Showing that the order of $xyx^{-1}$ equals the order of $y$ I have just started  out learning about groups and I am stuck on trying to solve the following question for a couple of hours now:

Let $G$ be a finite group and $x, y$ be in $G$. Show that order of $xyx^{-1}$ equals order of $y$.

The following is how far I was able to go:
To show that the order of $xyx^{-1}$ equals the order of $y$, I believe all I have to do is show the following:

Show that $(xyx^{-1})^m=y^n$.

$$\underbrace{(xyx^{-1})(xyx^{-1})(xyx^{-1})\cdots(xyx^{-1})}_{m\text{ times}} = (xy^mx^{-1}).$$
Now I have to show that $(xy^mx^{-1}) = y^n$, but I cannot do $xx^{-1}y^m=y^n$ because I don't know if the group is abelian or not.
Before you ask, this is not a homework question.
Thanks!
 A: So, if I understad correctly, you have shown that $(xyx^{-1})^m=xy^mx^{-1}$. Now, you have almost finished. You have two options:
If $m$ is the order of $y$, then $y^m=1$, and $(xyx^{-1})^m=xy^mx^{-1}=x1x^{-1}=1$, so the order of $xyx^{-1}$ is, at most, the order of $y$.
If $m$ is less than the order of $y$, then $y^m\neq 1$, and then $xy^mx^{-1}\neq 1$, since that would mean either $(1)$ $xy^{m}=x\Leftrightarrow y^m=1$, which you know it is not, or $(2)$ $y^mx^{-1}=x^{-1}\Leftrightarrow y^m=1$, which, again, isn't. Therefore, the order of $xyx^{-1}$ is no less than the order of $y$.
From here, if you know that $ord(xyx^{-1})\leq ord(y)$ and $ord(xyx^{-1})\geq ord(y)$, it must be that $ord(xyx^{-1})=ord(y)$.
A: It would be enough to show that $(xyx^{-1})^n = e$ iff $y^n = e$. This follows from noticing that $(xyx^{-1})^n = x y^n x^{-1}$.
Indeed if $y^n = e$ then $x y^n x^{-1}= x x^{-1} = e$. Conversely if $(xyx^{-1})^n = e$ then $x y^n x^{-1} = e$ so multiplying on the right by $x$ and on the left by $x^{-1}$ gives $y^n = x^{-1}x = e$ and we are done.
A: Note that conjugation by an element is a group automorphism of $G$. Now note that group automorphisms (and more generally isomorphisms!) preserve the order. 
