# Whether $Z(G) \subseteq G_1$ or not?

kindly tell me if I am correct or not.

Let $G_1$ and $G_2$ be two groups such that $|G_2|=p$ (a prime). Let $\varphi:G_2 \rightarrow Aut(G_1)$ be a group homomorphism defining the semidirect product $G=G_2 \ltimes_{\varphi} G_1$. Note that the center $Z(G_2 \ltimes_{\varphi} G_1)=(Z(G_1) \cap Fix(\varphi)) \times (Z(G_2) \cap Ker (\varphi))$. If $\varphi$ is not trivial (that is the semidirect product is not a direct product), then $Ker(\varphi)$ is trivial. This implies that $Z(G) \subseteq G_1$.

• For the sentence starting "note that" I get the containment $\supset$ but the other containment is not obvious. – hunter Dec 6 '13 at 17:36
• nevermind, both containments are clear. i'm sold. – hunter Dec 6 '13 at 17:38
• (to get $\supset$ i just check that what you write down is central. to get $\subset$ i consider special elements whose $G_2$, resp. $G_1$ part is $1$. It may help your reader to say something like this, but I agree that it's more or less clear.) – hunter Dec 6 '13 at 17:39
• @hunter For the proof see math.stackexchange.com/questions/243327/… – Vipul Kakkar Dec 6 '13 at 17:44
• The equality describing $Z(G)$ is not correct. Indeed, assume that $Z(G_2)$ is trivial, that the action is faithful but by inner automorphisms. Then $Z(G)$ is cyclic of order $p$ and sort of diagonally embedded in the semidirect product. – YCor Dec 7 '13 at 11:17