Let $G$ be a group. Suppose that the order of $G$ is finite and that $H$ is a subgroup of $G$. Is it true that an element of $G$ whose order divides the order of $H$ is in $H$?
Here is my attempt:
Let $|G|=n$ and $|H|=m$. Then $m|n$. Moreover, the order of every subgroup of $G$ divides the order of $G$ and if $g \in G$, then$|g|=|\langle g \rangle|$. Since in itself $H$ is a group (under the same operation in $G$), then the order of its element divides $|H|=m$. Thus, if $|g||m$ then $\langle g\rangle\leq H$ which means that $g \in H$.
Can you help me out with this one? I just can find a way put $g$ in $H$.