Proper subgroup of a finite $p$-group

Question

I am a beginner to $p$-groups and have the following question at hand:

Let $H$ be a proper subgroup of a finite $p$-group $G$. If $|H|=p^s$ , then there exists a subgroup of order $p^{s+1}$ containing $H$.

My thoughts:

I think Quotient groups will come handy here. But I am nit being able to use it. Can someone help?

Answer 1

Consider $N_G[H]$, the normalizer of $H$ in $G$. Since $H$ is a normal subgroup of $N_G[H]$, we can consider the factor group $N_G[H]/H$. Note that $p$ divides $\left | N_G[H]/H \right|$ so $N_G[H]/H$ has a subgroup of order $p$. This subgroup is of the form $K/H$ where $K$ is a subgroup of $N_G[H]$ containing $H$. Then $|K| = |H|(K:H) = p^sp = p^{s+1}$ as desired.

Answer 2

Define $G'=G/H$. $|G'|=|G|/|H|\implies G$ is a $p-$group. $\therefore\exists$ a normal subgroup of order $p$ in $G'$ (Say $K'$).

Let $\psi$ be the canonical projection $\psi:G\rightarrow G'$.

Define $$K=\psi^{-1}(K')=\{g\in G:gH\in K'\}$$ $\therefore K'=K/ H\implies |K|=|K'||H|=p\times p^s=p^{s+1}$

Is this proof correct?