Conjugation as an Automorphism: Proof and Applications

[Mr Davis 97]

Homework Statement

Let $G$ be a group and let $G$ act on itself by left conjugation, so each $g \in G$ maps $G$ to $G$ by $x \mapsto gxg^{-1}$. For fixed $g \in G$, prove that conjunction by $G$ is an automomorphism of G. Deduce that $x$ and $gxg^{-1}$ have the same order for all $x$ in $G$ and that for any subset $A$ of $G$, $|A| = |gAg^{-1}|$.

Homework Equations

The Attempt at a Solution

Here is how I approached the problem. Let $\sigma_g = gxg^{-1}$. I will take it as given that $\sigma_g$ is a bijection from $G$ to $G$, as I have already proved this in the general case before and don't feel like doing it again. Thus, we must only show that $\sigma_g$ is a homomorphism. Let $x_1, x_2 \in G$. Then $\sigma_g (x_1x_2) = g x_1 x_2 g^{-1} = g x_1 g^{-1} g x_2 g^{-1} = \sigma_g (x_1) \sigma_g(x_2)$. So $\sigma_g$ is an automorphism.

We can say that $|x| = |gxg^{-1}|$ because isomorphisms preserve order and we can say that $|A| = |gAg^{-1}|$ because $\sigma_g$ is injective and so the elements in $A$ will be mapped to distinct elements.

 

[fresh_42]

Homework Statement

Let $G$ be a group and let $G$ act on itself by left conjugation, so each $g \in G$ maps $G$ to $G$ by $x \mapsto gxg^{-1}$. For fixed $g \in G$, prove that conjunction by $G$ is an automomorphism of G. Deduce that $x$ and $gxg^{-1}$ have the same order for all $x$ in $G$ and that for any subset $A$ of $G$, $|A| = |gAg^{-1}|$.

Homework Equations

The Attempt at a Solution

Here is how I approached the problem. Let $\sigma_g = gxg^{-1}$.

$\sigma_g(x) = gxg^{-1}$

I will take it as given that $\sigma_g$ is a bijection from $G$ to $G$, as I have already proved this in the general case before and don't feel like doing it again. Thus, we must only show that $\sigma_g$ is a homomorphism. Let $x_1, x_2 \in G$. Then $\sigma_g (x_1x_2) = g x_1 x_2 g^{-1} = g x_1 g^{-1} g x_2 g^{-1} = \sigma_g (x_1) \sigma_g(x_2)$. So $\sigma_g$ is an automorphism.

We can say that $|x| = |gxg^{-1}|$ because isomorphisms preserve order ...

You can also just write the same line as before again: $(gxg^{-1})^n=gx^ng^{-1}$ and one equals $e=1$ if and only if the other does, too.

... and we can say that $|A| = |gAg^{-1}|$ because $\sigma_g$ is injective and so the elements in $A$ will be mapped to distinct elements.

You also need surjectivity as $G$ might not be finite, in which case (finite case) injectivity implies surjectivity and vice versa. Otherwise $A$ could be larger.

 

[fresh_42]

Correction:

in which case (finite case) injectivity implies surjectivity

is wrong. I confused this with vector spaces and dimensions. If we know nothing but that $A$ is a set, then even in a finite case, a proper subset is an injective embedding which is not surjective.