power of a product in non Abelian group

Question

To show $(ab)^n=a^n b^n$ is true, I use the commutative property which implies the group I'm working with is an Abelian group. Is there a way to show $(ab)^n=a^n b^n$ meets in non-abelian groups or it is simply false?

Answer 1

It is indeed "simply false" in non-abelian groups; indeed, the case $n = 2$, if true, asserts that

$(ab)(ab) = (ab)^2 = a^2 b^2 \Longrightarrow ab = ba, \tag 1$

which is of course essentially the definition of "abelian group".

Answer 2

It's not generally true that for any $a,b \in G, (ab)^2 = a^2b^2$ if $G$ is not abelian. For example take in $GL_2(\mathbb{R})$, the group of $2$-by-$2$ invertible matrices under matrix multiplication, $a = \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}$ and $b = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. We see that $ (ab)^2 = \begin{bmatrix}1 & -1 \\ 1 & 0\end{bmatrix}^2 = \begin{bmatrix} 0 & -1\\ 1 & -1 \end{bmatrix}. $

and

$ a^2b^2 = \begin{bmatrix} 1 & 2 \\ 0 & 1\end{bmatrix}\begin{bmatrix} -1 & 0\\ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & - 2 \\ 0 & - 1\end{bmatrix}. $

We see $(ab)^2 \neq a^2b^2$.