# Thread: Subgroup divides a Normal group

1. ## 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.

2. 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, $H\subseteq N$.

3. 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!

4. 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, $(G:N)|N| = |G|$.

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.