Specifically, I'm trying to solve the following problem:
Let $G$ be a multiplicative group of order $2^k$ where $k\geq1$. Show that $G$ has a proper subgroup $H<G$ containing the subset $S=\{g^2:g\in G\}$.
This may be a duplicate question, but all I've found here are answers that if $S$ is known to be a subgroup of $G$, then it's normal, or that if $G$ is abelian, then $S$ is a subgroup. But this question is more general. For example, if $G$ is the Klein-4 group, then $S$ is not a subgroup of $G$. I know that since $G$ is a $p$-group, it necessarily has subgroups of orders $2^j$ for all $j=1,\cdots,k$, which I assume is the key to this problem. I also know that $G$ has a non-trivial center, but that seems less relevant.
Any finite $p$-group is soluble. In particular its derived group $G'$ is a proper subgroup of $G$ when $G$ is non-trivial. Suppose then that $G$ is a nontrivial finite $p$-group.
When $p=2$, the group $G/G'$ is Abelian of $2$-power order, and so $G/G'$ has a subgroup of index $2$. This has the form $H/G'$ where $H$ has index $2$ in $G$. So all squares of elements of $G$ lie in the proper subgroup $H$.
For any $p$, $G$ will have a normal subgroup $H$ of index $p$ and all $p$-th powers will lie in $H$.