1. ## normal groups...

Hello... I'm having difficulties solving the next problem:
Let G be a group, and N<G is a normal sub-group of G.
I'm given that |N|=n, and [G:N]=m.
a) prove that to every g in G, g^m is an elemnt in N.
I've proved that.
b) Now say that gcd(m,n) = 1. Prove that N is the only sub-group of G with order n.

So, I'm having trouble with "b"... I don't seem to get anywhere.

And another question: I'm not given that G is finite, but is it not a consequence of "n" and "m" being finite? G is spanned by a finite number of cosets with a finite number of elements - I guess that means G cannot be infinite....

Thanks!!!

2. Originally Posted by aurora
Let G be a group, and N<G is a normal sub-group of G.
I'm given that |N|=n, and [G:N]=m.

b) Now say that gcd(m,n) = 1. Prove that N is the only sub-group of G with order n.
You are correct. If $|N|=n$ and $[G:N]=m$ then $|G|$ is finite, and furthermore, $|G|/|N| = [G:N]=m\implies |G| = nm$.

Let $H$ be a subgroup of order $n$. Since $N$ is normal subgroup of $G$ we can form a factor group $G/N$. The order of this factor group is $nm/n=m$. Let $a\in H$. And consider the coset $aN\in G/N$. Lagrange's theorem tells us that $\text{ord}(aN)$ divides $G/N=m$. But since $a\in H$ and $|H|=n$ it means $a^{n}=e$ (again by Lagrange's theorem). Hence, $(aN)^n = a^nN=N$. Thus, $\text{ord}(aN)$ divides $n$ (by the property of orders). But since $\gcd(n,m)=1$ it means $\text{ord}(aN)$ is forced to be $1$. Thus, $aN=N$ and so $a\in N$. This means $H\subseteq N$. And this completes the proof (that $H=N$ i.e. it is unique) because if it were the case that $H \subset N$ then it would mean $n=|H| < |N| = n$ which is impossible.

3. excellent! wonderful proof

Thank you.

Tomer.

4. Originally Posted by aurora
excellent! wonderful proof

Thank you.

Tomer.
Here is a related question (a converse statement if you wish to think of it that way). Let G be any group (even infinite) and it has a unique subgroup of index n prove that this subgroup is normal. (Nothing complicated if you approach it correctly).