Difference between revisions of "Conjugation"

To be an automorphism, it must be a bijection, which is to say it is both injective and surjective

Let $x\in G$ be given

Proof of injectivity

We wish to show that $c_x(y)=c_x(y')\implies y=y'$

Suppose $c_x(y)=c_x(y')$ then $xyx^{-1}=xy'x^{-1}$

$\implies xy=xy'$

$\implies y=y'$

So $c_x(y)=c_x(y')\implies y=y'$ is shown

Since $x\in G$ was arbitrary, we have shown all $c_x$ are injective

Proof of surjectivity

We wish to show that $\forall g\in G\exists y\in G[c_x(y)=g]$

Let $g\in G$ be given

• Note: we want $c_x(y)=g$ which is $xyx^{-1}=g\implies xy=gx\implies y=x^{-1}gx$
  • This is okay because:
    1. By hypothesis $x,g\in G$
    2. As $x\in G$ we know $\exists x^{-1}\in G$
    3. A group is closed under composition, so $x^{-1}gx\in G$ - which is a unique expression as the group is associative

       That is to say $(x^{-1}g)x=x^{-1}(gx)$

Choose $y=x^{-1}gx\in G$

Then $c_x(y) = xyx^{-1} = xx^{-1}gxx^{-1} = ege = g$

That is $c_x(y)=g$

Since $g$ was arbitrary we have shown for a given $x\in G$ that $c_x$ is surjective

Since $x$ was arbitrary we have shown that all $c_x$ are sujective

Thus all $c_x\in\text{Aut}(G)$ - as required

TODO: Not done yet