When is the set of $n$-th powers in a group a subgroup? [closed]

Question

Let $G$ be a non abelian group and $G_n=\{x^n | x\in G\}$ and n is integer. Is there a sufficient condition that makes $G_n$ be a subgroup of $G$ for arbitrary $n$?

Answer 1

One can construct examples of such groups which are not abelian via combinatorial group theory.

Clearly a group $G$ has this property for all $n$ if $\forall x, y\in G, \forall n, \exists z\in G$, $x^ny^n=z^n$.

For a group $G$, let us construct a group $G^R$ so that $G < G^R$ and for all $x,y\in G, \forall n$, there exists $z\in G^R$ such that $x^ny^n=z^n$.

To create such a group, we simply add a new element $g_{x,y,n}$ for each such pair $(x,y)\in G\times G, n\in \mathbb{N}$, so that $(g_{x,y,n})^n=x^ny^n$. If $x^ny^n$ has infinite order in $G$, then we see that this is an amalgamated free product with $\mathbb{Z}$, hence $G$ injects into $\langle G, g_{x,y,n} | (g_{x,y,n})^n=x^ny^n \rangle$. If $x^ny^n$ has order $m$, then we also assume that $g_{x,y,n}$ has order $nm$, and hence we also get a free amalgamated product of $G$ with $\mathbb{Z}/nm\mathbb{Z}$ over $\mathbb{Z}/m\mathbb{Z}$. Taking a union of these amalgamated products, we see that $G$ will inject in the group $$\langle G, g_{x,y,n} | (g_{x,y,n})^n=x^ny^n, x,y\in G\times G, n\in\mathbb{N} \rangle = G^R.$$

Now, iterate, taking $G_0=G, G_{i+1}=(G_i)^R$, and $\hat{G} = \underset{i}{\cup}\ G_i$. Any $x,y\in \hat{G}$ will lie in $G_i$ for some $i$, and hence $x^ny^n=z^n$ will have a solution $z\in G_{i+1} \subset \hat{G}$.

This shows that if you allow (presumably) infinitely generated groups, your class of groups can contain any given group as a subgroup. In particular, your condition does not give a variety of groups (such as the $n$-Abelian groups), which is to be expected given the quantifier formulation.