Subgroup divides a Normal group

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September 11th 2008, 11:40 AM
tttcomrader

Subgroup divides a Normal group

If G is a finite group, let H be a subgroup of G, and N a normal subgroup of G, and |N| and |(G:H)| are relatively prime, then show that H is a subset of N if and only if |H| divides |N|.

I'm pretty lost in this problem, would I have to use the property of module here?

Thank you.
September 11th 2008, 12:37 PM
ThePerfectHacker

Quote:

Originally Posted by tttcomrader

If G is a finite group, let H be a subgroup of G, and N a normal subgroup of G, and |N| and |(G:H)| are relatively prime, then show that H is a subset of N if and only if |H| divides |N|.

I'm pretty lost in this problem, would I have to use the property of module here?

Thank you.

This is kinda similar to this problem.

Since $N$ is normal subgroup form factor group $G/N$ .
Let $a\in H$ and consider $aN \in G/N$ .
Let $k$ be the order of $aN$ in $G/N$ .
By Lagrange's theorem we know $k | (G:N)$ .
Also $(aN)^{|H|} = (a^{|H|}N) = N$ thus $k||H|$ .

Now if $|H|$ divides $|N|$ then $(G:N)$ divides $(G:H)$ .
Thus, $k|(G:N) \implies k|(G:H)$
Thus, $k|H| \implies k||N|$ .
But $(G:H)$ and $|N|$ are relatively prime.
This forces $k=1$ .

Finally since the order of $aN$ in $G/N$ is one it must mean $a\in N$ .
Therefore, $H\subseteq N$ .
September 13th 2008, 02:29 PM
tttcomrader

The things I don't understand here are that, if $|H| | |N|$ , how do I know that $| (G:H) | | | (G:H)|$ ?

Also, which theorem states that $|G| / |N| = [G:N]$ ? Does it only works when N is a normal subgroup?
Thanks!
September 13th 2008, 05:26 PM
ThePerfectHacker

Quote:

Also, which theorem states that $|G| / |N| = [G:N]$ ? Does it only works when N is a normal subgroup?
Thanks!

There are $(G:N)$ cosets and each one has $|N|$ elements. Furthermore, each is disjoint and together (in union) give $G$ . Thus, $(G:N)|N| = |G|$ .

Quote:

Originally Posted by tttcomrader

The things I don't understand here are that, if $|H| | |N|$ , how do I know that $| (G:H) | | | (G:H)|$ ?

Since $|H| | |N|$ it means $\frac{|N|}{|H|}$ is an integer.
This means $\frac{(G:H)}{(G:H)} = \frac{|G|/|H|}{|G|/|N|} = \frac{|N|}{|H|}$ is an integer.