What is the center of a semidirect product?

Question

Let $G_1$ and $G_2$ be groups. Let $\varphi:G_2\rightarrow \operatorname{Aut}(G_1) $ be a group homomorphism defining the semidirect product $G_1 \rtimes G_2$. Determine the center $\operatorname{Z}(G_1 \rtimes G_2)$.

Answer 1

Denote by $\varphi_g$ the automorphism of $G_1$ associated with a $g\in G_2$. Take $z\in \operatorname{Z}(G_1\rtimes G_2)$ and write $z=xy$ for $x\in G_1$ and $y\in G_2$. Then, for every $g\in G_2$, we have $$gxy=\varphi_g(x)gy=xyg.$$ By equating the $G_1$ and $G_2$ parts in $gxy=\varphi_g(x)gy$, we see that $x=\varphi_g(x)$. Similarly, we equate the $G_1$ and $G_2$ parts in $gxy=xyg$ to obtain $gy=yg$. Since this is true for all $g\in G_2$, we may conclude that $x\in \operatorname{Fix}(\varphi)$ and $y\in \operatorname{Z}(G_2)$. With this in mind, we revisit our earlier computation: $$gxy=yxg=\varphi_y(x)yg=xyg$$ Thus we see that $x=\varphi_y(x)$, so $\varphi_y$ is the identity automorphism, and we have that $y\in \operatorname{Ker}(\varphi)$. Finally, taking $g\in G_1$, we have that $gxy=xyg$, from which we see that $x\in Z(G_1)$.

Putting it all together, we have $\operatorname{Z}(G_1\rtimes G_2) \subseteq \left(\operatorname{Z}(G_1)\cap\operatorname{Fix}(\varphi)\right)\rtimes \left(\operatorname{Ker}(\varphi)\cap \operatorname{Z}(G_2)\right),$ which is simply $\left(\operatorname{Z}(G_1)\cap\operatorname{Fix}(\varphi)\right)\times \left(\operatorname{Ker}(\varphi)\cap \operatorname{Z}(G_2)\right)$. The reverse inclusion is not true in general, as can be seen in YCor's answer below. So, all we can really say at this point is that $$\left(\operatorname{Z}(G_1)\cap\operatorname{Fix}(\varphi)\right)\times \left(\operatorname{Ker}(\varphi)\cap \operatorname{Z}(G_2)\right)\subseteq \operatorname{Z}(G_1\rtimes G_2)$$ I would be curious to see what additional description could bring this to an equality.

Answer 2

The general description of the centre of $G_2 \rtimes G_1$ in the last paragraph of @YCor's answer is not correct; there is one condition missing, namely that $g_2 \in \mathrm{Fix}(\varphi)$. (I spotted this while applying the description to the ribbon braid group $\mathbb{Z}^n \rtimes B_n$. The description as written would claim that its centre is $\mathbb{Z}^n \rtimes Z(B_n)$, whereas its centre is actually just $\Delta_{\mathbb{Z}} \times Z(B_n)$ where $\Delta_{\mathbb{Z}}$ denotes the diagonal in $\mathbb{Z}^n$.)

After adding this condition, the description simplifies slightly, so a correct description of the centre $Z(G_2 \rtimes G_1)$ of the semi-direct product $G_2 \rtimes G_1$ with respect to the action $\varphi \colon G_1 \to \mathrm{Aut}(G_2)$ is the set of pairs $(g_2,g_1)$ such that $g_1 \in Z(G_1)$, $g_2 \in \mathrm{Fix}(\varphi)$ and $\varphi(g_1) = \mathrm{Inn}(g_2)^{-1}$, where $\mathrm{Inn} \colon G_2 \to \mathrm{Inn}(G_2)$ sends an element $h$ to the inner automorphism $h - h^{-1}$.