On Cayley representation of groups

Question

Suppose $G$ is a group. Let’s define Cayley representation of $G$ as the homomorphism $\operatorname{Cay}$ from $G$ to $\operatorname{Sym}(G)$ that maps every element $a$ to the permutation $\phi_a: g \mapsto ag$.

Is it always true, that $\frac{N_{\operatorname{Sym}(G)}(\operatorname{Cay}(G))}{C_{\operatorname{Sym}(G)}(\operatorname{Cay}(G))} \cong \operatorname{Aut}(G)$? Here $N$ stands for normalizer and $C$ for centralizer.

For all finite groups of order up to $4$ this seems to be true, but what for other groups?

Answer 1

One way to approach this question is apply the isomorphism theorems, by looking for a surjective homomorphism $N_{\operatorname{Sym}(G)}(\operatorname{Cay}(G)) \mapsto \operatorname{Aut}(G)$ whose kernel is $C_{\operatorname{Sym}(G)}(\operatorname{Cay}(G))$.

Perhaps, by just following one's nose, one can start with an element $f \in N_{\operatorname{Sym}(G)}(\operatorname{Cay}(G))$ and use it to dream up an element of $\operatorname{Aut}(G)$.

Let's try it! Let $f : G \to G$ be a bijection having the property that for each $a \in G$ there exists $b \in G$ such that $f \circ \phi_a \circ f^{-1} = \phi_b$. This value of $b$ is well-defined, depending only on $f$ and $a$, because $\phi_b = \phi_{b'} \implies b=\phi_b(\text{Id}) = \phi_{b'}(\text{Id}) = b'$. We therefore get a function $\widehat f : G \to G$ depending only on $f$, by setting $\widehat f(a)=b$ whenever the equation $f \circ \phi_a \circ f^{-1} = \phi_b$ holds.

So could this function $\widehat f : G \to G$ be the one?

• Could this function $\widehat f : G \to G$ be a bijection?
• And could it also be a homomorphism?
• If so, could the function $N_{\operatorname{Sym}(G)}(\operatorname{Cay}(G)) \mapsto \operatorname{Aut}(G)$ defined by $f \mapsto \widehat f$ be a homomorphism?
• And if that is so, could the kernel of this homomorphism be equal to $C_{\operatorname{Sym}(G)}(\operatorname{Cay}(G))$?

I'll leave it to you to continue to follow your nose on these questions, it's kind of fun.

But there's still one last question:

• If all of the above are true, could the function $N_{\operatorname{Sym}(G)}(\operatorname{Cay}(G)) \mapsto \operatorname{Aut}(G)$ defined by $f \mapsto \widehat f$ be surjective?

To approach this last question requires you to go back into dreaming mode. Given an automorphism $\alpha : G \to G$, you need to construct some appropriate element $f \in \operatorname{Sym}(G)$, and then to prove a few things about that element: $f \in N_{\operatorname{Sym}(G)}(\operatorname{Cay}(G))$; and $\widehat f = \alpha$.

So I'll leave you with this: Can you take an arbitrary automorphism $\alpha : G \to G$, and use $\alpha$ to dream up some bijection of $G$ to try as the candidate for $f$?