Given a group $G$ of order 60 with 24 elements of order 5, 20 of order 3, and 15 of order 2, how do we find the sizes of centralisers of elements of $G$ without proving $G\simeq A_5$?

By considering Sylow subgroups I've managed to get the following bounds, but no better:

$4\leq\big|C_G(g_2)\big|\leq12$, $3\leq\big|C_G(g_3)\big|\leq6$, $5\leq\big|C_G(g_5)\big|\leq10$ (where $g_p\in G$ has order $p$ for each $p$).

I've read a solution which says

Since all non-trivial elements have prime order and $|G|=2^2.3.5$, $|C_G(g)|=$ 5 if $o(g)=5$, 3 if $o(g)=3$, 4 if $o(g)=2$ (all groups of order 4 are abelian).

... but I can't see how this follows!

Many thanks for any help with this!

  • 3
    $\begingroup$ A group of order $10$ that centralizes an element of order $5$ is necessarily abelian, and thus has an element of order $10$. Apply similar reasoning for the other cases. $\endgroup$
    – user641
    Apr 10 '13 at 12:56
  • $\begingroup$ @Steve D: Is this because every group of order 10 must be isomorphic to $C_{10}$ or $D_{10}$ (and therefore $C_{10}$), or is there a more general way to see it? $\endgroup$ Apr 11 '13 at 7:45
  • $\begingroup$ OK, got it now - we can just use the same argument as in Alexander Gruber's answer. $\endgroup$ Apr 11 '13 at 9:39

You've got $24$ elements of order $5$, $20$ of order $3$, and $15$ of order $2$, so including the identity, that's all $60$ elements of the group: every non-identity element has prime order.

Let $g$ have order $p$. Then surely $\langle g \rangle \leqslant C_G(g)$. If $|C_G(g)|$ has composite order, then there exists an element $h\in C_G(g)$ of order $q$ for some $q\not= p$. $g$ and $h$ commute, so $o(gh)=pq$, but contradicts that every element of $G$ has prime order. We conclude that any element $g$ of order $3$ or $5$ is self centralizing - that is, $\langle g \rangle = C_G(g)$ - and that if an element $g$ has order $2$, $|C_G(g)|=2$ or $4$.

In the latter case, let $g$ have order $2$, and note that $\langle g \rangle$ must be contained in some a Sylow $2$-subgroup $H$ of $G$. $|H_2|=4$, so either $\mathbb{Z}_4$ or $\mathbb{Z}_2\oplus \mathbb{Z}_2$. We know that $H_2=\mathbb{Z}_2\oplus \mathbb{Z}_2$ since $G$ has no elements of order four, but the group is abelian in either case, these are abelian, so in either case $g$ is centralized by all of $H_2$. Thus $|C_G(g)|=4$.

In view of the preceding proof, we can make the following generalization.

Proposition. If $G$ is a finite group with abelian Sylow subgroups in which every non-trivial element has prime power order, then for any $g\in G$, the order of $C_G(g)$ is the highest power of $o(g)$ dividing $|G|$.


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