How do I compute the normaliser of a group G, considered as a set, in the group of set bijections of G?

Question

Suppose G is a group and T(G) is the group of set bijections of G. I identify the elements of G as maps corresponding to the left multiplication by the chosen element. Then the normaliser of G in T(G) is the semi direct product of Aut(G) and G. How do I show this?

Answer 1

By definition, $\phi \in T(G)$ normalizes $G$ iff for every $g \in G$, there exists $\alpha(g) \in G$ such that $$ \phi \circ l_g \ \circ \phi^{-1} = l_{\alpha(g)}.$$ (Here $l_g$ denotes left multiplication by $g$.) This is equivalent to: For every $g,h \in G$, $$\phi(g \phi^{-1}(h)) = \alpha(g) h.$$ From here you should be able to work out the isomorphism to $Aut(G) \rtimes G$.