# If G has a normal subgroup of index p, prove that G has at least one element of order p.

I'm totally lost on this one.

If $G$ has a normal subgroup of index $p$, prove that $G$ has at least one element of order $p$.

EDIT:

Could you use Cauchy's Theorem?

Let $H$ be a normal subgroup of $G$, with index $p$ where $G$ is finite and $p$ is prime.

Then $(G:H)$ = $|G|$$/$$|H|$ = $p$, by the definition of index.

So $|G| = |H|*p$

By Cauchy's Theorem, if $G$ is a finite group, and $p$ is a prime divisor of $|G|$, then $G$ has an element of order $p$.

• You need that $G$ is finite. Otherwise, $(\mathbb Z, +)$ has a normal subgroup $p\mathbb Z$ of index $p$, but no element of order $p$. – Mathmo123 Dec 3 '14 at 12:54
• The edit is correct. – Ryan Dec 3 '14 at 13:28

## 4 Answers

The order of $G/N$ is prime, thus it is cyclic.

Not-so-overkill argument (compared to Cauchy's theorem):

Suppose $G$ is finite (since, as pointed out by Mathmo123, for $G$ infinite the statement does not hold). Then $G/H$ is of order $p$, hence cyclic, so you can choose a "generator modulo $H$", i.e. $g \in G$ such thet $\langle gH \rangle=G/H$.

Now, order of $g$ is easily seen to be a multiple of $p$. Hence, by taking suitable power of $g$, one obtains an element of order $p$.

• This was my answer, although without the extra steps. Did I get downvoted for not writing out all steps (which I would gladly give if asked)? – Ryan Dec 3 '14 at 14:18
• @Ryan: I am not the one who downvoted your answer, so I do not know. I did notice that you implied the same thing after I wrote the answer (I deleted it at first but then I decided to post it anyway; however, I can delete it once again if you wish). – Pavel Čoupek Dec 3 '14 at 14:50
• I assumed giving a (vital) hint would be a good answer, but I'm not to familiar with the site. Should I normally give full answers? I think your answer works better as a comment on mine, but I don't mind either way. – Ryan Dec 3 '14 at 14:54
• That I think usually depends on the type of question (sometimes the OP even specifies that he wants hints only). Since the OP had an idea how to solve it in the first place (using unnecessarily strong fact), I saw no harm in giving the full answer. (Probably someone who did not understand the hint you gave was under the impression that you did not read the question carefully, or something like that.} – Pavel Čoupek Dec 3 '14 at 14:58
• Thanks, next time I'll add 'Hint:..' or something. – Ryan Dec 3 '14 at 15:52

Take $x\notin N$. Since $G/N\simeq \mathbb{Z}/p\mathbb{Z}$, $x^p\in N$. Now, let $r$ be the order of $x$.

If $p$ doesn't divide $r$, $[x^r]\neq 0$ in $G/N$ because $G/N\simeq\mathbb{Z}/p\mathbb{Z}$ and $[x]\neq 0$, absurd.

Hence $p$ divides $r$: but then, $x^{r/p}\neq 0$ and has order $p$.

Hint: There are two other more or less standard proofs that I know of.

1. Induction on $|G|$. If $|G|=2$ or $3$, it's clear. Otherwise, first prove for abelian $G$ by choosing a nonidentity element and examining its order. Either $|x|$ or $|G/<x>|$ is divisible by $p$; use induction to find an element of order $p$; in the second case, a little work, but not much, is required to turn that into an element of $G$ rather than $|G/<x>|$. In the nonabelian case, if any element has centralizer a multiple of $p$ you are done by induction; otherwise, the class equation tells you that the center is a multiple of $p$ and again you are done by induction.

2. A slicker proof starts by examining the group action of $\mathbb{Z}/p$ by rotation on the set $X$ of all $p$-tuples $(g_1, g_2, \dotsc, g_p)\in G^p$ whose product is the identity. By the orbit-stabilizer theorem, every orbit has either size 1 or size $p$, so by divisibility arguments the number of size 1 orbits must be a multiple of $p$. Any element with a size $1$ orbit looks like $(g, g, \dotsc, g)$. But the number of size $1$ orbits cannot be zero, since $(e,e,\dotsc, e)$ has a size 1 orbit. So there must be some other element whose orbit has size 1, i.e. some $g\in G$ such that $g^p = e$.

• This gives a proof of Cauchy's theorem, which is a much stronger result than the one in the question. Knowing that $G$ has a normal subgroup of index $p$ is a stronger condition than knowing that $p\mid |G|$, and as such the proof is much simpler here – Mathmo123 Dec 3 '14 at 13:14
• @Mathmo123 True enough. I didn't read the question carefully enough. – rogerl Dec 3 '14 at 13:15