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Math Help - Normal subgroups and cosets

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    Normal subgroups and cosets

    Let G be a finite group and N a normal subgroup. Suppose that x is an element of G of prime order p with x not in N. Show that <x> \cap  N = 1 and that the cosets N, Nx,..., N x^{p-1} are distinct. Deduce that p divides the index [G:N].

    Now I have the ideas for the first two parts although I'm having trouble writing this concisely and properly. Also, I'm not sure where the last deduction comes into this - what am I missing?

    Finally (sketch thoughts for this bit will be fine!) - if simply x is not in N and m is the smallest integer for which x^m is in N, generalise the above result. What does this say about the quotient group G/N?
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    Quote Originally Posted by Amanda1990 View Post
    Let G be a finite group and N a normal subgroup. Suppose that x is an element of G of prime order p with x not in N. Show that <x> \cap  N = 1 and that the cosets N, Nx,..., N x^{p-1} are distinct. Deduce that p divides the index [G:N].
    Construct G/N, and consider xN. Since x\not \in N it means |xN|=k > 1. However, (xN)^p = x^pN = N, therefore k|p\implies k=p. Thus, the order of xN is p. This immediately implies that N,xN,...,x^{p-1}N are distinct cosets. If \left< x\right> \cap N \not = \{ 1\} it would mean that x^j \in N for some 1<j<p and so (xN)^j = N but that is a problem because we established that the order of xN must be p.
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    OK so this idea is very neat and elegant. In order to deduce that p divides the index [G:N] I'm guessing that we show p does not divide |N| (since we know p divides |G| and then we'll be done). I can't see how we can use what we already have to see this though.

    If x is not in N and m is the smallest integer for which x^m is in N, then it looks as if <x> \cap N = <x^m> and that  N, Nx...Nx^{m-1} are distinct cosets. Is this correct? And what is this supposed to say about G/N?? Thanks!
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    Quote Originally Posted by Amanda1990 View Post
    OK so this idea is very neat and elegant. In order to deduce that p divides the index [G:N] I'm guessing that we show p does not divide |N| (since we know p divides |G| and then we'll be done). I can't see how we can use what we already have to see this though.

    If x is not in N and m is the smallest integer for which x^m is in N, then it looks as if <x> \cap N = <x^m> and that  N, Nx...Nx^{m-1} are distinct cosets. Is this correct? And what is this supposed to say about G/N?? Thanks!
    Notice that \{N,xN,x^2N,...,x^{p-1}N\} = \left< xN\right> is a subgroup of G/N.
    Therefore, by Lagrange, |\left< xN\right>| divides [G:N].
    However, p = |\left< xN\right>|.
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