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 $n\_2$.



Groups with maximal element order 6
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.


