Let $a,b$ be in a group $G$. Show $(ab)^n=a^nb^n$ $\forall n\in\mathbb{Z}$ if and only if $ab=ba$.

Let $G$ be a group and $a,b\in G$. Show $(ab)^n=a^nb^n$ for all $n\in\mathbb{Z}$ if and only if $ab=ba$.

I don't known where to start. It seems trivially.

Solution 1:

Hint:

• Use $(ab)^{-1} = b^{-1}a^{-1}$ (you have $n \in \mathbb{Z}$, so it works for negative numbers too).
• Another approach would be to manipulate $abab = aabb$.

I hope this helps $\ddot\smile$

Solution 2:

$ab=ba\Rightarrow (ab)^n=\underbrace{ab.ab.ab\cdots ab}_{n\text{ times}}=a(ba)(ba)(ba)\cdots(ba).b=a^nb^n$ replacing $ba$ by $ab$.

Conversely $(ab)^n=a^nb^n,~~\forall n$, so $(ab)^2=a^2b^2$. Multiplying this equation on the right by $b^{-1}$ and on the left by $a^{-1}$, then we have get $ab=ba$.

Solution 3:

Groups satisfying the condition $(ab)^n = a^n b^n$ are known as $n$-abelian groups. The following is known:

Let $G$ be a group and suppose that there exist three consecutive positive integers $n$ for which $G$ is $n$-abelian. Then $G$ is a commutative group.

For a nice proof, see Howard E. Bell, Mathematics Magazine, Vol. 55, No. 3 (May, 1982).

Solution 4:

$(ab)^n=a^n b^n\implies (ba)^{n-1}=a^{n-1}b^{n-1}\implies b^{n-1}a^{n-1}=a^{n-1}b^{n-1}$.

Now put $n=2$.

This proves $G$ is abelian if $(ab)^n=a^n b^n$.

Proving that $(ab)^n= a^n b^n$ if $G$ is abelian is elementary.