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Normal subgroup

  1. Mar 29, 2009 #1
    1. The problem statement, all variables and given/known data
    If H is a normal subgroup of G and |H| = 2, show that H is a subgroup of Z(G).
    Is this true when |H| = 3?

    3. The attempt at a solution
    Since H is a normal subgroup of G and |H| = 2 = {1,a},
    a in Z(G), also aa = a2 = 1 in Z(G)
    aa-1 = a in Z(G). Therefore H is a subgroup of Z(G).
    I am not sure my approach is correct.
    and I have no idea next question when |H| =3, is it false? why??
  2. jcsd
  3. Mar 29, 2009 #2

    matt grime

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    In your attempt at a solution what is it that you're trying to show? What do the steps you've written mean in relation to what you're trying to show?

    In other words, what does it mean for H to be in Z(G)? Because at no point have you addressed what that is. In fact the first thing you assert is that a is in Z(G), but that is part of the result that you're trying to show.

    When you correctly prove this part, it will lead you to why it is false for |H|=3.
  4. Mar 29, 2009 #3
    To show that H is a subgroup of Z(G),
    1Z(G) in H
    whenever a,b in H then ab in H
    whenever a in H then a-1 in H.

    Since both H and Z(G) are normal in G, 1Z(G) is clearly in H.
    Since |H| = 2, H={1,a}. So 1,a in H then a1=1a in H
    also since aa = a2 = 1 = aa-1, so a = a-1 self-inverse.

    When |H|=3, we can't determine whether a2 in Z(G) or not.

    how about this approach??
  5. Mar 29, 2009 #4

    matt grime

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    I'll ask again: what is the definition of Z(G)? You have to show that H is a subgroup of Z(G), so at some point using the definition of Z(G) will be necessary. You have assumed that H is a subset of Z(G) without any justification for this.

    What you have shown is that if H<G, and H is a subset of K for K<G, then H<K. Well, that is trivially true and immaterial.
  6. Mar 29, 2009 #5


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    Here is the crucial point: you start by asserting that a is in Z(G). How do you know that is true? As matt grime said, you have not used the definition of Z(G). You have also not used the fact that H is a normal subgroup of G.

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