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.
 A: 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 (*).
A: You can say even more: the subgroup generated by the elements of order $N$ is characteristic.
Indeed, if $\varphi$ is an automorphism of $G$ and $x$ is an element of order $N$, then $\langle x\rangle=\{x^n:n\in\mathbb{Z}\}$ has $N$ elements and so also
$$
\varphi(\langle x\rangle)=\langle \varphi(x)\rangle
$$
has $N$ elements. Since $x\mapsto x^{-1}$ also preserves the order of elements, because $\langle x\rangle=\langle x^{-1}\rangle$, your $H$ can be described as
$$
H=\{h_1h_2\dots h_k: o(h_i)=N, k\ge0\}
$$
(where the empty product is $1$). Now,
$$
\varphi(h_1h_2\dots h_k)=\varphi(h_1)\varphi(h_2)\dots\varphi(h_k)
$$
is again a product of elements of order $N$.
A: if $h = g'^n$
$g*h*g^{-1} = g*g'^n*g^{-1} = g*g'*(g^{-1}*g)*g'*(g^{-1}*g)...*g^{-1} = (g*g'*g^{-1})^n $
