If $a$ and $b$ are in a group $G$ and $ab=ba$, show that $xax^{-1}$ commutes with $xbx^{-1}$ for any $x \in G$.

So I wrote:

WWTS: $\bf{xax^{-1} \times xbx^{-1}=xbx^{-1}\times xax^{-1} }$

Now, the problem I have is I don't know where to start. Let's say if I start with what is given:

ab=ba, then am I allow to multiply each side by x and x$^{-1}$ and use the associative law since this is a group. So for example:


$xabx^{-1}=xbax^{-1}$ and then by associative i can change:

$xax^{-1} b=xbx^{-1} a$ and multiply by x and x^-1 on the right

$xax^{-1} \times bxx^{-1}=xbx^{-1} \times axx^{-1}$

and use associative again

$xax^{-1} \times xbx^{-1}=xbx^{-1}\times xax^{-1}$

Any ideas?

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    $\begingroup$ your steps aren't right, you can't go from $xabx^{-1}$ to $xax^{-1}b$ (you do moves like this a couple times). all you know is that $a$ and $b$ commute, you know nothing about whether $x$ commutes with $a$ or $b$. additionally, you claim this is by the associative property, which it is not (it's commutative). $\endgroup$ – Tyler Feb 8 '15 at 23:54
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    $\begingroup$ oh so, I can't switch like this: $xabx^{-1}=xa(bx^{-1})=xax^{-1} b$ by associative law? $\endgroup$ – mika Feb 8 '15 at 23:57
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    $\begingroup$ Oh yes, makes sense. thanks! $\endgroup$ – mika Feb 8 '15 at 23:57
  • $\begingroup$ Yes, exchanging the letters would require commutativity besides associativity of the operation. $\endgroup$ – Berci Feb 8 '15 at 23:59
  • $\begingroup$ @mika Associativity is $(ab)c = a(bc)$ for all $a,b,c\in G$. You seem to mistake this for commutativity: $ab=ba$ for all $a,b\in G$. $\endgroup$ – AlexR Feb 9 '15 at 0:09

There's not much to it and you basically found it already:

$$\begin{align*} (x a x^{-1}) (x b x^{-1}) & = x a \underbrace{(x^{-1} x)}_{=e}bx^{-1} \\ & = x (ab) x^{-1} \\ & = x(ba)x^{-1} \\ & = xb(x^{-1} x)a x^{-1}\\ & = (xbx^{-1})(xax^{-1}) \end{align*}$$

Note that the group operation is associative, we use this a lot here.

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    $\begingroup$ where there some steps skipped from: $x(ba)x^{−1}=(xbx^{−1})(xax^{−1})$ $\endgroup$ – mika Feb 8 '15 at 23:53
  • $\begingroup$ @mika I'll add these, but it all boils down to $x^{-1}x = e$ (identity) and $a(bc) = (ab)c$ $\endgroup$ – AlexR Feb 8 '15 at 23:54
  • $\begingroup$ @mika After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. $\endgroup$ – AlexR Feb 9 '15 at 20:12
  • $\begingroup$ Alright! Thanks for the tip and solution! $\endgroup$ – mika Feb 9 '15 at 23:22

A very useful idea to such problems is that the map $$f(a)=xax^{-1}$$ is called "conjugation" and is an homomorphism from the group to itself. That is to say, we can prove, for all $a,b$ that: $$f(a)f(b)=f(ab).$$ This is very clear if you expand both expressions. Now, if we start with $$ab=ba$$ We can apply $f$ to both sides to get $$f(ab)=f(ba)$$ and then use that $f$ is a homomorphism to get $$f(a)f(b)=f(b)f(a)$$ which, if you expand, is the desired statement.

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    $\begingroup$ While nice, the OP seems to have a problem with the part that "can be shown", that $f$ is a homomorphism ;) $\endgroup$ – AlexR Feb 9 '15 at 0:08

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