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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?

Shaun
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Bhaskar Vashishth
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  • I don't think you are missing anything :-) I'm not sure what 'complete' means, but if it includes trivial center then I have no objections. Maybe you are thinking that the isomorphism $G\rtimes_\phi G \cong G\times G$ is given by the identity map $(a,b)\mapsto (a,b)$ but that is not the case. – Myself Sep 01 '14 at 17:37
  • complete means trivial centre and trivial outer automorphism group. all i am saying is if $G \rtimes_\phi G \cong G\times G$ then shouldn't $\phi$ be trivial homomorphism and not identity homomorphism as we take it for holomorph definition. – Bhaskar Vashishth Sep 01 '14 at 17:40
  • No because you are concluding too much from the $\cong$ sign. What it actually means is that there exists a map $\alpha : G\rtimes_\phi G\to G\times G$ such that ..., what you seem to think is that the specific map $G\rtimes_\phi G\to G\times G: (a,b)\mapsto (a,b)$ is an isomorphism. And that is not always the case. – Myself Sep 01 '14 at 17:46
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    Here is a hint: In the group $H = G \times_\phi A$, where $A = {\rm Aut}(G) \cong G$, we have $C_H(G) \cong G$ and $H = G \times C_H(G)$. – Derek Holt Sep 01 '14 at 19:11
  • I think as it is $G \rtimes_\phi G$ so definitely a copy of $G$ is normal in $G \rtimes_\phi G$ so both copies of $G$ are normal and thus it is direct product, as $K \rtimes_\phi H \cong K\times H$ iff $H_0$ & $K_0$ both are normal in $G$ where $H_0$ & $K_0$ denotes obvious copies in $K \rtimes_\phi H$.

    am i right here. i must be. after studying split extension whole night.:P

     – Bhaskar Vashishth Sep 02 '14 at 01:07
  • I think you are right, but I think you should avoid writing $G \rtimes_\phi G$, because it makes it unclear what you talking about when you refer to the subgroup $G$. This group has at least three subgroups isomorphic to $G$. I might write out a full solution later. – Derek Holt Sep 02 '14 at 07:50

1 Answers1

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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.

Shaun
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Derek Holt
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