Normalizer of G viewed as a subgroup of S_G If we take $G$ to be a group, we can view it a subgroup of $S_G$ by Cayley's theorem (we're considering the left action).
Then the centralizer of $G$ in $S_G$ is all bijections $f$ such that $f(xy) = xf(y)$ and it's not hard to see that this is in fact the right translations. In other words $C(G) = G$.
But how does one go about computing the normalizer N(G)? I keep reading that it should be $G \rtimes Aut(G)$ but I don't see where that's coming from. How does one end up with a semidirect product?
Any insights would be useful. Thanks!
 A: Consider the map from $N(G)$ to $Aut(G)$ that assigns to every element the automorphism that it induces. The kernel of this map is those normalizing elements that do not induce any nontrivial automorphism, i.e. $C(G)$.
As it is easy to describe $Aut(G)$ also as a subgroup of $N(G)$, we get a semidirect product structure $C(g)\rtimes Aut(G)$.
As you noted, we have that $C(G)$ is isomorphic to $G$, but it is not equal to $G$, that is the $G$ in this semi direct product is different from the actual $G$. 
A: First, to establish $C(G)\cong G$, let $\sigma\in C(G)$. Then $\sigma(1)=g$ for some $g\in G$, and 
$$\sigma(x)=\sigma\lambda_x(1)=\lambda_x\sigma(1)=\lambda_x(g)=xg$$
where (as indicated above $\lambda_x\in S_G$ is left multiplication by $x$, which is the image of $G$ in $S_G$). This implies that $\sigma=\rho_g$ is right multiplication by $g$ and $C(G)\cong G$.
To see that $N(G)\cong G\rtimes\mathrm{Aut}(G)$, we note that both $G$ and $\mathrm{Aut}(G)$ are subgroups of $N(G)$, with $G$ normal. Since $G\cap\mathrm{Aut}(G)=\{1\}$ (left multiplication by $x$ is not an automorphism unless $x=1$), we have immediately that $$G\rtimes\mathrm{Aut}(G)\cong G\mathrm{Aut}(G)\leq N(G).$$
We will now prove equality. For convenience we will write $x$ for $\lambda_x$. Let $\sigma\in N(G)$ and $x\in G$. Then, $\sigma x={\phi_\sigma(x)}\sigma$ for some $\phi_\sigma\in S_G$. In fact, $\phi_\sigma\in \mathrm{Aut}(G)$ since
\begin{align*}
{\phi_\sigma(xy)}\sigma=\sigma{xy}={\phi_{\sigma}(x)}\sigma y=\phi_\sigma(x){\phi_{\sigma}(y)}\sigma.
\end{align*}
Now, define $\Phi:N(G)\to \mathrm{Aut}(G)$ by $\Phi(\sigma)=\phi_\sigma$. Then $\Phi$ is clearly a homomorphism since
$$
\phi_{\sigma\tau}(x)(\sigma\tau)=(\sigma\tau)x=\sigma\phi_{\tau}(x)\tau
=\phi_\sigma(\phi_\tau(x))\sigma\tau.
$$
Moreover, for $\sigma\in\mathrm{Aut}(G)$, $\phi_\sigma=\sigma$, so the map is surjective. The kernel of this map is $C(G)$, so $$|N(G)|=|C(G)||\mathrm{Aut}(G)|=|G||\mathrm{Aut}(G)|=|G\rtimes\mathrm{Aut}(G)|.$$
Finally, identifying $N(G)=G\rtimes\mathrm{Aut}(G)$, we have that 
$$\ker{\Phi}=\{(g,\iota_{g^{-1}})\mid g\in G\}$$
where $\iota_x$ is the inner automorphism defined by $x$. This follows from the fact that $\rho_g=\lambda_g\iota_{g^{-1}}$ which is the explicit identification of $G$ with $C(G)$ in $S_G$.
