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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.

Noah Snyder
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Jarbas Dantas Silva
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4 Answers4

7

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.

DonAntonio
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  • ok...develop the 4) – Jarbas Dantas Silva Oct 16 '12 at 15:52
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    @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 $G$. – Mikasa Oct 16 '12 at 16:12
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    @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-Álvarez 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? –  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
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    (1)(2)(3) are actually Strong Cayley theorem. See a complete one there. –  Mar 01 '18 at 07:45
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I think another approach in light of Don's answer can be:

Lemma: Suppose $G$ 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 $N_G(H)$ be $G$ itself and this means that $H\vartriangleleft G$.

Mikasa
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  • I'm missing something, but isn't $H < N_G(H)$ always? – Jason DeVito - on hiatus Oct 16 '12 at 16:18
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    @JasonDeVito: dear prof. I wanted to say $H\neq N_G(H)$. If my way is not right please tell me. Thanks. – Mikasa 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 - on hiatus Oct 16 '12 at 16:26
  • Babak's solution applies always for finite nilpotent groups, just as mine is just a rephrasing of "A finite group is nilpotent iff any maximal (proper, of course) subgroup is normal and of index a prime" – DonAntonio Oct 16 '12 at 16:31
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    @Jason: join me in my campaign to squash the convention that $\subset$ allows equality! :-) – Mariano Suárez-Álvarez Oct 29 '12 at 19:32
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    @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 - on hiatus Oct 29 '12 at 23:08
  • Hello, dear friend! $\quad\ddot\smile\quad$ – amWhy Mar 21 '13 at 16:04
  • @amWhy: When I came here, you are in the Heaven I think. :-) – Mikasa Mar 21 '13 at 19:20
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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.

Community
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Mikko Korhonen
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0

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 $G = MM = M$.

Mikko Korhonen
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