Thread: 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!!!

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 and then is finite, and furthermore, .

Let be a subgroup of order . Since is normal subgroup of we can form a factor group . The order of this factor group is . Let . And consider the coset . Lagrange's theorem tells us that divides . But since and it means (again by Lagrange's theorem). Hence, . Thus, divides (by the property of orders). But since it means is forced to be . Thus, and so . This means . And this completes the proof (that i.e. it is unique) because if it were the case that then it would mean which is impossible.

excellent! wonderful proof

Thank you.

Tomer.

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).