TITLE: Divisibility of group exponents when the subgroup has finite index.
QUESTION [7 upvotes]: Let $G$ be a group (not necessary finite) and $H$ a subgroup of $G$ of index $n$ such that exp $(H)<+\infty$ .

Show that $$\exp(G)<+\infty$$ and $$\exp(G)\mid\exp(H)\cdot n.$$

Remarks.

The case when $H$ is normal is a consequence of the group structure on the cosets.
If $g\in G$, in general $g^n\not\in H$, for example $G=S_5$, $H=S_4$ and $g=(12)(345)$.

REPLY [4 votes]: It has been proved already in the comments that ${\rm exp}(G)$ is finite, so let's assume that. Let $g \in G$. By factorizing $|g|$, $n$ and ${\rm exp}(H)$ into prime powers, we can easily reduce to the case when $|g|$ is a power of a prime $p$.
In the permutation action of $G$ on the cosets of $H$, $g$ maps to a product of cycles each of length a power of $p$. Let $p^a$ be the highest power of $p$ that divides $n$. Then some cycle of $g$ in this action must have length $p^b$ with $b \le a$, and so $g^{p^a}$ fixes a point and hence lies in a conjugate of $H$. So $|g^{p^a}|$ divides ${\rm exp}(H)$ and we are done.