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  1. #1
    Member elninio's Avatar
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    Let G be a group. For all a,b in G, Prove that (ab)^n = a^nb^n for all n positive integers if and only if ab=ba.
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  2. #2
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by elninio View Post
    Let G be a group. For all a,b in G, Prove that (ab)^n = a^nb^n for all n positive integers if and only if ab=ba.
    Have you tried using induction? Base case is ab = ba \iff abab = aabb - take it away, maestro ...
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  3. #3
    Member elninio's Avatar
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    Aha! I'm not used to using induction in a modern algebra class.
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  4. #4
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by elninio View Post
    Aha! I'm not used to using induction in a modern algebra class.
    Here's a good chance to get used to it! You'll need it a lot.

    Give it a go, if you get into trouble come back and I'll try and give you some more hints. If you need to brush up on your proof by induction, then do just that first.
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  5. #5
    Member elninio's Avatar
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    Its been over two years since I've done induction in number theory.

    Here is my write-up:

    ab=ba => abab=aabb

    Assume n=k holds such that (ab)^k=a^k * b^k

    n=k+1 yields (ab)^(k+1) = a^(k+1) * b^(k+1)

    Which implis ab(ab)^k = a(a)^k * b(b)^k

    Which implies (ab)^n = a^n * b^n
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  6. #6
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by elninio View Post
    Its been over two years since I've done induction in number theory.

    Here is my write-up:

    ab=ba => abab=aabb

    Assume n=k holds such that (ab)^k=a^k * b^k

    n=k+1 yields (ab)^(k+1) = a^(k+1) * b^(k+1)

    Which implis ab(ab)^k = a(a)^k * b(b)^k

    Which implies (ab)^n = a^n * b^n
    No, you want to prove (ab)^{(k+1)} = a^{(k+1)} b^{(k+1)}, not assume it.

     (ab)^{(k+1)} = (ab)^k (ab) = a^kb^k ab = a^kb^{k-1} b a b = a^kb^{k-1} a b b etc. etc.

    It's a bit of a long process, but you get the gist ...
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  7. #7
    Member elninio's Avatar
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    Call me slow, but I understand what is going on, however I dont understand how this will eventually prove that
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  8. #8
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by elninio View Post
    ...
    Call me slow, but I understand what is going on, however I dont understand how this will eventually prove that
    You can progressively shuffle that rightmost a through that k's worth of b's until it's adjacent the a^k on the left. Then all the a's will be together and all the b's will be together, and you're left with what you're trying to prove.

    There may be a quicker way of doing it, but when I first put this together for the ProofWiki site, I first proved by induction that if a commutes with b then a commutes with b^n for all n, which would make the above somewhat cleaner.
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