Show that if $ab$ has finite order $n$, then $ba$ also has order $n$. - Fraleigh p. 47 6.46.

Question

Attempted proof based on here and yahoo.

Given: $a,b$ elements of $G$, and $ab$ has finite order $n$.

Hence $\color{magenta}{|ab| = n} \iff (ab)^n = e$.

I must prove that $n$ is the smallest positive integer such that $(ba)^n = e.$ Notice:

$\begin{align}(\color{darkorange}{a}\color{darkcyan}{b})^n &= e \implies \color{darkorange}{a}\color{darkcyan}{b}(ab)^{n-1} = e \implies \color{darkcyan}{b}(ab)^{n-1}\color{darkorange}{a} = e \implies (ba)^n = e \end{align} $
$\implies |ba| \le n$.

Now I must prove that there's no positive integer $m$ such that $m<n$ and $(ba)^m = e.$
For a proof by contradiction, suppose $|ba|<\color{magenta}{n}$. Then we could apply the same reasoning to find that $ |ab| ≤ |ba| < \color{magenta}{|ab|}$, which is absurd.
So $|ba| = |ab| = n. \quad \blacksquare$

Questions

1. Whence does $ |ab| ≤ |ba| $ arise, in the second last line?
2. What's the intuition of the claim proved?
3. How do you prognosticate that $|ba|<\color{magenta}{n}$ is impossible, and that you should prove it by contradiction?

Answer 1

You have seen why $(ab)^n=e$ implies $(ba)^n=e$, but what you need to realize is that this is symmetric. The same logic shows that $(ba)^n=e$ implies $(ab)^n=e$!

The same is true for minimal $n$, so it follows that their orders must agree.

Answer 2

Since $x\mapsto bxb^{-1}$ is an automoorphism, you have implications both ways: $$(ab)^n=e\iff b(ab)^nb^{-1}=e\iff (ba)^n=e$$

Answer 3

Hint: conjugate elements have the same order.

Answer 4

Key Idea $\ $ Isomorphisms preserve all "group-theoretic" properties, which includes the order of an element $\,g.\,$ Indeed, the order of $\,g\,$ is the order (cardinality) of the cyclic group generated by $\,g.\,$ But an isomorphic image of a group has the same order (cardinality).

Your is the special case of a conjugation isomorphism $\ g\mapsto bgb^{-1},\, $ with inverse $\ g\mapsto b^{-1}gb.$

All cyclic permutations arise by conjugation, e.g. $\,a_3 a_4\cdots a_n\,a_1a_2 = (a_1 a_2)^{-1}(a_1 a_2 \cdots a_n) (a_1 a_2)$