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Math Help - normal groups...

  1. #1
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    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!!!
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  2. #2
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    Quote Originally Posted by aurora View Post
    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.
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  3. #3
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    excellent! wonderful proof

    Thank you.

    Tomer.
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  4. #4
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    Quote Originally Posted by aurora View Post
    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).
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