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Let $G$ be a finite $p$-group and $K$ be a normal subgroup. I want to show that there exists a normal subgroup $N$ of $G$ such that $N \leq K$ and $[K:N]=p$. I tried in this way: from Sylow's theorem, there exists a normal series $G=G_0 \rhd G_1 \rhd \cdots \rhd G_a=\{e\}$ such that $|G_i/G_{i+1}|=p$, then $K=K \cap G_0 \geq K \cap G_1 \geq \cdots \geq K \cap G_a=\{e\}$ and $[K \cap G_i : K\cap G_{i+1}] \leq p$. Therefore there exists some $i$ such that $[K:K \cap G_i]=p$. The problem is, I cannot conclude that $K \cap G_i$ is normal in $G$. Is this approach wrong? How can I prove this?

(Note. In fact, I'm interested in only one case: when $K=G''$ (the commutator subgroup of the commutator subgroup))

Therefore..
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Hints:

Using the all-powerful theorem that says that a finite $\;p$- group has a non-trivial center, prove by induction on $\;n\;$ that any group of order $\;p^n\;$ has a normal subgroup of order $\;p^k\;$ , for all $\;0\le k\le n\;$ .

This solves at once your question...

Added as I didn't notice the requirement $\;N\lhd G\;$ also..."

A little change of plans: Using the above theorem show there exists a series as the one you mention but s.t. $\;G_i\lhd G\;\;\forall\,i\;$ , and then it follows at once that $\;K\cap G_i\lhd G\;$, too.

DonAntonio
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  • Hmm then K have a normal subgroup N with order |K|/p, but how can I prove that N is normal "in G"? – Therefore.. Apr 27 '14 at 17:05
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    You're right, it doesn't follow so smoothly as I thought (no wonder: I didn't see the condition $;N\lhd G;$ ...). I added some stuff to fix this, @Guest – DonAntonio Apr 27 '14 at 17:23
  • Still confused ;( If we set $G_i$'s to satisfy $G_i \lhd G$, then $G_0 \rhd G_1 \rhd \cdots$ cannot be satisfied.. since this series is obtained by Sylow theorem. – Therefore.. Apr 27 '14 at 17:27
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    @Guest, and that's why I remarked "using the above theorem": there exists $;G_1\lhd G;$ s.t. $;|G_1|=p;$ . Then, there exists $;\overline{G_2}\lhd G/G_1;$ s.t. $;|\overline{G_2}|=p;$ , and by the Correspondence theorem there exists $;G_2\lhd G;$ s.t. $;\overline{G_2}=G_2/G_1;$...and etc.! – DonAntonio Apr 27 '14 at 17:42
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    Oh finally I got it. Thank you @DonAntonio (and Mikko for another proof.) – Therefore.. Apr 27 '14 at 17:46
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Let $K$ be a nontrivial normal subgroup of $G$. If $K$ has order $p$, we can pick $N = 1$, so assume that $K$ has order $\geq p^2$.

Now $K \cap Z(G)$ is nontrivial, so it contains a subgroup $P$ of order $p$. Then $P$ is normal in $G$. If $K$ has order $p^2$ we may pick $N = P$, otherwise repeat this argument for $G/P$ and $K/P$.

Mikko Korhonen
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