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Thread: 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 $\displaystyle |N|=n$ and $\displaystyle [G:N]=m$ then $\displaystyle |G|$ is finite, and furthermore, $\displaystyle |G|/|N| = [G:N]=m\implies |G| = nm$.

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