# embedding a proper normal subgroup of a p-group

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 $$1\leq i , $$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?

• 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. – Andreas Caranti May 17 at 11:46
• That should work also. I was just trying to see if the proof of normality was correct in my attempt. Much appreciated – kishan17 May 18 at 19:08