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Math Help - Subgroup divides a Normal group

  1. #1
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    Subgroup divides a Normal group

    If G is a finite group, let H be a subgroup of G, and N a normal subgroup of G, and |N| and |(G:H)| are relatively prime, then show that H is a subset of N if and only if |H| divides |N|.

    I'm pretty lost in this problem, would I have to use the property of module here?

    Thank you.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    If G is a finite group, let H be a subgroup of G, and N a normal subgroup of G, and |N| and |(G:H)| are relatively prime, then show that H is a subset of N if and only if |H| divides |N|.

    I'm pretty lost in this problem, would I have to use the property of module here?

    Thank you.
    This is kinda similar to this problem.

    Since N is normal subgroup form factor group G/N.
    Let a\in H and consider aN \in G/N.
    Let k be the order of aN in G/N.
    By Lagrange's theorem we know k | (G:N).
    Also (aN)^{|H|} = (a^{|H|}N) = N thus k||H|.

    Now if |H| divides |N| then (G:N) divides (G:H).
    Thus, k|(G:N) \implies k|(G:H)
    Thus, k|H| \implies k||N|.
    But (G:H) and |N| are relatively prime.
    This forces k=1.

    Finally since the order of aN in G/N is one it must mean a\in N.
    Therefore, H\subseteq N.
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  3. #3
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    The things I don't understand here are that, if  |H| | |N| , how do I know that  | (G:H) | | | (G:H)| ?


    Also, which theorem states that  |G| / |N| = [G:N] ? Does it only works when N is a normal subgroup?
    Thanks!
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  4. #4
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    Also, which theorem states that  |G| / |N| = [G:N] ? Does it only works when N is a normal subgroup?
    Thanks!
    There are (G:N) cosets and each one has |N| elements. Furthermore, each is disjoint and together (in union) give G. Thus, (G:N)|N| = |G|.

    Quote Originally Posted by tttcomrader View Post
    The things I don't understand here are that, if  |H| | |N| , how do I know that  | (G:H) | | | (G:H)| ?
    Since |H| | |N| it means \frac{|N|}{|H|} is an integer.
    This means \frac{(G:H)}{(G:H)} = \frac{|G|/|H|}{|G|/|N|} = \frac{|N|}{|H|} is an integer.
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