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Math Help - Groups G and G', M is a normal subgroup of G'

  1. #1
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    Groups G and G', M is a normal subgroup of G'

    We are given:
    1. h: G -> G' is a group homomorphism.
    2. M is a normal subgroup of G'.
    3. N is the set of x in G such that h(x) is in M.

    Prove that N is a normal subgroup of G.

    It's not that hard for me to prove that N is a subgroup of G--that just comes from the fact that h is a group homomorphism. But I really have no idea how to prove that N is not just a subgroup but a normal subgroup. In fact, these normal subgroup proofs are just plain difficult for me right now. Please help.
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    Re: Groups G and G', M is a normal subgroup of G'

    You need to show that yxy^{-1}\in N for all y\in G and x\in N. What happens when you apply the definition of N to yxy^{-1}?
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    Re: Groups G and G', M is a normal subgroup of G'

    Quote Originally Posted by emakarov View Post
    You need to show that yxy^{-1}\in N for all y\in G and x\in N. What happens when you apply the definition of N to yxy^{-1}?
    Do I just take h(yxy^-1), which clearly is an element in M, use the homomorphism property to split it up into h(y)h(x)h(y^-1), and I'm pretty much done?

    EDIT: Wait...do I need to use the fact that x being in N is equivalent to h(x) being in M?
    Last edited by phys251; July 14th 2013 at 04:38 PM.
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    Re: Groups G and G', M is a normal subgroup of G'

    Pretty much. You obviously also need to use the assumption that M is normal, which implies that h(yxy^(-1)) = h(y)h(x)h(y)^(-1) ∈ M since h(x) ∈ M and therefore yxy^(-1) ∈ N.
    Thanks from phys251
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    Re: Groups G and G', M is a normal subgroup of G'

    Quote Originally Posted by emakarov View Post
    Pretty much. You obviously also need to use the assumption that M is normal, which implies that h(yxy^(-1)) = h(y)h(x)h(y)^(-1) ∈ M since h(x) ∈ M and therefore yxy^(-1) ∈ N.
    So h(y)h(x)h(y)^(-1) ∈ M immediately leads to yxy^-1 ∈ N, right? If so, I'm done.
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    Re: Groups G and G', M is a normal subgroup of G'

    Quote Originally Posted by phys251 View Post
    EDIT: Wait...do I need to use the fact that x being in N is equivalent to h(x) being in M?
    Yes, you need to use both directions of z ∈ N ⇔ h(z) ∈ M (which holds by definition of N): left-to-right for z = x and right-to-left for z = yxy^(-1).

    Quote Originally Posted by phys251 View Post
    So h(y)h(x)h(y)^(-1) ∈ M immediately leads to yxy^-1 ∈ N, right?
    Yes.
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    Re: Groups G and G', M is a normal subgroup of G'

    Awesome. Thanks!
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