TITLE: An infinite group $G$ and $\forall x\in G, x^n=e$
QUESTION [7 upvotes]: Let $G$ be an infinite group and $n\in \mathbb N$. If for any infinite subset $A$ of $G$ there is $a\in A$ such that $$a^n=e,~~~~(e=e_G)$$ then prove that for every element $x\in G$ we have $x^n=e$.

This question was asked me and honestly I could not find any way to break it well. I worked on the order of an element $x$ to show that every element is of finite order but it seemed not to complete a solid approach. Thanks for the time!

REPLY [3 votes]: Here is a slightly (but only very slightly) different answer:
If $X$ is the set of elements $x$ satisfying $x^n = 1$ then the complement of $X$ is finite by the assumed property.
So if $y$ is some element in the complement, we must have that $N_G(\langle y\rangle)$ has finite index in $G$ (since this index is the number of conjugates of $\langle y\rangle$.
Also, $C_G(\langle y\rangle)$ must be finite, as mentioned in one of the other answers.
Finally, we just need something related to these where we have control of the size. For this purpose, we can use the general fact that for any group $G$ and any $y\in G$, $N_G(\langle y\rangle)/C_G(\langle y\rangle)$ is finite (this is a nice little exercise).
Combining the above yields the desired contradiction.