Show that for any $g,x \in G$ $|x^{-1} g x | = |xgx^{-1}| = |g|$

Question

If G is a group, show that for any $g,x \in G$ $|x^{-1} g x | = |xgx^{-1}| = |g|$

My problem is that I know my solution to be too simple. My basic thought is:

Assume that $g,x \in G$. and then note that $|x^{-1} g x | = | \frac{1}{x} g x| = | \frac{x}{x} g| = | 1 \cdot g| = |g|$. QED

I dont believe it to be this simple however, and would love to hear someone explain what more I need to do in this situation. I imagine that the key part that I'm missing is that I might need to show how for other cases, where

Answer 1

As pointed out in the comments, your approach seems to rely on $G$ being commutative, which is not assumed in the problem.

Here is an outline of a different approach: First show that if $n$ is a positive integer then $$(xgx^{-1})^n=xg^nx^{-1}.$$ From this you can conclude that $(xgx^{-1})^n=e$ if and only if $g^n=e$, which implies that $g$ and $xgx^{-1}$ have the same order.

Answer 2

There is a reason why in group theory one never writes $\frac{1}{x}$ for $x^{-1}$: this way it's more difficult to make errors like yours.

You're assuming that $gx=xg$, but this is generally not true for groups; it can hold for two particular elements, but not for any two elements. Groups where you can do that are called abelian. But if $G$ is abelian, then $x^{-1}gx=x^{-1}xg=g$, so the statement is completely trivial.

A simple proof of your assignment relies on the fact that, for any $x\in G$, the map $\varphi_x\colon G\to G$ defined by $$ \varphi_x(g)=x^{-1}gx $$ is an isomorphism.

It's a general fact that if $\alpha\colon G\to G'$ is an isomorphism of groups, then the order of $g\in G$ is the same as the order of $\alpha(g)$.