Deduce $\lvert ab\rvert = \lvert ba\rvert$ for all $a,b\in G$ where $G$ is a group.

Question

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.

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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 $a,b\in G$.

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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 $\lvert g^{-1}xg\rvert$.

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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.

Answer 1

Hint: Find $g$ such that $g^{-1}(ab)g=ba$.

Answer 2

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 $\,|ab| = |ba|.$

Answer 3

Here's a more direct proof. If $\text{ord}(ab)=n$ is the order of $ab$, then

$$ \underbrace{(ab)(ab)...(ab)(ab)}_{n\text{ times}}=a\underbrace{(ba)(ba)..(ba)}_{n-1\text{ times}}b=1$$ $$\begin{align}(ba)^{n-1}&=(ba)^{-1} \\ (ba)^n&=1 \end{align} $$

So $\text{ord}(ba)\leq\text{ord}(ab)$. By symmetry, $\text{ord}(ba)\geq\text{ord}(ab)$, so $\text{ord}(ba)=\text{ord}(ab)$.