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

I'm afraid the answer by Alexander is wrong. The given subgroup is indeed contained in the center of the semidirect product, but the reverse inclusion can fail.

Simple example: let $G_1$ be a center-free group, with an element $x$ of prime order $p$ (for instance, pick $G_1$ non-abelian of order 6 and $p=2$ or 3). Let the cyclic group $G_2$ of order $p$ act on $G_1$ so that the generator $c$ of order $p$ acts by conjugation by $x$: $\varphi(c^n)(g)=x^ngx^{-n}$. Then $\varphi:G_2\to\mathrm{Aut}(G_1)$ is injective, and hence the subgroup $(Z(G_1)\cap \mathrm{Fix}(\varphi))\times(Z(G_2)\cap\mathrm{Ker}(\varphi))$ is trivial. But the center of the semidirect product $G=G_2\rtimes G_1$ is not trivial: if the law of the latter is given by $(g_2,g_1)(h_2,h_1)=(g_2\varphi(g_1)(h_2),g_1h_1)$, then the element $(x,c^{-1})$ is central (thus the center $Z$ is cyclic of order $p$ and $G$ is direct product of $G_1$ and $Z$).

A correct description of the center of an arbitrary semidirect product should involve the kernel of the map $\varphi':G_2\to\mathrm{Out}(G_1)$, but it looks a bit complicated.

Edit: the center can be described as follows: let $f$ be the canonical map $G_2\to \mathrm{Inn}(G_2)=G_2/Z(G_2)$ and $s$ the canonical map $\mathrm{Ker}(\varphi')\to\mathrm{Inn}(G_2)$. Then the center of $G$ is the set of pairs $(g_2,g_1)\in G_2\rtimes G_1$ such that $g_1\in s^{-1}(f(\mathrm{Fix}(\varphi)))\cap Z(G_1)$, and $f(g_2)=s(g_1)^{-1}$.

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 $h - h^{-1}$.

Answer 3

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.