Let $G$ is a $p-$group and $H$ a proper subgroup of $G$, $|H|=p^s$. Show that $G$ has a subgroup $K$ such that $H \subset K$ and $|K|=p^{s+1}$.
I try to find a subgroup $K$ of $G$, $|K|=p^t$ with some $t>s$. Let $K=\langle g,H\rangle$ with some $g\in G,g\not\in H$, we have ${|g|=p^n,|H|=p^s}\Rightarrow|K|=p^n\times p^s$. Is this true? Can you help me!
For this we can use that a $p$-group has a non-trivial center and proceed by induction on the order. If the subgroup does not contain the center then its normalizer does and the normalizer is therefore a strictly larger subgroup. If it does contain the center then you can mod out by a central subgroup of order $p$ and use the induction.