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Let A, B and H be groups with $\theta_{1}$ an injective homomorphism from H to Aut(A) while $\theta_{2}$ is an homomorphism from H to Aut(B).

We know there is a subgroup of Aut(A $\times$ B) which is isomorphic to Aut(A) $\times$ Aut(B) consisting of those automorphisms of A $\times$ B of the form f(ab) = g(a)h(b) where g is an automorphism on A and h an automorphism on B.

Let $\theta$: H $\rightarrow$ Aut(A $\times$ B) be defined as $\theta$(h) = $\theta_{1}$(h) $\theta_{2}$(h).

Assume A $\times_{\theta_{1}}$ H is complete. Does anyone have lead on the necessary and sufficient conditions for the group (A $\times$ B) $\times_{\theta}$ H to also be complete?

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    $\begingroup$ You start by saying that $H$ maps to $\mathrm{Aut}(N)$; but I do not see any definition of $N$ anywhere... Is that meant to be $\mathrm{Aut}(A)$? $\endgroup$ Mar 10, 2014 at 19:29

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Complete=(centerless and all automorphisms are inner). So it's certainly necessary that

The action of $H$ on the center of $B$, $Z(B)$, has no nontrivial fixed points,

as any such fixed point would lie in the center of $(A \times B) \rtimes_{\theta} H$.

Regarding the condition about automorphisms, it's a bit harder to say. For example, if we knew that $A \times B$ was characteristic in $(A \times B) \rtimes_{\theta} H$, then the structure of the automorphism group has been worked out fairly explicitly (for a recent version see, e.g., Curran, 2008). In this case, the automorphism group is isomorphic to the group of matrices

$$\left(\begin{array}{cc} \alpha & \delta \\ 0 & \eta \end{array}\right)$$

such that $\alpha \in Aut(A \times B)$, $\eta \in Aut(H)$, and $\delta\colon H \to A \times B$ is a derivation relative to $\eta$, that is, for all $h,h' \in H$, $\delta(hh') = \delta(h) \delta(h')^{\eta(h)}$. Because $H$ preserves the direct product structure of $A \times B$ by assumption, if we write $\delta(h) = (\delta_A(h), \delta_B(h)) \in A \times B$ (with the obvious notation), then we find that $\delta_A \colon H \to A$ must be a derivation with respect to $\eta$ and $\delta_B \colon H \to B$ must be as well.

Now, if we make the further assumption that $A,B$ are centerless and share no common direct factors, then $Aut(A \times B) = Aut(A) \times Aut(B)$, and we may write the above matrix as

$$\left(\begin{array}{cc} \alpha_A \alpha_B & \delta_A \delta_B \\ 0 & \eta \end{array}\right) $$

then we find that $\left(\begin{array}{cc} \alpha_A & \delta_A \\ 0 & \eta \end{array}\right) \in Aut(A \rtimes H)$ and $\left(\begin{array}{cc} \alpha_B & \delta_B \\ 0 & \eta \end{array}\right) \in Aut(B \rtimes H)$. So in this case, certainly a necessary condition is that

$B \rtimes H$ itself is complete.

I believe that this condition and the one above are also sufficient in this case

Now, if we drop the conditions that imply that $Aut(A \times B) = Aut(A) \times Aut(B)$ the situation becomes a bit more complicated, but you can probably work out the necessary and sufficient condition. (It is not hard, using the above notation, to work out what inner automorphisms look like.)

However, if you drop the condition that $A \times B$ is characteristic then it becomes quite a bit more complicated, and I think you may get stuck quickly trying to find necessary and sufficient conditions.

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