If $G$ is complete, then the holomorph of $G$ is isomorphic to $G\times G$.

Question- If $G$ is complete, then the holomorph of $G$ is isomorphic to $G\times G$.

I am studying semidirect products for the first time, and in some notes I found this exercise. As far as I know about this problem, if $G$ is a group then let $H={\rm Aut}(G)$ and let $\phi:H \to{\rm Aut}(G) $ be the identity map, i.e. all elements go to itself, then Holomorph of $G$ is $G \rtimes_\phi{\rm Aut}(G)$. Now if $G$ is complete then its outer automorphism group is identity, then $G \cong{\rm Aut}(G)$ so Holomorph is $G \rtimes_\phi G$ but $G \rtimes_\phi G \cong G\times G$ when $\phi$ is the trivial homomorphism i.e. everything goes to $1$.

So what am I missing here?

Let $G$ be a group with $Z(G)=1$ and $I = {\rm Inn}(G)$. So $I = \{c_g : g \in G\}$ with $c_g: h \mapsto ghg^{-1}$, and the map $g \mapsto c_g$ is an isomorphism $G \to I$.

Let $S = G \rtimes I$ be the semidirect product. So, for $g_1,g_2,h_1,h_2 \in G$, we have

$$(g_1,c_{h_1})(g_2,c_{h_2}) = (g_1c_{h_1}(g_2),c_{h_1h_1}) = (g_1h_1g_2h_1^{-1},c_{h_1h_1}).$$

As usual, we identify $G$ with the subgroup $\{ (g,1) : g \in g \}$ of $S$.

Let $C = \{ (g^{-1},c_g) : g \in G \} \le S$. Then, using the above multiplication rule, it is routine to check that $C \le C_S(G)$, and the map $g \mapsto (g^{-1},c_g)$ is an isomorphicm $G \to C$.

We also see easily that $S = GC$ and (since $Z(G)=1$) $G \cap C = 1$, so $S = G \times C \cong G \times G$. In fact $Z(G)=1$ implies that $C = C_S(G)$.

More generally, if $Z(G) \ne 1$ and $I = {\rm Inn}(G)$, then $G \rtimes I$ is a central product of two copies of $G$ with their centres amalgamated.