Prove that $C_g(x) = gxg^{-1}$ is an isomorphism Let $G$ be a group and $g \in G$. Define $C_g : G \to G$ as $C_g(x) = gxg^{-1}$
Prove that $C_g$ is an isomorphism.
Homomorphism:
$C_g(xy) = gxyg^{-1} = gx(gg^{-1})yg^{-1} = (gxg^{-1})(gyg^{-1}) = C_g(x)C_g(y)$
Injective:
Assume $C_g(x) = C_g(y)$
$$gxg^{-1} = gyg^{-1}$$
Left multiply by $g^{-1}$ and right multiply by $g$ to get $x=y$
Surjective:
Let $C_g(x) = y$ for some $y$.
Therefore, $y = gxg^{-1} \therefore x = g^{-1}yg \in G$ since it is the multiplication of elements of a group and, by closure it is part of a group.
I feel that something is wrong on the part about proving surjectivity. 
 A: The proof of $C_g$ being a homomorphism is good. In order to show bijectivity it's easier to find the inverse map.
Note that
$$
C_g(C_h(x))=C_g(hxh^{-1})=g(hxh^{-1})g^{-1}=(gh)x(gh)^{-1}=C_{gh}(x)
$$
so that $C_g\circ C_h=C_{gh}$.
Also $C_1$ is the identity map. Thus $C_{g^{-1}}$ is the inverse map of $C_g$.
A: You almost got the surjectivity right, and the idea is exactly the right one.  You are kind of working backwards, which is good, but when writing it down the way you did, it is not really a proof.  The rest looks good.  Let me show you how I would write it down:
Surjectivity means to show that that $\operatorname{im}C_g = G$, or in other words that for any $y \in G$ there exists an $x \in G$ such that $C_g(x) = y$.  Your Ansatz started with a given $y \in \operatorname{im}C_g$, which doesn't help us to show that $\operatorname{im}C_g = G$.
So, to show surjectivity we shall start with some arbitrary $y \in G$.  We then have $ x:= g^{-1}yg \in G$, and we compute $C_g (x) = C_g (g^{-1}yg) = gg^{-1}ygg^{-1} = y$, which shows that $ y \in \operatorname{im}C_g$.  Since $y \in G$ was arbitrary, we have shown that $ \operatorname{im}C_g = G$, in other words we have shown surjectivity.
A: You proof about surjectivity is fine, except for this point: "...Let $C_g(x) = y$ for some $y$....": no, you have to prove that for every $y\in G$ there is some $x\in G$ such that $G_g(x)=y$, which is what you have correctly done.
