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?

  • 1
    $\begingroup$ Well, think about this. If the subgroup, under conjugation by an element, maps to another subgroup, there must be an element that is mapped to that isn't in the first subgroup. But, by group properties, that element must have order dividing m. But that is a contradiction for obvious reasons $\endgroup$ – Don Thousand Jun 2 at 21:20
  • $\begingroup$ Hmm, why must it have order dividing $m$? This is similar to my other question here, but looks like $n^m$ is not necessarily $e$. $\endgroup$ – 0xd34df00d Jun 2 at 21:55
  • $\begingroup$ Actually, yes, as a counter-example (or I'm terribly wrong and know nothing): take $G = S_3$, $m = 2$, then $H = G$ and $H$ is thus clearly normal, but taking $n$ to be the composition of two "swaps" (which is a rotation) and $g = e$ shows that the order $3$ of the mapped element surely does not divide $m$. $\endgroup$ – 0xd34df00d Jun 2 at 22:10
  • $\begingroup$ Your approach is quite reasonable, but I don't see the need for the "dummy multipliers". $\endgroup$ – Robert Lewis Jun 9 at 23:08


$n = \displaystyle \prod g_i^{k_i} \in H, \; \vert g_i \vert = m, \tag 1$


$g \in G \; \text{arbitrary}, \tag 2$

we have

$gng^{-1} = g \left ( \displaystyle \prod g_i^{k_i} \right ) g^{-1} = \prod (gg_i^{k_i}g^{-1}) = \prod (gg_ig^{-1})^{k_i}; \tag 3$


$\vert gg_ig^{-1} \vert = \vert g_i \vert = m; \tag 4$

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

$g^{-1}hg \in H, \; \forall g \in G; \tag 8$


$h = gg^{-1}hgg^{-1} \in gHg^{-1}, \tag 9$

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


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


Finish the argument now.

  • $\begingroup$ You are right, I misread. I shall edit at once. $\endgroup$ – DonAntonio Jun 2 at 21:27
  • $\begingroup$ This is probably a stupid question now, but if $h \in H$, why does necessarily $h^n = e$? That's obviously the case for abelian groups (but every subgroup there is normal anyway), but it's not immediately obvious to me for arbitrary non-commutative groups (and this is were I also got stuck when I tried considering $(gng^{-1})^n$ in the original notation to see what happens). $\endgroup$ – 0xd34df00d Jun 2 at 21:34
  • $\begingroup$ @0xd34df00d It doesn't have to have order $\;n\;$ . What I'm trying to show is that the group $\;H\;$ is generated by a normal set of elements, meaning: conjugating them we stay inside the set. $\endgroup$ – DonAntonio Jun 2 at 21:40
  • $\begingroup$ After doing something similar I guess I see what you're saying, thanks! $\endgroup$ – 0xd34df00d Jun 2 at 22:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.