Is it true that an element of a group whose order divides the order a subgroup is an element of the subgroup 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 $H$.
 A: This is not true, let $G=\mathbb{Z}_2\oplus \mathbb{Z}_2$ and $H=\mathbb{Z}_2\oplus \left\{0\right\}$ and $z= (\bar{0},\bar{1})$. The order of $z$ is 2 and the order of $H$ is 2, but $z\notin H$.
EDIT: To answer your comments: One can show that any group of order $175$ is abelian. Moreover, $G\cong \mathbb{Z}_7\oplus \mathbb{Z}_{25}$ or $G\cong \mathbb{Z}_{7}\oplus \mathbb{Z}_5\oplus \mathbb{Z}_5$ or $G\cong \mathbb{Z}_{175}$. Assuming that $|H|=25$ and $|g|=5$ is still not enough to conclude that $g\in H$. Indeed, if $G\cong \mathbb{Z}_{7}\oplus \mathbb{Z}_5\oplus \mathbb{Z}_5$ this can still fail!
A: This is NOT true: $H=\langle (12)\rangle$ as subgroup of $S_3$. Take the element to be $(2 3)$. This has order $2$ and $|H|=2$, but $(2 3) \notin H$.
Remark Inspired by the post: for a finite group $G$ we define  a non-trivial subgroup $H$, an attractor of $G$, having the property that for any $g \in G$ with o$(g) \mid |H|$, $g \in H$. It is easy to see that such an $H$ must be characteristic (that is $\alpha[H]=H$ for every $\alpha \in Aut(G)$), and if $K$ is a subgroup with $H \cap K \neq 1$, then $H \cap K$ is an attractor of $K$. It would be interesting for example for $p$-groups to determine the structure of these attractors. Also, if $G$ has a normal Sylow-subgroup, then this is an attractor.
