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Math Help - Abstract Algebra help

  1. #1
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    Abstract Algebra help

    Hello,
    I need help with the following problem:

    Let G be a finite group, H and N are subgroups of G such that N is normal, and |N| and [G:N] are relatively prime. Show that if H is contained in N, then |H| divides |N|.

    I would really help anyone's help on this. Thank you in advance
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by anlys View Post
    Hello,
    I need help with the following problem:

    Let G be a finite group, H and N are subgroups of G such that N is normal, and |N| and [G:N] are relatively prime. Show that if H is contained in N, then |H| divides |N|.

    I would really help anyone's help on this. Thank you in advance
    This seems like an easy application of Lagrange's Theorem to me; the order of a subgroup of will always divide the order the the group. Here, if you forget the blurb about N normal and stuff and just look at the last sentence, then N is the group and H is the subgroup, so...
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  3. #3
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    Quote Originally Posted by Swlabr View Post
    This seems like an easy application of Lagrange's Theorem to me; the order of a subgroup of will always divide the order the the group. Here, if you forget the blurb about N normal and stuff and just look at the last sentence, then N is the group and H is the subgroup, so...
    Hello Swlabr,
    Thank you for your response. In regards to your comment, I was thinking about using the Lagrange's theorem too. So if we assume that H is contained in N, we would have to show that H is a subgroup of N. If this is true, then we can conclude that |H| divides |N|, right? But, I already tried to prove that H is a subgroup of N, and I am kinda stuck now.
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by anlys View Post
    Hello Swlabr,
    Thank you for your response. In regards to your comment, I was thinking about using the Lagrange's theorem too. So if we assume that H is contained in N, we would have to show that H is a subgroup of N. If this is true, then we can conclude that |H| divides |N|, right? But, I already tried to prove that H is a subgroup of N, and I am kinda stuck now.
    H is a subgroup of N because it is also a subgroup of G and because it is a subset of N. You need to use both of these facts.

    What have you done to prove it so far?
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  5. #5
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    Quote Originally Posted by Swlabr View Post
    H is a subgroup of N because it is also a subgroup of G and because it is a subset of N. You need to use both of these facts.

    What have you done to prove it so far?
    Oh, yeah, you're right. It is quite straightforward! I didn't use both of the facts before. Thank you so much for your help!
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