It has been awhile since I took Abstract Algebra and decided to go back through Dummit and Foote. However, I have come across a problem I can't seem to figure out.
If $x$ and $g$ are elements of the group $G$, prove that $\lvert x\rvert = \lvert g^{-1}xg\rvert$. Deduce that $\lvert ab\rvert = \lvert ba\rvert$ for all $a,b\in G$.
Let $\lvert x\rvert = n$. Then \begin{align*} x^n & = (g^{-1}xg)^n\\ & = \underbrace{(g^{-1}xg)\cdots (g^{-1}xg)}_{n\text{ times}}\\ & = g^{-1}x^ng\\ & = g^{-1}eg\\ & = g^{-1}g\\ & = e \end{align*} Thus, $\lvert g^{-1}xg\rvert = n = \lvert x\rvert$. Now, suppose $\lvert x\rvert = \infty$ and $\lvert g^{-1}xg\rvert = n$. Then $$ g^{-1}x^ng = e\Rightarrow gg^{-1}x^ngg^{-1} = geg^{-1}\Rightarrow x^n = e $$ which is a contradiction. That is, if $\lvert x\rvert = \infty$, then so does $\lvert g^{-1}xg\rvert$.
The problem is I can't seem to deduce $\lvert ab\rvert = \lvert ba\rvert$. I imagine it isn't difficult, but I just don't see it.