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Math Help - prove by induction

  1. #1
    No one in Particular VonNemo19's Avatar
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    prove by induction



    Let G be a group, and a,b\in{G}. For any positive integer  n we define a^n by

    a^n=\underbrace{aaa...a}_{\text{n factors}} .


    Now prove by induction:
    If ab=ba, then (ab)^n=a^nb^n

    *edit: Please, no spoilers. Just a nudge is all I need. Thanks.
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  2. #2
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    Re: prove by induction

    So we know ab=ba. The base case n=1 is trivial. We need to carry out the inductive step. The inductive hypothesis is that (ab)^n=a^nb^n. We need to prove that (ab)^{n+1}=a^{n+1}b^{n+1}. How can we see here something that appears in the inductive hypothesis? When we see it, we can use ab=ba (more than once).
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  3. #3
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    Re: prove by induction

    the trick is to see how you're going to get from the "n-step" to the "n+1-step".

    let's see if we can prove (ab)^2 = a^2b^2 from ab = ba.

    (ab)^2 = (ab)(ab) = a(ba)b do you see it coming...? we can just switch the middle pair.

    (ab)^2 = a(ba)b = a(ab)b = (aa)(bb) = a^2b^2.

    now try it for (ab)^3, and see if you can generalize.....
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