Orders of centralizers $C_G(g)$ in a group of order 60? 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!
 A: 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 $|G|$.

