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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."

Shaun
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user861776
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    No element in any group has "uncountably infinite order". If an element of a group has so-called infinite order, the cyclic subgroup it generates always has countably many elements. There is only one kind of "infinite order" that an element can have, and if we have to put a size to it (which isn't a common thing to do), countable is the most reasonable. – Arthur Apr 06 '21 at 08:30
  • This makes sense to me. Thank you. Does the proof look ok? – user861776 Apr 06 '21 at 08:40
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    I do not like the following (near the end) "which is impossible if $|ab|=+\infty$" and think "and hence the order of $ab$ is also finite" is better. – ancient mathematician Apr 06 '21 at 08:50
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    This proof can be made to work, so fine. But I think that the following is more illuminating: (i) prove $|g^{-1}xg|$ divides $|x|$ whenever $|x|$ is finite (ii) deduce that $x$ and $g^{-1}xg$ have the same order always and (iii) corollary $ab$ and $ba=a^{-1}(ab)a$ have the same order. – ancient mathematician Apr 06 '21 at 08:54
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    In any group $G$, for every $a,b\in G$, the elements $ab$ and $ba$ are conjugate to each other. –  Apr 06 '21 at 09:06
  • @ancientmathematician Thank you for the feedback. Can I ask why you don't like that language? Is it because I am suggesting, in effect, that $+\infty$ is a number? Other than that, is the proof mathematically correct? – user861776 Apr 06 '21 at 09:18
  • I have seen many of these alternate approaches, but I wanted to get feedback specifically on my proof above. I haven't seen any other answers prove it this way. – user861776 Apr 06 '21 at 10:30
  • @user861776: in reply to your question, yes that is my objection. Otherwise I think it's all fine. – ancient mathematician Apr 06 '21 at 10:54

2 Answers2

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Consider $\lvert ab\rvert=n$. Then

$$\begin{align} (ab)^n&=\underbrace{(ab)\dots (ab)}_{n\text{ times}}\\ &=e\\ &=a^{-1}ea\\ &=a^{-1}\underbrace{(ab)\dots (ab)}_{n\text{ times}}a\\ &=\underbrace{(ba)\dots (ba)}_{n\text{ times}}\\ &=(ba)^n, \end{align}$$

so the order of $ba$ divides $n$; similarly, $n$ divides the order of $ba$. Hence $\lvert ab\rvert =\lvert ba\rvert$.

This happens because conjugation by an element of a group is an automorphism of that group.

Shaun
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I propose this variant,a bit simpler, in my opinion:

It is enough to prove that, if $(ab)^n=e$, then $(ba)^n=e$, because the reverse implication is valid permuting the variables $a$ and $b$.

Indeed, $\:(ba)^n\color{red}b=b(ab)^n=be=e\color{red}b$, which implies $\:(ba)^n=e$ by the simplification rule in a group.

Bernard
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