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Let $G$ and $H$ be two groups and $\phi: H \rightarrow Aut(G)$ be a group homomorphism . Then we know $$Inn(G\times H)=Inn(G)\times Inn(H).$$ What about $Inn(G \rtimes H)$?

I think (not sure) we have $Inn(G)\times Inn(H)\subseteq Inn(G \rtimes H)$. Do we can determine $Inn(G \rtimes H)$?

Thank you

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You are right about doubting the inclusion. As $\mathop{Inn}(G) = G/\mathop{Z}(G)$ you will have to determine (I change notation!) the center of $G = N\rtimes H$, which is generated by the central elements of $N$ that are centralized by $H$ and by the central elements of $H$ that centralize $N$. An example where the inclusion becomes nonsense is $N = Z_3^2$ and $H = Q_8$ with the faithful action of the quaternion group. –  j.p. Jan 13 '13 at 11:30
See also Question 243327 about the centre of semidirect products. –  Derek Holt Jan 13 '13 at 12:46
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1 Answer 1

up vote 2 down vote accepted

As I established here, letting $\varphi:H\rightarrow \text{Aut}(G)$ define $G\rtimes H$, we have $$Z(G\rtimes H) = \left(Z(G)\cap\text{fix}(\varphi)\right)\times \left(\text{ker}(\varphi)\cap Z(H)\right).$$ so $$\text{Inn}(G\rtimes H)\cong \frac{G\rtimes H}{ \left(Z(G)\cap\mbox{fix}(\varphi)\right)\times \left(\text{ker}(\varphi)\cap Z(H)\right)}.$$

In general this is not isomorphic to $\text{Inn}(G)\times\text{Inn}(H)\cong \left(G/Z(G)\right)\times \left(H/Z(H)\right)$, as you can see that $Z(G)\cap \text{fix}(\varphi)$ may be proper in $Z(G)$, and likewise for $Z(H)\cap \text{ker}(\varphi)$.

In particular, we are modding out by the parts which are not only central, but unaffected by $\varphi$, so we expect any nontrivial action of $\varphi$ to be preserved in the quotient. This suggests we may want to look at $\text{Inn}(G)\rtimes \text{Inn}(H)$, where the semidirect product is defined by inducing $\varphi$ up to the inner automorphism groups. Let's see how these are related.

Let $\text{Inn}(G)\rtimes \text{Inn}(H)$ be defined by $\overline{\varphi}:\text{Inn}(H)\rightarrow \text{Aut}(\text{Inn}(G))$ such that $$\overline{\varphi}(\theta_h)=\Theta_{\theta_h}\hspace{18pt}\text{where}\hspace{18pt}\Theta_{\theta_h}:\theta_g\mapsto\theta_{\varphi(h)(g)}.$$ Let's take a look at $\chi:\text{Inn}(G\rtimes H)\rightarrow \text{Inn}(G)\rtimes \text{Inn}(H)$. Write $\theta\in \text{Inn}(G\rtimes H)$ by $\theta_{gh}$, where $g\in G$ and $h\in H$; similarly for $\theta\in \text{Inn}(G)\rtimes \text{Inn}(H)$ we write $\theta_g \theta_h$, where $\theta_g\in \text{Inn}(G)$ and $\theta_h\in \text{Inn}(H)$. Then we can define $\chi(\theta_{gh})=\theta_{g}\theta_{h}$. Now take arbitrary $\theta_{g_1h_1},\theta_{g_2h_2}\in \text{Inn}(G\rtimes H)$.

$$\begin{eqnarray*} \chi(\theta_{g_1h_1})\chi(\theta_{g_2h_2})&=&\theta_{g_1}\theta_{h_1}\theta_{g_2}\theta_{h_2}\\ &=&\theta_{g_1}\overline{\varphi}(\theta_{h_1})(\theta_{g_2})\theta_{h_1}\theta_{h_2}\\ &=&\theta_{g_1}\theta_{\varphi(h_1)(g_2)}\theta_{h_1}\theta_{h_2}\\ &=&\theta_{g_1\varphi(h_1)(g_2)}\theta_{h_1h_2}\\ &=&\chi(\theta_{g_1\varphi(h_1)(g_2)h_1h_2})\\ &=&\chi(\theta_{g_1h_1g_2h_2})\\ &=&\chi(\theta_{g_1h_1}\theta_{g_2h_2}) \end{eqnarray*}$$

Now what about $\text{ker}(\chi)$? We have that $\chi(\theta_{gh})=\theta_{g}\theta_{h}=\text{id}_{\text{Inn}(G)\rtimes \text{Inn}(H)}$ if and only if $g\in Z(G)$ and $h \in Z(H)$. On the other hand $\theta_{gh}=\text{id}_{\text{Inn}(G\rtimes H)}$ if and only if $$gh\in Z(G\rtimes H)=\left(Z(G)\cap\text{fix}(\varphi)\right)\times \left(\text{ker}(\varphi)\cap Z(H)\right).$$ So in fact there are nontrivial $\theta_{gh}$ in the kernel of $\chi$ whenever $Z(G)\cap \text{fix}(\varphi)$ is proper in $Z(G)$, and likewise whenever $Z(H)\cap \text{fix}(\varphi)$ is proper in $Z(H)$.

Clearly $\chi$ is surjective, so indeed we have that $\text{Inn}(G\rtimes H)$ can be bigger than $\text{Inn}(G)\rtimes \text{Inn}(H)$ - in other words we can gain inner automorphisms through taking a semidirect product! These stem from the nontrivial action of the restriction of $\varphi$ to $Z(G)\rtimes Z(H)$ (which is the kernel of $G\rtimes H \rightarrow \text{Inn}(G)\rtimes \text{Inn}(H)$).

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@ Alexander Gruber :Do we can say $fix(\varphi)$ is the set $\{g\in G\mid \alpha(g)=g \forall \alpha\in Aut(G)\}$? –  maryam Jan 14 '13 at 11:00
$\text{fix}(\varphi)=\{g\in G|\varphi(h)(g)=g \forall h\in H\}.$ –  Alexander Gruber Jan 14 '13 at 17:27
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