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Math Help - HH = H, etc.

  1. #1
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    HH = H, etc.

    "H is a subgroup of G, a is in H. Show that Ha = H."

    Is this as simple as showing that since H is a group, all operations are closed within that group? Or is there more to it than that? If it's the latter, I'm stuck.
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    Re: HH = H, etc.

    Quote Originally Posted by phys251 View Post
    "H is a subgroup of G, a is in H. Show that Ha = H."
    Is it clear to you that Ha=\{x*a:~x\in H\}~? (of course * is the operation in G)

    Is it clear that Ha\subseteq H~? Why?

    Can you show H\subseteq Ha~?
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    Re: HH = H, etc.

    Quote Originally Posted by Plato View Post
    Is it clear to you that Ha=\{x*a:~x\in H\}~? (of course * is the operation in G)
    Right, that's just the definition of Ha.

    Is it clear that Ha\subseteq H~? Why?
    Because of closure?

    Can you show H\subseteq Ha~?
    So the trick is to show that the sets contain each other. Is H\subseteq Ha again a direct result of closure, or is there more to it than that?
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    Re: HH = H, etc.

    Quote Originally Posted by phys251 View Post
    So the trick is to show that the sets contain each other. Is H\subseteq Ha again a direct result of closure, or is there more to it than that?
    \\\text{If }h\in H \text{ is it true that }h*a^{-1}\in H~?\text{ WHY?}\\\text{So is it true that }(h*a^{-1})*a\in Ha~?
    Last edited by Plato; July 21st 2013 at 05:50 PM.
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    Re: HH = H, etc.

    Quote Originally Posted by Plato View Post
    \\\text{If }h\in H \text{ is it true that }h*a^{-1}\in H~?\text{ WHY?}\\\text{So is it true that }(h*a^{-1})*a\in Ha~?
    Wait, this may be simpler than I realize. Since h*a always maps to an element in H, via the properties of groups, HH always maps to H.
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