In group $G$, $xyx^{-1}$ and $y$ have the same order. I am trying to prove the result that in a group $G$, for any $x,y \in G$, the order of $xyx^{-1}$ is the same as the order of $y$. This is what I have.

Suppose the order of $y$ is finite, call it $n$. So $n$ is the smallest positive integer so that $y^n = e$. If $n = 1$, $y^1 = e$, so $xyx^{-1} = xex^{-1} = xx^{-1} = e$. If $n \geq 2$, we have
$$(xyx^{-1})^n = (xyx^{-1})(xyx^{-1}) \ldots (xyx^{-1}) = xy(x^{-1} x)y(x^{-1} x) \ldots (x^{-1} x)(yx^{-1}) = xy^n x^{-1} = xx^{-1} = e.$$
Suppose instead $(xyx^{-1})$ has order $n$, so $n$ is the smallest positive integer so that $(xyx^{-1})^n = e$. If $n = 1 $, then $xyx^{-1} = e$. Multiplying by $x^{-1}$ on the left and $x$ on the right gives $y = x^{-1} x = e$, so $y$ has order $1$. If $n \geq 2$, we have:
\begin{align*}
(xyx^{-1})(xyx^{-1}) \ldots (xyx^{-1}) & = e \\
xy(x^{-1} x)y(x^{-1} x)y(x^{-x} x) \ldots (yx^{-1}) & = e \\
xy^n x^{-1} = e 
\end{align*}
Multiplying by $x^{-1}$ on the left and $x$ on the right gives $y^n = x^{-1} x = e$, so $y$ has order $n$.

The first thing I am having difficulty with is proving that this $n$ is the smallest $n$. The second thing is, I can't find a better and more rigorous way to do it than the $\ldots$ proof. I can't get induction to work.

Suppose the order of $y$ is infinite. So $y^n \neq e$ for any natural number $n$. For contradiction, suppose $xyx^{-1}$ has finite order, so for some $m$,
$$(xyx^{-1})^m = e.$$
By the same algebra above, we get $y^m = e$, a contradiction to $y$ having infinite order.


Suppose the order of $xyx^{-1}$ is infinite, so $(xyx^{-1})^n \neq e$ for all natural numbers $n$. If $y$ has finite order $m$, then by the first set of work, we get $(xyx^{-1})^m = e$, a contradiction to the order being infinite.

 A: Well, you can prove that much more directly, just noticing that for every $k \in \mathbb{N}$, you have the equivalence
$$(xyx^{-1})^k=e \text{ }\Longleftrightarrow \text{ }y^k=e$$
A: Hint:  One fact about automorphisms is that they preserve the order of elements.  This applies in particular to any inner automorphism.  So perhaps just prove that $g\mapsto xgx^{-1}$ is an automorphism, for any $x\in G$.
A: Lemma. $(xyx^{-1})^n=xy^nx^{-1}$ for every $n\in \Bbb N$.
Proof. Induction on $n$. The claim holds for $n=1$; let it hold for $n$ (inductive hypothesis); then, for $n+1$ we get: $(xyx^{-1})^{n+1}=(xyx^{-1})^nxyx^{-1}\space\stackrel{(i.h.)}{=}\space xy^nx^{-1}xyx^{-1}=\space xy^neyx^{-1}=xy^nyx^{-1}=xy^{n+1}x^{-1}$. $\space\space\Box$
Corollary 1. $o(xyx^{-1})\mid o(y)$.
Proof. By the lemma, $(xyx^{-1})^{o(y)}=xy^{o(y)}x^{-1}=xex^{-1}=xx^{-1}=e$. $\space\space\Box$
Corollary 2. $o(y)\mid o(xyx^{-1})$.
Proof. By the lemma, $e=(xyx^{-1})^{o(xyx^{-1})}=xy^{o(xyx^{-1})}x^{-1}$, whence $y^{o(xyx^{-1})}=x^{-1}ex=x^{-1}x=e$. $\space\space\Box$
By the corollaries 1 and 2, $o(xyx^{-1})=o(y)$.
