Proving subgroup generated by elements of a given order is normal Aluffi II.7.7 suggests proving the following: let $G$ be a group, $m$ a positive integer, and let $H \subseteq G$ be the subgroup generated by all elements of order $m$ in $G$. Prove that $H$ is normal.
In other words, given arbitrary $n = \prod g_i^{k_i}, |g_i| = m$ we need to show that $\forall g \in G : g n g^{-1}$ can also be represented as a product of some elements of order $m$. And that's about it: I'm stuck after that.
I've noted that a function $n \mapsto gng^{-1}$ is an automorphism, so $|n| = |gng^{-1}|$. I'm not sure how to conclude based on that that $gng^{-1}$ belongs to $H$, though.
So what would be a reasonable way to prove the claim?

Ok, I guess I solved it. Given the representation of $n$ as above, it's sufficient to note that if $|g_i| = n$, then $g g_i g^{-1}$ also has order $n$, so $g^{-1}ng$ also has a representation by elements in $H$ (after adding sufficiently many dummy multipliers of the form $g^{-1}g$ in the representation above). Does this sound reasonable?
 A: With
$n = \displaystyle \prod g_i^{k_i} \in H, \; \vert g_i \vert = m, \tag 1$
and
$g \in G \; \text{arbitrary}, \tag 2$
we have
$gng^{-1} = g \left ( \displaystyle \prod g_i^{k_i} \right ) g^{-1} = \prod (gg_i^{k_i}g^{-1}) = \prod (gg_ig^{-1})^{k_i}; \tag 3$
but
$\vert gg_ig^{-1} \vert = \vert g_i \vert = m; \tag 4$
thus $gng^{-1}$ is a product of elements of order $m$, and as such,
$gng^{-1} \in H; \tag 5$
this shows
$gHg^{-1} \subset H; \tag 6$
now for
$h \in H \tag 7$
by (6) we have
$g^{-1}hg \in H, \; \forall g \in G; \tag 8$
then
$h = gg^{-1}hgg^{-1} \in gHg^{-1}, \tag 9$
which shows that in fact equality binds in (6); thus,
$gHg^{-1} = H, \tag{10}$
that is, $H$ is normal in $G$.  $OE\Delta$.
Proving subgroup generated by elements of a given order is normal
A: I really didn't understand that representation of $\;n\;$ as product and the $\;g\,'$s, but it is way easier: you want
$$H_n:=\left\langle\;g\in G\;|\;ord(g)=n\;\right\rangle$$
Now you can either use that the order of any element in a group equals the order of its image under any automorphism of that group and conjugation is an automorphism, or directly: if $\;h\in G\,,\,\,x\in G\;$ , then
$$\left(x^{-1}hx\right)^n=x^{-1}h^nx=x^{-1}x=1$$
Finish the argument now.
