An infinite group $G$ and $\forall x\in G, x^n=e$ 
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! 
 A: Note that $G$ must be a torsion group: if $x \in G$, then either $\langle x \rangle $ is finite, or $\langle x \rangle$ is infinite. In the latter case $A=\langle x \rangle -\{1\}$ is an infinite set, and hence $x^{kn}=1$ for some integer $k$. In both cases, $x$ has finite order.
Now consider the subset $S$ of $G$ defined by $S=\{x \in G: x^n \neq 1\}$. Assume that $S$ is non-empty. We are going to derive a contradiction from that. Observe that $S$ is a normal subset (i.e. it is closed under conjugation) and, it must be finite. For if it were infinite, then there is an $x \in S$ with $x^n=1$, contrary to the definition of $S$! Now we can apply the Dietzmann Lemma: $N:=\langle S \rangle$ is a finite normal subgroup of $G$.
Let $g \in C_G(N)$ and assume $g \notin N$, so that $g^n=1$. Let $x \in S$. Then $(gx)^n=g^nx^n=x^n\neq1$. Hence $gx \in S \subseteq N$, and $x \in N$, hence $g \in N$, a contradiction, so after all $g \in N$. This means $C_G(N) \subseteq N$. Finally use the fact that $G/C_G(N) \hookrightarrow \text{Aut}(N)$, which is a finite group. So index$]G:C_G(N)]$ is finite, and we conclude that $G$ must be finite, a contradiction.
A: 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.
A: Let $H$ be the subset of elements such that $x^n = e$. You may check and see that this is a subgroup. Also, the difference set $A = G \setminus H$ must be finite (it cannot be infinite by the given condition). Therefore $H$ is an infinite subgroup. But by something like Lagrange's theorem (think about it...), the size of the the complement of a subgroup must be an integer multiple of the size of the subgroup, and particularly, if $H$ is infinite, then $A$ must be infinite or empty. But $A$ is finite, so it is empty.
EDIT: Oops, looks like my unwritten proof that $H$ is a subgroup only works if $G$ is Abelian, so maybe this answer is incorrect. (I'll leave it here, anyway - maybe someone can fix it.)
A: Notice that $G$ has no elements of infinite order. Suppose there exists $y \in G$ with $|y| \nmid n$. Replacing $y$ with a power of $y$, we may assume $\text{gcd}(|y|, n) = 1$. If $y^G = \{gyg^{-1} \mid g \in G\}$ is infinite, we are done (since $|y| = |gyg^{-1}|$ for any $g \in G$). 
On the other hand, if $|y^G| = |G : C_G(y)| < \infty$, then $|C_G(y)| = \infty$, so there exists an infinite sequence of distinct elements $a_1, a_2, \ldots$ in $C_G(y)$ with $a_i^n = e$ for all $i$. But $|a_iy| = \text{lcm}(|y|, |a_i|) \nmid n$ for all $i$, so $\{a_iy \mid i = 1, 2, \ldots\}$ is an infinite set of elements, none of whose orders divide $n$.
