MHB Normal subgroups (Sylow Theorem)

[mathgirl1]

Let G be a group of order pm where p is a prime and p > m. Suppose H is a subgroup of order p. Show that H is normal in G.

There is a very similar problem

Let |G| = p^nm where p is a prime and n \ge 1, p > m. Show that the Sylow p-subgroup of G is normal in G.

Proof:
Let n_p be the number of Sylow p-subgroups of G. By the 3-d Sylow Subgroup Theorem we know that n_p \mid m and that n_p \equiv 1 (mod p). Since m < p and n_p \mid m, it follows that 1 \le n_p \le m &lt; p. Since we also know that n_p \equiv 1 (mod p), it follows that n_p=1. Let P be a Sylow p-subgroup of G. Since for every g \in G, g^{-1}Pg is also a Sylow p-subgroup of G and since n_p=1, it follows that for every g \in G that g^{-1}Pg=P. Hence P is normal in G, as claimed.

So this is the same problem that I need to solve with n=1. Is there a way to solve this without Sylow Theorems or Sylow subgroups? We barely covered that so not sure we can use it to prove this problem.

Also, why is g^{-1}Pg also a Sylow p-subgroup of G since P is?

Any help is appreciated. I would like to solve this problem without Sylow theorems but not sure where to start cause I am already stuck in this thinking.

Thanks!

 

[ThePerfectHacker]

You have shown above that there is exactly one subgroup of order $p$. Let us call it $P$. Furthermore, this is a Sylow $p$-subgroup because $p$ is the largest power of $p$ which divides $|G|=pm$. Thus, by the third theorem all Sylow $p$-subgroups are conjugate. Consider $gPg^{-1}$. Show that this subgroup has order $p$, and so, it is a Sylow $p$-subgroup. But since there is only one such subgroup it follows that $gPg^{-1} = P$. Since this is true for all $p$ we conclude that $P$ is invariant under conjugation, so it is a normal subgroup.

 

[johng1]

Hi,
You asked for a different proof that doesn't use the Sylow theorems. Here is one. The idea of representing (finding a homomorphism) a group G on the cosets of a subgroup is worth remembering.