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Math Help - Prove or disprove that K is a subgroup of G

  1. #1
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    Prove or disprove that K is a subgroup of G

    Let H be a subgroup of G, let a be a fixed element of G, and let K be the set of all elements of the form a*h*a^(-1), where h E H. That is, K = {x E G | x= a*h*a^(-1) for some h E H}
    Prove or disprove that K is a subgroup of G
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by rainyice View Post
    Let H be a subgroup of G, let a be a fixed element of G, and let K be the set of all elements of the form a*h*a^(-1), where h E H. That is, K = {x E G | x= a*h*a^(-1) for some h E H}
    Prove or disprove that K is a subgroup of G
    Is aha^{-1}bhb^{-1} necessarily of this form? Of course not!

    So, think any non-abelian group and try to find a counter-example. Start with S_3.
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    Quote Originally Posted by Swlabr View Post
    Is aha^{-1}bhb^{-1} necessarily of this form? Of course not!

    So, think any non-abelian group and try to find a counter-example. Start with S_3.
    You misread the problem. a is fixed
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by FancyMouse View Post
    You misread the problem. a is fixed
    Ooooh, that's boring!

    You then just verify that such a set is, in fact, a subgroup.

    Is (aha^{-1})(aga^{-1}) \in K?

    What do you (rainyice) think (hah^{-1})^{-1} is?
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  5. #5
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    Quote Originally Posted by Swlabr View Post
    Ooooh, that's boring!

    You then just verify that such a set is, in fact, a subgroup.

    Is (aha^{-1})(aga^{-1}) \in K?

    What do you (rainyice) think (hah^{-1})^{-1} is?


    I think it will become ha^(-1)h^(-1)
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by rainyice View Post
    I think it will become ha^(-1)h^(-1)
    Precisely, and so you have your result!
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  7. #7
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    Quote Originally Posted by Swlabr View Post
    Precisely, and so you have your result!

    so it isn't a subgroup?
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  8. #8
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    It is. Swlabr had a typo there. He meant aha^{-1}, and your result should be ah^{-1}a^{-1}, which shows that every element in K has an inverse.
    In fact K is just a conjugate subgroup of H. You'll know more about it once you learned group action on sets and Sylow theorems.
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