Given the group $G$ and $h,g \in G$ prove that $|gh|=|hg|$ [duplicate]

Solution 1:

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 ^_^

Solution 2:

Since $y=gxg^{-1}$, we have

$$\begin{align} y^n&=\underbrace{(gxg^{-1})\dots(gxg^{-1})}_{n\text{ times}}\\ &=\underbrace{g\cdot x\cdot (g^{-1}g)\cdot\dots\cdot (g^{-1}g)\cdot x \cdot g^{-1}}_{n\text{ times }x}\\ &=gx^ng^{-1}, \end{align}$$

so, if $x^n=e$, then $y^n=e$, and vice versa (by the inverse of conjugation).