Prove that if $G$ is a group and $H$ is a subgroup of $G$ generated by all elements of order $N$ in $G$, then $H$ is a normal subgroup of $G$. I've tried proving that $ghg^{-1}\in H$ ($\forall g \in G$), but I don't see how the special property of $H$ guarantees this.
Any insight? I've turned away from it to work on other things, and it's consistently stumped me.

Let $A=\{z\in G:o(z)=N\}$. Then $H=\langle A\rangle$ and  thus if $h\in H$ then 
$$
h=z_1^{e_1}\dots z_m^{e_m}\tag{*}
$$ where $e_{i}=+1$ or $-1$, $z_i\in A$ and $m$ an arbitrary natural number. So take $h\in H$, $g\in G$ and show that $ghg^{-1}$ will be of the form (*).
