Here is the question I am trying to answer:
Show that, if $G$ is Abelian, then the set $Tor(G)$ of torsion elements is a subgroup.
Definition: In a group $G, x \in G$ is a torsion element if $x$ is of finite order.
Here is the answer to the question:
My question is:
Why the group should be Abelian to say that $(xy)^{mn} = x^{mn} y^{mn}$?
On
Because $(xy)^n$ in general is equal to $xyxyxy\cdots xy $ ($n$ copies). If the group is not abelian, you may not combine the x's and y's together
On
Here's a counterexample in a non-abelian group, the invertible $2 \times 2$ matrices with real entries, $GL(2,\mathbb R)$. Let $$ A = \pmatrix{-1 & 3\cr 0 & 1\cr},\ B = \pmatrix{0 & 1\cr 1 & 1\cr}$$ Both $A$ and $B$ have order $2$, but $$ AB = \pmatrix{3 & -1\cr 1 & 0\cr}$$ which does not have finite order. Indeed $AB$ has an eigenvalue $\lambda = (3 + \sqrt{5})/2 > 1$, so for any positive integer $n$, $(AB)^n$ has an eigenvalue $\lambda^n$ which can't be $1$, and thus $AB \ne I$. Thus the product of two torsion elements in a nonabelian group need not be torsion.
To be clear: it is possible for the equality to just happen to hold for some values of $m$ and $n$, or some choice of elements $x$ and $y$, in a given group. It's just that the assertion would need to be justified for those specific elements/values/group.
But if $G$ is abelian, then the equality will always hold. More generally,
Proposition. Let $G$ be a group, and let $x,y\in G$. Then $xy=yx$ if and only if $(xy)^k = x^ky^k$ for every integer $k$.
Proof. If $(xy)^k=x^ky^k$ holds for all integers $k$, then taking $k=2$ we have $xyxy=(xy)^2 = x^2y^2 = xxyy$. From $xyxy=xxyy$, multiplying by $x^{-1}$ on the left and $y^{-1}$ on the right, we get $yx=xy$.
(Alternatively, taking $k=-1$ you have $y^{-1}x^{-1} = (xy)^{-1} = x^{-1}y^{-1}$, and taking inverses on both sides we get $xy = yx$).
Conversely, suppose that $xy=yx$. We proceed by induction. If $k=0$ or $k=1$, the equality clearly holds. Assume the equality holds for $k$. We have $$\begin{align*} (xy)^{k+1} &= (xy)^k(xy)\\ &= (x^ky^k)xy &&\text{(by the induction hypothesis)}\\ &= x^k(y^kx)y\\ &= x^kxy^ky &&\text{(since }x\text{ commutes with }y\text{)}\\ &= x^{k+1}y^{k+1}. \end{align*}$$ Finally, note that $xy=yx$ implies, by taking inverses, that $y^{-1}x^{-1}=x^{-1}y^{-1}$. So for $k\gt 0$, we have $$\begin{align*} (xy)^{-k} &= ((xy)^{-1})^k\\ &= (y^{-1}x^{-1})^k\\ &= (y^{-1})^k (x^{-1})^k &&\text{(applying the }k\gt 0\text{ case to }x^{-1} \text{ and }y^{-1}\text{)}\\ &= y^{-k}x^{-k}\\ &= x^{-k}y^{-k} &&\text{(since }x^{-1}\text{ and }y^{-1}\text{ commute)} \end{align*}$$ proving the result for all integers $k$.