## In a p-group, every proper subgroup of minimal index is normal

Let $p$ be a prime and $k$ a positive integer. Prove that if $G$ is a group of order $p^k$ then every subgroup of index $p$ is normal in $G$. Deduce that every group of order $p^2$ has a normal subgroup of order $p$.

Let $G$ be a group of order $p^k$ and $H \leq G$ a subgroup with $[G : H] = p$. Now $G$ acts on the conjugates $gHg^{-1}$ by conjugation, since $g_1g_2 \cdot H = (g_1g_2)H(g_1g_2)^{-1} = g_1(g_2Hg_2^{-1})g_1^{-1} = g_1 \cdot (g_2 \cdot H)$ and $1 \cdot H = 1H1 = H$. Moreover, under this action we have $H \leq \mathsf{stab}(H)$. By a previous theorem, we have $[G : \mathsf{stab}(H)][\mathsf{stab}(H):H] = [G:H] = p$, a prime.

If $[G : \mathsf{stab}(H)] = p$, then $[\mathsf{stab}(H):H] = 1$ and we have $H = \mathsf{stab}(H)$; moreover, $H$ has exactly $p$ conjugates in $G$. Let $\varphi : G \rightarrow S_p$ be the permutation representation induced by the action of $G$ on the conjugates of $H$, and let $K$ be the kernel of this representation. Now $K \leq \mathsf{stab}(H) = H$. By the first isomorphism theorem, the induced map $\overline{\varphi} : G/K \rightarrow S_p$ is injective, so that $|G/K|$ divides $p!$. Note, however, that $|G/K|$ is a power of $p$ and that the only powers of $p$ that divide $p!$ are $1$ and $p$. So $[G:K]$ is $1$ or $p$. If $[G:K] = 1$, then $G = K$ so that $gHg^{-1} = H$ for all $g \in G$; then $\mathsf{stab}(H) = G$ and we have $[G:\mathsf{stab}(H)] = 1$, a contradiction. Now suppose $[G:K] = p$. Again by a previous theorem we have $[G:K] = [G:H][H:K]$, so that $[H:K] = 1$, hence $H = K$. Again, this implies that $H$ is normal so that $gHg^{-1} = H$ for all $g \in G$, and we have $[G:\mathsf{stab}(H)] = 1$, a contradiction. Thus $[G:\mathsf{stab}(H)] \neq p$.

If $[G : \mathsf{stab}(H)] = 1$, then $G = \mathsf{stab}(H)$. That is, $gHg^{-1} = H$ for all $g \in G$; thus $H \leq G$ is normal.

Now let $G$ be a group of order $p^2$. By Cauchy’s Theorem, $G$ contains an element $x$ of order $p$. Then $[G:\langle x \rangle] = p$, hence $\langle x \rangle$ is normal.

• saigonbettas  On December 7, 2010 at 8:28 pm

Do you mean “stab(H)” is stabilizer of H? And how we can say if [G:stab(H)]=p, then [stab(H):H]=1?

• nbloomf  On December 7, 2010 at 10:34 pm

1. Yes.

2. Since $[G:H] = p$ is prime and because $H \leq \mathsf{stab}(H)$ we have $[G : \mathsf{stab}(H)][\mathsf{stab}(H):H] = [G:H] = p$.

• Hanh  On January 8, 2011 at 9:17 am

Hey,

Your blog is really interesting because it contains so many examples and results about groups and rings which I am studying.

I also have a following problem, but I haven’t solved it yet. I hope you can give me an idea for that.

Problem: Let G be a group (maybe not finite) and H is a subgroup of G such that $x^2\in H$ for all $x\in G.$ Prove that $H$ is normal in G.

Hanh.

• nbloomf  On January 9, 2011 at 12:36 am

Try this:

Let $xh \in xH$. Note that $xhx^{-2} \in xH$, since $x^{-2} \in H$. Now $(xhx^{-2})^2 = xhx^{-2}xhx^{-2} \in H$, so that $xhx^{-1} \in H$. Thus $xh \in Hx$, and we have $xH \subseteq Hx$. Similarly, $Hx \subseteq xH$. Thus $xH = Hx$ for all $x \in G$, and $H \leq G$ is normal.

• Hanh  On January 9, 2011 at 9:22 am

Wow! Great!
Thank you very much for the solution. It has taken me for 3 months to struggle the problem, but I did not know how to do that.
Best,
Hanh.

• TG  On November 23, 2011 at 8:19 am

We can use corollary 5 in sec 4.2 straight away right?

• nbloomf  On November 23, 2011 at 10:28 pm

Indeed you can.

• TG  On November 24, 2011 at 1:12 am