If $xy=yx$ in group $G$, then $(xy)^n=x^ny^n$

Question

I am showing that if for some group $G$, $xy=yx$ for every $x, y \in G$ then

$$(xy)^n=x^ny^n.$$

I claim this holds by induction on $n$. So base case if $n=1$, we have

$$(xy)^1=xy=x^1y^1$$

as needed. Then suppose for $k$ that we have

$$(xy)^k=x^ky^k$$

And we want to show

$$(xy)^{k+1}=x^{k+1}y^{k+1}$$

But we can expand

\begin{align} (xy)^{k+1} &= (xy)^k(xy) && \text{def of exponent} \\ &=x^ky^kxy && \text{by inductive hypotheses} \\ &=x^kxy^ky && \text{as $xy=yx$ thus $x$ commutes with all $k$ $y$'s}\\ &=x^{k+1}y^{k+1} && \text{def of exponent} \end{align}

and we are done by induction. The only step I am shaky about is the third equality. Do I separately need to show if $xy=yx$ then $y^kx=xy^k$ as well or?

Answer 1

You can show $y^kx=xy^k$ by induction. Another trick is to take an $x$ and $y$ off each end so that $(xy)^{n+1}=x(yx)^ny=x(xy)^ny=x(x^ny^n)y=x^{n+1}y^{n+1}$.

Answer 2

To finish the inductive step, $y^kx = y^{k-1} (yx) = (y^{k-1} x)y = xy^{k-1}y=xy^k$.

As this is a low quality question, I will add my two cents into how to think about these operations. Since groups are associative and commutative, we can just rearrange the variables in any order we want.