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This is part of Exercise 2.1.15 of F. M. Goodman's "Algebra: Abstract and Concrete".

Let $G$ be a group. Suppose $(ab)^2=a^2b^2$ for all $a, b\in G$. Show that $(ab)^n=a^nb^n$ for all $a, b\in G$, $n\in\mathbb{N}$.

My Attempt:

I'm using induction on $n$. Let $a, b\in G$.

Obviously $(ab)^1=ab=a^1b^1$ so the result holds for $n=1$.

Assume $(ab)^r=a^rb^r$ for some $r\in\mathbb{N}$.

Consider when $n=r+1$: we have $$\begin{align} (ab)^{r+1}&=(ab)^rab \\ &=a^rb^rab \\ &=a^{r+1}a^{-1}b^rab, \end{align}$$ but I don't know where to go from here.

  • 4
    Hint: Use the initial identity to prove $ab = ba$. – 2017-02-26
  • 0
    From the equation $(ab)^2=a^2b^2$ premultiply by $a^{-1}$ & post multiply by $b^{-1}$ ... observe that $a$ & $b$ commute ... – 2017-02-26
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    Follow the comment of quasi and then use the result in http://math.stackexchange.com/questions/492332/let-a-b-be-in-a-group-g-show-abn-anbn-forall-n-in-mathbbz-if-an?rq=1 – 2017-02-26

3 Answers 3

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Following your approach, you can try a different factorization:

$$ (ab)^{r+1}=(ab)(ab)^r=aba^rb^r. $$ Then, insert a $bb^{-1}=e$ just after the first $a$ in $a^r$. This gives an $abab$ to which you can apply the original equality. $$ aba^rb^r=aba(bb^{-1})a^{r-1}b^r=(abab)b^{-1}a^{r-1}b^r=(a^2b^2)b^{-1}a^{r-1}b^r=a^2ba^{r-1}b^r. $$ Continue by using (a second) induction to move the $b$ past the $a^r$, one factor at a time.

6

Since $abab=aabb$ you can get $ba=ab$, that is $a,b$ commute. Now you can rearrange any product between them.

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Since $abab = a^2 b^2$, then $ab = b^{-1} a b^2$ and $a^r b^r ab = a^r b^r b^{-1} a b^2 = a^r b^{r-1} a b^2$, and you can "shift" the $a$ successively to obtain the expression $a^{r+1}b^{r+1}$