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Math Help - Commutator, Abelian

  1. #1
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    Commutator, Abelian

    A commutator in a group is any element [x, y]=xyx^{-1}y^{-1} for x, y \in G. If K is subgroup of G, prove K contains all the commutators of G iff K \triangleleft G and G/K is abelian.
    Attempt
    ( \Rightarrow) K \triangleleft G is trivial. To show G/K is abelian, take g_1 +K and g_2+K and show they commute, how do I do this?
    ( \Leftarrow) This is the harder direction. I don't see how to prove this way.

    Thanks for any help.
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  2. #2
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    Quote Originally Posted by canberra1454 View Post
    A commutator in a group is any element [x, y]=xyx^{-1}y^{-1} for x, y \in G. If K is subgroup of G, prove K contains all the commutators of G iff K \triangleleft G and G/K is abelian.
    Attempt
    ( \Rightarrow) K \triangleleft G is trivial. To show G/K is abelian, take g_1 +K and g_2+K and show they commute, how do I do this?
    ( \Leftarrow) This is the harder direction. I don't see how to prove this way.

    Thanks for any help.
    ( \Rightarrow) If K contains all the commutators of G, then K \triangleleft G and G/K is abelian.

    Assume K contains all the commutators of G. Then xyx^{-1}y^{-1} \in K for all x,y \in G. It follows that (xy)(yx)^{-1} \in K. Thus, (xy)K = (yx)K for all x,y \in G. Therefore, G/K is abelian.
    We remain to show that K is normal. To show that K is normal, we need to show that for all g \in G,  gKg^{-1} = K. Let x^{-1}y^{-1}xy \in K for x,y \in G. Then, we need to check g(x^{-1}y^{-1}xy)g^{-1} is in K. g(x^{-1}y^{-1}xy)g^{-1} = gx^{-1}y^{-1}x(g^{-1}yy^{-1}g)yg^{-1}=((gx^{-1})y^{-1}(gx^{-1})^{-1}y)(y^{-1}gyg^{-1}), which is in K. Thus, K \triangleleft G.

    ( \Leftarrow) If K \triangleleft G and G/K is abelian, then K contains all the commutators of G.

    Assume K \triangleleft G and G/K is abelian. Then, (xy)K = (yx)K for all x,y \in G. It follows that (xy)(yx)^{-1} = xyx^{-1}y^{-1} \in K. Thus, K contains all the commutator of G.
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  3. #3
    Super Member Gamma's Avatar
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    Quote Originally Posted by aliceinwonderland View Post
    We remain to show that K is normal. To show that K is normal, we need to show that for all g \in G,  gKg^{-1} = K. Let x^{-1}y^{-1}xy \in K for x,y \in G. Then, we need to check g(x^{-1}y^{-1}xy)g^{-1} is in K. g(x^{-1}y^{-1}xy)g^{-1} = gx^{-1}y^{-1}x(g^{-1}yy^{-1}g)yg^{-1}=((gx^{-1})y^{-1}(gx^{-1})^{-1}y)(y^{-1}gyg^{-1}), which is in K. Thus, K \triangleleft G.
    I was working on this problem and noticed a slight error, but an easy fix. You assumed that the element you were testing the conjugation by g with was in the commutator subgroup (G'), but we only know that K contains the commutator subgroup. So the correct proof should go something like this.

    Let k\in K and for any g\in G we consider gkg^{-1} and show it is in K.

    gkg^{-1}k^{-1}\in G' \subset K

    As K is a subgroup, it is closed under multiplication, so
    gkg^{-1}k^{-1}k = gk^{-1}g^{-1} \in K.

    g and k were arbitraty, so gKg^{-1} \subset K \Rightarrow K \triangleleft G as desired.
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  4. #4
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    Quote Originally Posted by Gamma View Post
    I was working on this problem and noticed a slight error, but an easy fix. You assumed that the element you were testing the conjugation by g with was in the commutator subgroup (G'), but we only know that K contains the commutator subgroup. So the correct proof should go something like this.

    Let k\in K and for any g\in G we consider gkg^{-1} and show it is in K.

    gkg^{-1}k^{-1}\in G' \subset K

    As K is a subgroup, it is closed under multiplication, so
    gkg^{-1}k^{-1}k = gk^{-1}g^{-1} \in K.

    g and k were arbitraty, so gKg^{-1} \subset K \Rightarrow K \triangleleft G as desired.
    hmm, since K contains all the commutators of G, isn't the below one in my previous post an enough argument to show that K is normal in G?

    "To show that K is normal, we need to show that for all g \in G,  gKg^{-1} = K. Let x^{-1}y^{-1}xy \in K for x,y \in G. Then, we need to check g(x^{-1}y^{-1}xy)g^{-1} is in K."

    I think the only step I missed is to show that K is indeed a subgroup of G, but it is trivial.
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  5. #5
    Super Member Gamma's Avatar
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    What if K-G'\not = \emptyset? Not every element in K needs to be commutator is what my concern is.
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  6. #6
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    Quote Originally Posted by Gamma View Post
    What if K-G'\not = \emptyset? Not every element in K needs to be commutator is what my concern is.
    aha, I read the question again and I found that K is a subgroup of G by hypothesis. So, K can be a subgroup of G that contains G'.

    Thanks for pointing out my mistake.
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  7. #7
    Super Member Gamma's Avatar
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    Bingo! I was trying to do the exact same thing you did when I was going through example problems, but thought something was fishy since that is basically exactly how I proved G' was normal in G, so I thought something was up. It is kind of a cute proof IMHO.

    Seems like a fairly common algebra question, so just wanted to clean it up in case someone like myself was searching for inspiration for it online.
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