Prove $|xax^{-1}| = |a|$ 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.
 A: 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?
A: 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. 
A: 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
$$(xax^{-1})^n =e \Rightarrow xa^nx^{-1} = e \Rightarrow a^n = x^{-1}(xa^nx^{-1})x = x^{-1}ex = x^{-1}x = e$$
