# prove by induction

• Oct 23rd 2011, 06:39 AM
VonNemo19
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.
• Oct 23rd 2011, 06:56 AM
ymar
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).
• Oct 23rd 2011, 09:43 AM
Deveno
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.....