1. ## Groups

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.

2. Originally Posted by elninio
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 ...

3. Aha! I'm not used to using induction in a modern algebra class.

4. Originally Posted by elninio
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.

5. 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

6. Originally Posted by elninio
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 ...

7. ...
Call me slow, but I understand what is going on, however I dont understand how this will eventually prove that

8. Originally Posted by elninio
...
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.