TITLE: Let G be a group and let $a\in G$. Define $F_a:G\mapsto G$ via $F_a(x)=axa^{-1}$ for all $a\in G$. Prove that $F_a$ is an isomorphism from G onto G
QUESTION [5 upvotes]: This is what I have for my proof: 
$F_{a^{-1}}(F_{a}(x))=F_{a^{-1}}(axa^{-1})=a^{-1}axa^{-1}a=(a^{-1}a)x(a^{-1})(a)=1x1=x$
$F_{a}(F_{a^{-1}}(x))=F_{a}(a^{-1}xa)=aa^{-1}xaa^{-1}=(aa^{-1})x(aa^{-1})=1x1=x$
$\Rightarrow F_{a}$ has a two sided inverse 
$\Rightarrow F_a$ is invertible 
$\Rightarrow F_a$ is one-to-one and onto G 
So I need to establish that $F_a$ preserves that group operations to show its an isomorphism and since $F_a$ is a map between two groups we only need to show that it preserves that group product
Let $x,y\in G$ 
$F_a(xy)=axya^{-1}=ax1ya^{-1}=ax(a^{-1}a)ya^{-1}=(axa^{-1})(aya^{-1})=F_a(x)F_a(y)$
$\Rightarrow F_a$ preserves the group product 
$\therefore F_a$ is an isomorphism from G onto G
Do I need to add anything else or is that exactly what I need?

REPLY [2 votes]: Or; let $a\in G$ fixed and $x,y\in G$ be arbitrary elements. So:
$$x=y\Longleftrightarrow axa^{-1}=aya^{-1}$$ that means your map is well defined and it is one-one. Assume $g\in G$ so with taking $x_0=a^{-1}ga$ we have $F_a(x_0)=g$. This means that the function is onto. The way of proving that $F_a$ is a homomorphism is exactly as you did.