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 $\operatorname{Z}(G_1 \rtimes G_2)$.
Abstract Algebra – What Is the Center of a Semidirect Product?
abstract-algebragroup-theorysemidirect-product
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I find it helpful to distinguish between the inner semidirect product and the outer semidirect product. (Similiar to the distinction between the inner direct sum and the outer direct sum of vector (sub)spaces.)
Given a group $G$ and two subgroups $H$ and $N$ of $G$, the group $G$ is called the inner semidirect product of $H$ and $N$ with $N$ normal, if we have $NH = G$ and $H \cap N = 1$, and $N$ is normal in $G$. We then write $$ G = N \rtimes H \,. $$ (This notion of an inner semidirect product does not depend on any homomorphism.)
Given on the other hand any two groups $N$ and $H$ and a group homomorphism $\theta \colon H \to \operatorname{Aut}(N)$, the corresponding outer semidirect product $N \rtimes_\theta H$ is defined as the set $N \times H$ together with the multiplication $$ (n_1, h_1) \cdot (n_2, h_2) := (n_1 \theta(h_1)(n_2), h_1 h_2) \,. $$ Note that this group $N \rtimes_\theta H$ very much depends on $\theta$.
Now what is the connection between these two?
If $G$ is a group and $H$ and $N$ are two subgroups of $G$ such that $G = N \rtimes H$, then $N$ is normal in $G$. The conjugation action of $H$ on $G$ does therefore restrict to an action of $H$ on $N$. From this we get a group homomorphism $$ \theta \colon H \to \mathrm{Aut}(N) \quad\text{given by}\quad \theta(h)(n) = hnh^{-1} \,. $$ It can then be checked that the map $$ N \rtimes_\theta H \to G = N \rtimes H, \quad (n,h) \mapsto n \cdot h $$ is a group isomorphism. So every inner semidirect product can be seen as an outer semidirect product via the conjugation action of $H$ on $N$.
If on the other hand $N$ and $H$ are two groups and $\theta \colon H \to \operatorname{Aut}(N)$ a group homomorphism, then we can regard both $N$ and $H$ as subgroups $\widehat{N}$ and $\widehat{H}$ of the outer semidirect product $N \rtimes_\theta H$ via the inclusions \begin{alignat*}{2} N &\to N \rtimes_\theta H \,, &\quad n &\mapsto (n,1) \,, \\ H &\to N \rtimes_\theta H \,, &\quad h &\mapsto (1,h) \,. \end{alignat*} It then follows for those subgroup $\widehat{N}$ and $\widehat{H}$ that both $\widehat{N} \cdot \widehat{H} = \widehat{N} \rtimes_\theta \widehat{H}$ and $\widehat{N} \cap \widehat{H} = 1$, and that the subgroup $\widehat{N}$ is normal in $N \rtimes_\theta H$. The group $N \rtimes_\theta H$ is therefore the inner semidirect product of its subgroups $\widehat{N}$ and $\widehat{H}$, with $\widehat{N}$ normal.
PS: So given any two groups $H$ and $N$ there may exist multiple outer semidirect products $N \rtimes_\theta H$ because we can choose different group homomorphisms $\theta \colon H \to \operatorname{Aut}(N)$. But if we are given an inner semidirect product $G = N \rtimes H$ (i.e. we are given the group $G$ and its subgroups $H$ and $N$ such that $G = N \rtimes H$) then we are also implicitely given a group homomorphism $\theta \colon H \to \operatorname{Aut}(N)$ with $G \cong N \rtimes_\theta H$: The homomorphism $\theta$ is hidden in the group structure of $G$ and can be retrieved via the conjugation action of $H$ on $N$.
Forget about the actual construction of the semidirect product for now. I argue that the semidirect product is important because it arises naturally and beautifully in many areas of mathematics. I will list below many examples, and I urge you to find a few that interest you and look at them in detail.
