Suppose $G$ - is a group, for each $n \in \mathbb{N}$ $G^n = \{g^n| g \in G\}$. Does the statement that $G^n$ is finite always imply, that $\langle G^n \rangle$ is finite?
What I have managed to find: If $G^n$ is finite, then G is a torsion group: Suppose $g \in G$ is an element of infinite order. That means, that for each $k \in \mathbb{N}$ elements $g^kn$ are different. Thus we have constructed a countable subset of $G^n$. But $G^n$ is finite. Contradiction.
From that it is easily deduced, that $\langle G^n \rangle$ is a finitely generated torsion group. That gives us nothing, however, as by Adyan-Novikov theorem infinite finitely-generated torsion (and even periodic!) groups actually DO exist.
And I do not know how to proceed further.
Any help will be appreciated.
Yes.
Let $G$ be a group. Its FC-center is the union of all finite conjugacy classes, or equivalently the set of elements whose centralizer has finite index.
In any group $G$, the FC-center $F(G)$ has the property that its set $T(G)$ of torsion elements is a locally finite subgroup, and $F(G)/T(G)$ is torsion-free abelian.
Now in your setting, $G$ is torsion (by Derek's comment), and hence $F(G)=T(G)$. Since $G^n\subset T(G)$ is finite, it generates a finite subgroup.