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Background

In the book of Judson's book on abstract algebra, Sylow's First Theorem is proved by first invoking the class equation and then considering the case where $p$ can/cannot divide $[G:C_G(g)]$ ($C_G(g)$ is the centraliser subgroup $g$ in $G$). But it is not clear what happens if $p$ cannot divide $[G:C_G(g)]$ when $C_G(g)$ is the whole group.

Question

If $G$ is an abelian group, and thus $C_G(g)$ is the whole group for all $g$ in $G$, the induction hypothesis can no longer require that $C_G(g)$ has a group of order $p^r$ since the order of the subgroup is the same as the group itself, then how can Sylow's First Theorem be proved? I understand that Cauchy theorem for abelian groups will say $G$ has a subgroup of order $p$, but then $G$ is now divisible by $p^r$, which is not necessarily a prime number, how should I prove that it also has a subgroup of order $p^r$ (although it definitely has a subgroup of order $p$ by Cauchy)?

The question is now resolved. It turns out that when $G$ is abelian, $G$ = $Z(G)$, and hence $|Z(G)|$ is now divisible by $p$, and the second part of Judson's proof will take care of this case.