Subgroups of $p$-group containing element of order $p$

Question

I know that every group of order $p^n$ admits a subgroup of order $p^k,\ k=1,\dots,n$, where $p$ is a prime and $n\ge 1$. However, is it true that given an element $g$ of order $p$ we can choose $H$ of order $p^k$ such that $p\in H$? I have no clue on how to approach this, and would appreciate some help.

Answer 1

Let $G$ be a group of order $p^n$, $p$ a prime.

Proposition. If $K$ is a subgroup of $G$ and $|K|=p^k$, $0\leq k\lt n$, then there exists a subgroup $H$ of $G$ such that $K\leq H$ and $[H:K]=p$.

Proof. Since $K$ is a proper subgroup of $G$, we know that it is properly contained in its normalizer, $N$. Now consider $N/K$; this is a nontrivial $p$-group, so by Cauchy's Theorem, it contains an element $xK$ of order $p$. The subgroup $\langle xK\rangle$ corresponds, under the quotient map $N\to N/K$, to a subgroup $H$ of $N$ that contains $K$, with $[H:K]=|\langle xK\rangle| = p$. This proves the theorem. $\Box$

Now apply the theorem by starting with $K=\langle g\rangle$, lather, rinse, and repeat.