6

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.

dustin
  • 8,241

3 Answers3

7

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

lhf
  • 216,483
  • 1
    From above, we have that $\lvert ab\rvert = \lvert g^{-1}(ab)g\rvert$. \begin{align}
    \lvert ab\rvert & = \lvert g^{-1}(ab)g\rvert\tag{Let $g = b^{-1}a$}\
    & = \lvert ba^{-1}(ab)b^{-1}a\rvert\
    & = \lvert ba\rvert
    \end{align
    }
    – dustin Dec 31 '14 at 00:46
  • 1
    $g = b^{-1}$ will do, though $g = b^{-1}a$ also works. – Geoff Robinson Dec 31 '14 at 00:50
3

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

Bill Dubuque
  • 272,048
2

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 $\text{ord}(ba)=\text{ord}(ab)$.

Jack M
  • 27,819
  • 7
  • 63
  • 129