Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $p$ be a prime number and $G$ a finite group where $|G|=p^n$, $n \in \mathbb{Z_+}$. Show that any subgroup of index $p$ in it is normal in $G$. Conclude that any group of order $p^2$ have a normal subgroup of order $p$, but without using the Sylow theorems.

share|improve this question
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative "conclude" to be rude when asking for help; please consider rewriting your post. –  Noah Snyder Oct 16 '12 at 14:44
add comment

4 Answers

I think another approach in light of Don's answer can be:

Lemma: Let $G$ is a $p$-group and $H<G$ then $H\lneqq N_G(H)$.

Here we know that $[G:H]=p$ then $H$ is a proper subgroup of $G$. So the lemma tells us in this group we have $H$ as a proper subgroup of its normalizer in $G$. In fact our conditions make $N_G(H)$ to be $G$ itself and this means that $H\vartriangleleft G$.

share|improve this answer
I'm missing something, but isn't $H < N_G(H)$ always? –  Jason DeVito Oct 16 '12 at 16:18
@JasonDeVito: dear prof. I wanted to say $H\neq N_G(H)$. If my way is not right please tell me. Thanks. –  B. S. Oct 16 '12 at 16:24
I think it's just a matter of notational convention. I was taught things like $H< G$ and $H\subset G$ all allow equality, even though the notation is a bit misleading. I wouldn't be surprised if there are other notational conventions. Sorry for my misunderstanding. Now that I do understand, +1! (And you don't have to call me prof, "Jason" is fine :-) ) –  Jason DeVito Oct 16 '12 at 16:26
@Jason: join me in my campaign to squash the convention that $\subset$ allows equality! :-) –  Mariano Suárez-Alvarez Oct 29 '12 at 19:32
@Mariano: I personally use $\subseteq$ and $\subsetneq$ and never use $\subset$, but I've also seen $\subset$ enough to know that one should be cautious. –  Jason DeVito Oct 29 '12 at 23:08
show 3 more comments

You only need the following

Lemma:: If $\,G\,$ is a finite group and $\,p\,$ is the smallest prime diving $\,|G|\,$ , then any subgroup of index $\,p\,$ is normal in $\,G\,$ .

Proof (highlights): Let $\,N\leq G\,\,,\,\,[G:N]=m\,$ , and define an action of $\,G\,$ on the set of $\,X\,$ of left cosets of $\,N\,$ by $\,g\cdot(xN):=(gx)N\,$ :

1) Check the above indeed is a group action on that set

2) Check that the given action induces a homomorphism $\,\phi:G\to S_X\cong S_m\,$ with kernel

$$\ker\phi=\bigcap_{x\in G}N^x\,,\,\,\,N^x:=x^{-1}Nx $$

(the above kernel is also called the core of $\,N\,$)

3) Check that $\,\ker\phi\,$ is the greatest subgroup of $\,G\,$ normal in $\,G\,$ which is contained in $\,N\,$

4) Now apply the above to the case $\,m=p=\,$ the smallest prime dividing the order of the group.

share|improve this answer
ok...develop the 4) –  Jarbas Dantas Silva Oct 16 '12 at 15:52
@JarbasDantasSilva: Not only $[G:\ker\phi]$ divides $|G|=p^n$, but also divides $p!=|S_p|$. So beacuse the $p$ is the smallest prime number dividing the order of the group,we have $[G:\ker\phi]=p$. $\ker\phi\subset N$ so $\ker\phi=H$. But $\ker\phi$ is a normal subgroup of $G$. –  B. S. Oct 16 '12 at 16:12
@JarbasDantasSilva, it is generally best if you do not give orders to people here. Instead of now develop 4, something like «I tried to do 4 and got stuck precisely here when doing this and that» is much, much bett6er! –  Mariano Suárez-Alvarez Oct 29 '12 at 19:31
@Don: I am completely confused. You havent assumed the nature of m anywhere in the proof, so what does p being the smallest prime have to do with the proposition that N will be normal? –  ramanujan_dirac Jan 21 '13 at 18:29
@ramanujan_dirac, of course I've used, and in a rather very essential way, the number $\,m\,$ in my proof: it appears in both in (2) and (4)...is this what you were refering to? Of course, a subgroup $\,N\,$ is normal iff $\,N=N^x\,$ for all $\,x\in G\,$ ...so I do see a pretty direct relation. –  DonAntonio Jan 21 '13 at 18:35
add comment

One more solution. This one I saw in an old paper (1895) by Frobenius (from here).

We proceed by induction. The case $n = 1$ is clear. Let $H$ be a subgroup of index $p$, ie. $H$ has order $p^{n-1}$. Since $p$-groups have nontrivial center, there exists $x \in Z(G)$ of order $p$. If $x \in H$, then $H/\langle x\rangle \trianglelefteq G/\langle x\rangle$ by induction and thus $H \trianglelefteq G$. If $x \not\in H$, then $G = H\langle x \rangle$ and $H \trianglelefteq G$ since $x$ is central.

share|improve this answer
add comment

This is a bit ad-hoc, but I thought up one more elementary solution.

Let $H$ be a subgroup of index $p$. Suppose that $H$ is not normal. Then there exists $g \in G$ such that $g^{-1}Hg \neq H$. Thus $G$ is equal to the product $H(g^{-1}Hg)$, but this is in contradiction with the following fact.

If $M$ is a subgroup of $G$ and $g \in G$ such that $G = Mg^{-1}Mg$, then $M = G$.

Proof: Since $g \in Mg^{-1}Mg$, we get $g \in M$ and thus $G = MM = M$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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