# prove by induction

• Oct 23rd 2011, 05:39 AM
VonNemo19
prove by induction

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

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

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

*edit: Please, no spoilers. Just a nudge is all I need. Thanks.
• Oct 23rd 2011, 05:56 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 $\displaystyle (ab)^n=a^nb^n$. We need to prove that $\displaystyle (ab)^{n+1}=a^{n+1}b^{n+1}$. 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, 08:43 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 $\displaystyle (ab)^2 = a^2b^2$ from ab = ba.

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

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

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