a formula for the center of the semidirect product of two groups for a given homomorphism is given. I also wonder whether the formula is correct or not.

The definition of the semidirect product shows that $(n_1,h_1)$ and $(n_2,h_2)$ commute if and only if $h_1h_2=h_2h_1$ and $n_1\phi_{h_1}(n_2)=n_2\phi_{h_2}(n_1)$, where $\phi:H\rightarrow Aut(N)$ is a homomorphism.

The first condition is easy ($h_1$ and $h_2$ commute), but I have no idea how to use the second equation to check whether the formula from the forum is correct or not.

Wikipedia mentions the special homomorphism sending $h$ to $\phi(h)=\phi_h$ with $\phi_h(n)=hnh^{-1}$. Can the formula for the center of the semidirect product be simplified in this case ?

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    $\begingroup$ The formula is not correct in general. For example ler $G = S_3 \rtimes C_2$ where a generator of $C_2$ induces an inner automorphism of $S_3$. Then $G \cong S_3 \times C_2$ so $|Z(G)|=2$, but the formula gives the trivial subgroup. $\endgroup$ – Derek Holt Apr 13 '16 at 13:25

The rule for the action of $H$ on $N$ when $\phi(H)$ is a subgroup of ${\rm Aut}(N)$, and we form the semidirect product $NH$ via $\phi$ is that conjugation by $h \in H $ in the semidirect product induces the automorphism $\phi(h)$ on $N$.

Set $G = NH$. Then $H \cap Z(G) \leq Z(H)$, and for each $h \in H \cap Z(G)$, $\phi(h)$ must be the trivial automorphism of $N$, since conjugation of $N$ by $h$ is the trivial automorphism of $N$. Conversely, these conditions together imply that $h \in Z(G)$.

Hence it is true that $H \cap Z(G) = Z(H) \cap {\rm ker} \phi$.

Now if $n \in N \cap Z(G)$, then $n \in Z(N)$, and $\phi(h)$ fixes $n$ for each $h \in H$, since conjugation by any such $h$ fixes $n$. Conversely, any $n \in Z(N)$ fixed by all $\phi(h)$'s is in $Z(G)$.

Thus $N \cap Z(G) = Z(N) \cap {\rm Fix} (\phi(H))$, where ${\rm Fix}(A)$ denotes the set of all fixed points of a group of automorphisms $A$.

This shows that $(Z(N) \cap {\rm Fix}(\phi(H))) \times (Z(H) \cap {\rm ker} \phi) \subseteq Z(G)$.

I leave it to you to check whether the reverse inclusion holds ( following Derek Holt's comment). It remains to check when an element of the form $nh$ with $n \in N$ and $h \in H$ both non-identity can be in $Z(G)$.

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    $\begingroup$ But the reverse inclusion is not always true! When elements of $H$ induce inner automorphisms of $N$, you can get diagonal type elements of $Z(G)$. $\endgroup$ – Derek Holt Apr 13 '16 at 13:21
  • $\begingroup$ @DerekHolt : OK, thanks, I should have worded it differently. I left it to Peter to check the "diagonal" elements, and had not worked out the details myself. $\endgroup$ – Geoff Robinson Apr 13 '16 at 13:38

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