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

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

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


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

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