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So, I have a question that says for any group elements $a$ and $x$, prove that $|xax^{-1}| =|a|$, where $|\cdot|$ is the order of an element.

I know that $xx^{-1} = e$ and $ae=a$, but I'm not sure how to get the $x$'s on the same side.

Currently, for my proof, which is not much, I have:

Let $n = |xax^{-1}|$, then $(xax^{-1})^n = e$.

Honestly, I have no idea where to go from here, or if what I've got is even right. I would really appreciate just a point in the right direction.

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3 Answers

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Hint: Generally, when you want to prove that $o(a)=o(b)$ (assuming they're finite) in a group $G$, you show that

$$b^{o(a)} = a^{o(b)} = e$$

If you prove this, then it implies that $o(a) \mid o(b)$ and $o(b) \mid o(a)$. Since the order of an element is positive, that implies $o(a)=o(b)$.

Edit: As you figured, $(xax^{-1})(xax^{-1})=xa^2x^{-1}$. What happens exactly is this

$$(xax^{-1})(xax^{-1})=xa(x^{-1}x)ax^{-1}=xa(e)ax^{-1}=xa^2x^{-1}$$

If we calculate $(xax^{-1})(xax^{-1})(xax^{-1})$, we see the same pattern. Each time we multiply $xax^{-1}$ by itself, the term $x\cdot x^{-1}$ appears which is canceled. So, you can conclude that $(xax^{-1})^n=xa^nx^{-1}$.

Now, if you plug in $o(a)$ instead of $n$, you get $(xax^{-1})^{o(a)}=xa^{o(a)}x^{-1}=x(e)x^{-1}=e$. Hence, $o(xax^{-1}) \mid o(a)$. Conversely, $$xa^{o(xax^{-1})}x^{-1}=(xax^{-1})^{o(xax^{-1})}=e$$ $$xa^{o(xax^{-1})}x^{-1}=e \implies a^{o(xax^{-1})}=x^{-1}x=e$$

This shows that $o(a) \mid o(xax^{-1})$. Since they're both positive, $o(a) = o(xax^{-1})$

Can you handle the infinite case now?

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By induction, one has $$ (xax^{-1})^n=xa^nx^{-1},\tag{1} $$ which implies that the order of the element $y:=xax^{-1}$ must be at most the order of $a$. But the order of $y$ is at least the order of $a$ since if $|y|=n$, then by (1) one has $$ a^n = x^{-1}(xa^nx^{-1})x=x^{-1}y^nx=e.\tag{2} $$

If the order of $a$ is infinite, then, (2) shows that the order of $y$ cannot be finite.

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An abstract proof could be:

$$A_x: a \mapsto xax^{-1}$$ is a group automorphism. So, $ord(a) = ord(xax^{-1})$

A proof "on foot" could be:

It is enough to show $$a^n = e \Leftrightarrow (xax^{-1})^n = e$$. But, that's trivial since

$$(xax^{-1})^n = \underbrace{(xax^{-1}) \cdots (xax^{-1})}_{n} = xa^nx^{-1}$$ So, $$a^n = e \Rightarrow (xax^{-1})^n = xa^nx^{-1} = xex^{-1} = xx^{-1}=e$$. And $$(xax^{-1})^n =e \Rightarrow xa^nx^{-1} = e \Rightarrow a^n = x^{-1}(xa^nx^{-1})x = x^{-1}ex = x^{-1}x = e$$

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