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I'd appreciate input on this proof. Does it successfully prove the result?

"Let $P$ be a finite $p$-group for some prime $p$, and let $q$ be a divisor of $|P|$. Show that $P$ has a normal subgroup of order $q$."

We will use induction on $n$, where $|P|=p^n$. Clearly the result is true if $|P|=p^1$. Now assume the statement is true for all groups of order $p^k$ where $k<n$.

Since $P$ is a $p$-group it has nontrivial center, and $Z(P)$ is also a $p$-group. It follows that $P$ has a normal subgroup of order $p$, say $N$. Form the quotient group $P/N$, which has order $p^{n-1}$, and therefore has a normal subgroup of order $q$ for each divisor $q$ of $p^{n-1}$ by the induction hypothesis.

We can show that $P$ has a normal subgroup of index $i$ for $1 \le i \le p^n$. Clearly this is true for $i=1,p^n$, and $|P:N|=p^{n-1}$. Consider the canonical homomorphism $\pi:P \to P/N$, which is surjective. Then by the Correspondence Theorem, $\pi$ and $\pi^{-1}$ are inverse bijections between subgroups of $P/N$ and subgroups of $P$ containing $N$, that respect normality and index. Then since $P/N$ (by the induction hypothesis) has a normal subgroup of index $p \le i \le p^{n-2}$, so does $P$, and the result follows.

Thanks.

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  • $\begingroup$ looks good to me. $\endgroup$ – hunter Dec 6 '13 at 15:13
  • $\begingroup$ @hunter: Good, thanks. $\endgroup$ – Alex Petzke Dec 7 '13 at 1:28

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