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I am trying to prove this claim: Let $G$ be a finite $p$-group and $H$ be a proper normal subgroup of order $p^k$,then $H$ can be embedded into a normal subgroup of order $p^{k+1}$.

Here is my attempt:

Proof: Let $|G|=p^n$, $H\unlhd G$, $|H|=p^k$ where $0\leq k < n$. Then $|G/H|>1$, hence $Z(G/H)>1$ and since $G/H$ is a p-group, $p||Z(G/H)|$. Therefore by Cauchy's theorem, $\exists gH\in Z(G/H)$ such that $o(gH)=p$. Let $K=\langle g\rangle$, then $K\cap H=\{1\}$ since if $x\in K\cap H$, then $x=g^i$ for some $1\leq i \leq p$, but for $1\leq i <p$, $g^i\notin H$ since $o(gH)=p$, hence $i=p$ and $K\cap H =\{1\}$. Now consider $KH$, clearly $H\leq KH$ and $|KH|=p^{k+1}$.

It remains to show $KH\unlhd G$. Take $x\in G$, then $\forall 1\leq i \leq p$, $xg^iHx^{-1}=(xHg^iH)x^{-1}=(g^iH xH)x^{-1}=g^ixHx^{-1}=g^iH \subseteq KH$, noting $g^i\in Z(G/H)$ since $g\in Z(G/H)$ and $xHx^-1=H$ since $H\unlhd G$, thus $KH\unlhd G$. $\square$

Is this proof correct?

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  • $\begingroup$ Looks correct, but overcomplicated. Rather, consider that the very same argument that shows that a non-trivial finite $p$ group has a non-trivial centre yields that if $H$ is a non-trivial normal subgroup of the finite $p$-group $G$, then $H$ intersects $Z(G)$ non-trivially. So exactly as in your argument there is a subgroup $N$ of order $p$ such that $N \le H \cap Z(G)$. Now pass to the quotient group $G/N$ and argue by induction on $k$, say. $\endgroup$ – Andreas Caranti May 17 at 11:46
  • $\begingroup$ That should work also. I was just trying to see if the proof of normality was correct in my attempt. Much appreciated $\endgroup$ – kishan17 May 18 at 19:08

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