If a group G has order p^n, show any subgroup of order p^n-1 is normal

[demonelite123]

If a group G has order $p^n$, show that any subgroup of order $p^{n-1}$ is normal in G.

i have no idea how to start this. i know that to show that a subgroup N is normal in G, i need that $gNg^{-1} = N$. so i start with any subgroup N of order $p^{n-1}$ but i have no idea how to continue.

this problem appears in the section before the sylow theorems are introduced so i can't use them. i know that for p-groups, the center is nontrivial and has prime power order. also in this same section, Cauchy's theorem was introduced which says if p divides the order of a group then that group has an element of order p. these concepts were introduced fairly recently to me so this may be why i am having trouble.

can someone give me a hint or 2 to continue? thanks

 

[micromass]

Let H be your subgroup of order p^n-1. Consider the action of G on the set $X=\{Hg~\vert~g\in G\}$ by multiplication on the right. This determines a group morphism

$$G\rightarrow Sym(X)$$

with kernel N. The idea is to prove that N=H. Prove this in the following steps:

1) $N\subseteq H$
2) [G:N]=p[H:N]
3) G/N is a subgroup of Sym(X), so [G:N] divides p!
4) [H:N]=1.

 

[Deveno]

don't you actually want a left-action, so that you have a homomorphism, rather than an anti-homomorphism? i mean, usually mappings are written on the left, so we have x(gH), rather than (Hg)x (or Hg.x, if you prefer). it's a small (minor) issue, but usually actions are taken to be left-actions, because of the way we normally compose functions.

 

[micromass]

don't you actually want a left-action, so that you have a homomorphism, rather than an anti-homomorphism? i mean, usually mappings are written on the left, so we have x(gH), rather than (Hg)x (or Hg.x, if you prefer). it's a small (minor) issue, but usually actions are taken to be left-actions, because of the way we normally compose functions.

It would indeed make more sense. But I'm so used to right actions and right cosets that I can't help it