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.

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  • 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


    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.

    Thank you in advance.


    • 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.

  • 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

        :-), Thanks for the reply.

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