I have a small question regarding the semidirect product. Consider a group $G$ which is the semidirect product $\mathbb{Z}_3 \ltimes (\mathbb{Z}_5 \times \mathbb{Z}_5)$ (internal semidirect product). Let $\phi\colon\mathbb{Z}_3\to\operatorname{Aut}(\mathbb{Z}_5 \times \mathbb{Z}_5)$. There will be $\phi_0 ,\phi_1 ,\phi_2$ corresponding to $\bar{0},\bar{1},\bar{2}$ of $\mathbb{Z}_3$, right?

  • $\begingroup$ Can you compute the automorphism group of $\Bbb Z_5 \times \Bbb Z_5$? A fact you will need once you have done that is that the image of a generator of $\Bbb Z_3$ must have order dividing the order of the codomain. $\endgroup$ – Lukas Kofler Oct 11 '18 at 9:45
  • $\begingroup$ I don't understand. Do you mean that this semidirect product can not exist? $\endgroup$ – Buddhini Angelika Oct 11 '18 at 9:49

Since Aut$(\mathbb{Z}_5)=C_4$, Aut($\mathbb{Z}_5 \times \mathbb{Z}_5$) $\cong C_4\times C_4$. The only action $\mathbb{Z}_3$ can have on $\mathbb{Z}_5 \times \mathbb{Z}_5$ is the trivial one. Thus you get a direct product in the end.

  • $\begingroup$ Thanks a lot @Jianping but when I compute automorphisms of order 3 in the automorphism group of $\mathbb{Z}_5 \times \mathbb{Z}_5$, in GAP i get several automorphisms... $\endgroup$ – Buddhini Angelika Apr 17 at 16:11

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