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I am reading the following proof of Sylow's First Theorem given in Herstein's Abstract Algebra:

Suppose $G$ is a group of order $p^n m$ where $p$ a prime and $p$ does not divide $m$. Then $G$ has a subgroup of order $p^n$.

Herstein proceeds by induction with the base case being trivial. For the induction step he assumes that the result is true for all groups $H$ such that $|H| < |G|$.

Supposing the result is false for $G$, by the induction hypothesis this means that $p^n$ cannot divide $|H|$ for any subgroup $H$ of $G$ if $H \neq G$. In particular if $a \notin Z(G)$, then $C(a) \neq G$ so $p$ must then divide the order of $|G|/|C(a)|$, where $C(a)$ denotes the centraliser of $a$.

However what happens if we have a group $G$ of order $2^2\cdot5$ with class equation 1 + 4 + 5 + 5 + 5? The center of the group is trivial and it is not true that $2$ divides 5. How is this not in contradiction with what Herstein is saying above?

Thanks.

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Not clear how your argument has any bearing on Herstein's, but if $|G/C(a)|=5$, then $|C(a)|=4$, so you have the subgroup of order $4$.

Herstein's argument is by contradiction. Yes, there exists groups of order $p^nm$ with proper subgroups $H$ such that $|G|/|H|$ is not divisible by $p$ - indeed, he's proving, by contradiction, that there is always such an $H$.

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    I must be blind you are absolutely right.2011-10-23