TITLE: Deduce $\lvert ab\rvert = \lvert ba\rvert$ for all $a,b\in G$ where $G$ is a group.
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: The key idea is that $\,ab\,$ and $\,ba\,$ are conjugate, i.e. $\, ab = g^{-1} ba\, g\,$ for $\, g =\, \ldots\, $ But conjugation is always a group isomorphism. $ $ Here it restricts to the subgroup isomorphism $\ \langle ab\rangle\cong \langle ba\rangle.\,$ Thus both groups have the same order, hence their generators have the same order $\,|ab| = |ba|.$