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In recent days I am studying semidirect product of groups and I have come up with the following question which has already been answered here (From semidirect to direct product of groups), but I can't understand its solution from the comments, can you please clarify me with step by step solutions?

Let $G = N \rtimes_{\varphi} H$. If there exists a homomorphism $f: G \rightarrow N$ which is the identity on $N$, then is it true that $G$ is the direct product of $N$ and $H$?

Priya Sarkar
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  • You should ask two separate questions. Also the first question relates to split exact sequences in groups, look it up – julio_es_sui_glace Mar 13 '24 at 20:18
  • @ julio_es_sui_glace I understand your point, I am sorry, let me remove my second question then, can you tell me the answer of the first question, also if you kindly give some references where I can find the first answer then I will be so helpful. Actually, I have searched a lot about my first question, but I could not find its answer in split exact sequence, can you please provide me the solution? – Priya Sarkar Mar 13 '24 at 20:24
  • @DietrichBurde How do you show that $H \subseteq ker(f)$? – Priya Sarkar Mar 13 '24 at 23:55
  • @PriyaSarkar We can't because it is not necessarily true. Your concerns are justified. – Derek Holt Mar 14 '24 at 08:18
  • Since the statement referred to in the linked post is wrong, or at least unclear in its meaning, this question should not be closed. – Derek Holt Mar 14 '24 at 09:36
  • @Dietrich Burde. Rather than downvoting the post, people should read and understand the post properly, in stead of jumping around some unnecessary comments and trusting them to be true without even looking at it. – Priya Sarkar Mar 14 '24 at 14:03

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Since the question concerns equality rather than isomorphism, it seems to me that, with the standard convention for identifying subgroups of $N \rtimes_\phi H$ with $N$ and $H$, the answer to the question is no.

Let $H=N$ be any nonabelian group, and let $\phi$ be the action of $H$ on $N$ by conjugation.

Then there is homomorphism from $G = N \rtimes_\phi H$ to $N$ which is the identity on $N$ and has kernel $\{(h,h^{-1}) : h \in H \}$, but $G$ is not the direct product of its subgroups $\{(h,1): h \in H \}$ and $\{(1,n): n \in N\}$, which are the subgroups that are customarily identified with $H$ and $N$ in the semidirect product.

Of course we do have $G \cong N \times H$, but that was not the question.

Derek Holt
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