counterexample looking up all groups with order $<1000$. That is the reason why I'd like to prove it. $\endgroup$
    – waveman
    Feb 10, 2021 at 15:59
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    $\begingroup$ There are very few non-solvable groups with this property. I have thought of three so far: $S_5$, $S_6$, and a group with structure $2^4:A_5$ - there appear to be no simple examples. $\endgroup$
    – Derek Holt
    Feb 10, 2021 at 17:00
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    $\begingroup$ Let $E$ be a Sylow $2$-subgroup. Then $E$ must be elementary abelian. It can be regarded as a $\mathbb{Z}/2$-linear representation of the group $Q=N_G(E)/E$, which has order $3^\beta 5^\gamma$, so it splits as a direct sum of irreducible representations of $Q$. For each such irreducible subrepresentation $V$, the $Q$-equivariant endomorphism ring $F$ is a finite field of order $2^k$ for some $k$, and $V$ has dimension $1$ over $F$, so the action of $Q$ factors through a homomorphism from $Q$ to the group $F^\times$, which is cyclic of order $2^k-1$. $\endgroup$ Feb 10, 2021 at 17:27
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    $\begingroup$ @dodd Are you suggesting that, on the other hand, GAP the store chain is able to tell us anything about infinitely many groups? :-) $\endgroup$ Feb 10, 2021 at 17:50
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    $\begingroup$ Sorry, I was thinking for a moment that all orders should divide 6. If order 4 is allowed then $E$ need not be elementary abelian. $\endgroup$ Feb 10, 2021 at 18:22

1 Answer 1

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Let $P$ be a Sylow $3$-subgroup of $G$ (which we may assume to be nontrivial) and consider the conjugation action of $P$ on the set $X$ of all elements of order six in $G$. Remove from $X$ all of its $P$-fixed points. The resulting set $Y$ is a union of non-trivial $P$-orbits and therefore $|Y|$ divisible by three. As $y^{-1} \in Y$ whenever $y \in Y$, we see that $|Y|$ is even and so divisible by six.

It remains to count the $P$-fixed points in $X$, that is, the elements of order six in $C_G(P)$. Your conditions on $G$ force that $C_G(P)$ is the direct product of $Z(P)$ (an elementary abelian $3$-group) and a (possibly trivial) elementary abelian $2$-group. If $|C_G(P)|=2^a3^b$, then the number of elements of order six in $C_G(P)$ is $(2^a-1)(3^b-1)$, and your claim follows.

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