I'm trying to solve exercise 2.4.3 in Artin's Algebra:
Let $a$ and $b$ be elements of a group $G$. Prove that $ab$ and $ba$ have the same order.
Here is my attempt. I would appreciate any and all feedback. I would like for this proof to be as rigorous as possible, so please let me know if I have misstated something or left something out.
We first treat the case where $|ab|$ and $|ba|$ are finite. Call the former $n$ and and the latter $m$. If $n = 1$, the result is obvious since $ab = e$, so $a = b^{-1}$ and, similarly, $ba = e$. Suppose $n \geq 2$. We then have: \begin{align*} e = (ab)^n = a(ba)^{n-1} b, \end{align*} so multiplying by $a^{-1}$ on the left and $b^{-1}$ on the right, we obtain \begin{align*} a^{-1} b^{-1} = (ba)^{n-1}. \end{align*} Multiplying by $ba$ on the left, we obtain \begin{align*} (ba)^n = e. \end{align*} Therefore, $|ba| = m \mid n$. By interchanging labels, we obtain that $n \mid m$, so $n = m$, i.e., $|ab| = |ba|$. Next, I claim that if $ab$ has infinite order, then $ba$ has infinite order and vice versa. If $|ba|$ were finite, call it $n$, then we have $|ab| \mid n$, which is impossible if $|ab| = +\infty$, so $|ba|$ must also be infinite. Analagously, if $ba$ has infinite order, $ab$ does as well.
How does this look? I assume when we say that an element has "infinite order," there's no need to distinguish between countable and uncountable infinities since this is only a definition for "there exists no power that gets us back to the identity."