I'm reading group action in textbook Algebra by Saunders MacLane and Garrett Birkhoff.
I have a problem of understanding the last sentence:
Since conjugation is an automorphism, any two conjugate elements have the same order.
Assume $x,y \in G$ are conjugate, then they are equivalent. As such, $gxg^{-1} = y$ for some $g \in G$. This means $gx = yg$. From here, I could not get how $x,y$ have the same order.
Could you please elaborate on this point?
Best Answer
Mac Lane and Birkhoff are saying that it's not obvious (at least not directly) that $x$ and $gxg^{-1}$ have the same order. But once we know that $x \mapsto gxg^{-1}$ is an automorphism then it becomes obvious, since all automorphisms preserve order.
To see why, let $\varphi : G \to G$ be an automorphism. Then let $x \in G$ have order $n$, and let $\varphi x$ have order $m$. Now
$$(\varphi x)^n = \varphi (x^n) = \varphi e = e$$ So $m$ divides $n$.
Similarly,
$$(\varphi^{-1} \varphi x)^m = \varphi^{-1}((\varphi x)^m) = \varphi^{-1} e = e$$ And $n$ divides $m$ too, so they must be equal.
There is also a direct computational proof for the conjugation isomorphism. It is basically the exact same proof as above, but writing $gxg^{-1}$ everywhere I wrote $\varphi$ above. I encourage you to try to prove it yourself!
I hope this helps ^_^