Before doing that let me just give the following extra motivation: say you have a group $G$, and you find two subgroups $H,K$ such that every element of $G$ can be uniquely written as a product $hk$ for $h \in H$, $k \in K$. In other words, you have a set-theoretic bijection between $G$ and $H \times K$. Then certainly you'd want to understand $G$ by studying its smaller components $H$ and $K$. One way to achieve this would be to find a suitable group structure on $H \times K$ intertwining the structures of $H$ and $K$ such that the above bijection becomes a group isomorphism. This can be done; however doing things in this generality becomes rapidly tedious. If instead we restrict our attention to such decompositions with $H$ normal in $G$, the problem becomes much more manageable. In this case we have what we call a split exact sequence $$1 \to H \to G \to K \to 1,$$ and $G$ is called a semidirect product of $H$ and $K$. An existence and uniqueness theorem gives us all the possible semidirect products one can obtain from $H$ and $K$ through the group of homomorphisms from $K$ to $\operatorname{Aut}H$. Note that $K = \mathbb{Z}/2$ appears often in practice, because this guarantees normality of $H$. Now here are some examples:
The symmetric group $S_n = A_n \rtimes \mathbb{Z}/2$. The exact sequence is $$1 \to A_n \to S_n \xrightarrow{\mathit{sign}} \mathbb{Z}/2 \to 1.$$
The dihedral group $D_n = \mathbb{Z}/n \rtimes \mathbb{Z}/2$. The exact sequence is $$1 \to \mathbb{Z}/n \to D_n \xrightarrow{\mathit{det}} \mathbb{Z}/2 \to 1.$$
The infinite dihedral group $D_\infty = \mathbb{Z} \rtimes \mathbb{Z}/2$. The exact sequence depends on your explicit construction. You may take $$1 \to \mathbb{Z} \to \mathbb{Z}/2 * \mathbb{Z}/2 \to \mathbb{Z}/2 \to 1$$ or $$1 \to \mathbb{Z} \to A(1,\mathbb{Z}) \to \mathbb{Z}/2 \to 1,$$ where $A(1,\mathbb{Z})$ is the group of affine transformations of the form $x \mapsto ax + b$, where $a \in \{ \pm 1 \} \cong \mathbb{Z}/2$ and $b \in \mathbb{Z}$.
Many matrix groups, thanks to the determinant map. For example, $G = \operatorname{GL}(n,\mathbb{F})$, $O(n,\mathbb{F})$ and $U(n)$ have respective subgroups $H = \operatorname{SL}(n,\mathbb{F}),\operatorname{SO}(n,\mathbb{F}),\operatorname{SU}(n)$ and $K = \mathbb{F}^\times,\mathbb{Z}/2,U(1)$.
The fundamental group of the Klein bottle is $G = \langle x,y \mid xyx = y \rangle$. This is just the nontrivial semidirect product of $\mathbb{Z}$ with itself. Interestingly, the other (trivial) semidirect product $\mathbb{Z}^2$ is the fundamental group of the other closed surface of Euler characteristic $0$, namely the torus.
The affine group $A(n,\mathbb{F}) = \mathbb{F}^n \rtimes \operatorname{GL}(n,\mathbb{F})$. Its elements are transformations $\mathbb{F}^n \to \mathbb{F}^n$ of the form $x \mapsto Ax + b$, with $A$ an invertible matrix and $b$ a translation vector. The exact sequence is $$1 \to \mathbb{F}^n \to A(n,\mathbb{F}) \xrightarrow{f} \operatorname{GL}(n,\mathbb{F}) \to 1$$ where $f$ forgets the affine structure (the translation part).
The hyperoctahedral group $O(n,\mathbb{Z})$ is the group of signed permutation matrices. We have two decompositions $O(n,\mathbb{Z}) \cong \operatorname{SO}(n,\mathbb{Z}) \rtimes \mathbb{Z}/2$ and $O(n,\mathbb{Z}) \cong (\mathbb{Z}/2)^n \rtimes S_n$. In the corresponding exact sequences the surjective map is respectively the determinant homomorphism and the "forget all the signs" homomorphism.
Best Answer
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 $f(g_2)=s(g_1)^{-1}$.