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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.

Atsina
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  • If $G$ is the Klein-4 group, then $S = {e}$ which is certainly a subgroup. But you actually already knew that $S$ was a subgroup in this case, as you correctly noted that if $G$ is abelian, then $S$ is a subgroup. – Arthur Jan 02 '20 at 16:26
  • I believe you are mistaken concerning the Klein group. Every element in the Klein group has order 2, so the set of squares is the identity element. – davidlowryduda Jan 02 '20 at 16:26
  • You both are right...careless mistake on my part. Still though, in general, $S$ can't be assumed to be a subgroup. Product of squares isn't necessarily a square. I'm sure there's a good example demonstrating this, probably using the free group, but I'm not sure off the top of my head. – Atsina Jan 02 '20 at 16:28

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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 $H$.

Angina Seng
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  • Could you elaborate on why any finite $p$-group is solvable? I'm referencing Isaacs' "Algebra: A Graduate Course" textbook and not finding this information. The only thing that seems relevant is Theorem 8.3 that says "A group $G$ is solvable iff $G^{(n)}=1$ for some $n$, but I'm not sure how to understand this. – Atsina Jan 02 '20 at 16:52
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    https://math.stackexchange.com/questions/1960432/proof-that-all-p-groups-are-solvable @Atsina – Angina Seng Jan 02 '20 at 16:55