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Let $\varphi_1:K\rightarrow\text{Aut}(H)$ and $\varphi_2:K\rightarrow\text{Aut}(H)$ be two homomorphisms. Suppose that the semidirect products constructed by these homomorphisms are isomorphic by an isomorphism $\varphi$ such that $\varphi(H)=H$. Does $\ker\varphi_1\cong\ker\varphi_2$?. The result is true if $H$ is a non trivial p-group and $K$ is a non trivial q-group, where $p$ and $q$ are distinct prime numbers. However, this is not true in general, so are there additional conditions under which the statement is true?.

Thank you very much.

نورالدين سنانو
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    Can you include why you think this doesn't hold in general? I think I have a proof for all semidirect products, but I'm worried I missed something. – Jacob Manaker Jul 15 '21 at 01:39
  • @ Jacob Manaker. Thank you for your comment. It is very common for $\ker\varphi_1=\ker\varphi_2=1$ amongst non-isomorphic semi-direct products. But you can put your proof for discussion. – نورالدين سنانو Jul 15 '21 at 02:11
  • Writing it out in detail, I did miss something; sorry. – Jacob Manaker Jul 15 '21 at 02:53
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    Captain Lama argues that if the kernels of homomorphisms $\varphi_i:K\rightarrow\operatorname{Aut}(H)$, i=1,2, are isomorphic, then the semidirect products $H\rtimes_{\varphi_i}H$ are not necessarily isomorphic. @نورالدين سنانو, how does this statement relate to your problem? Thank you. – kabenyuk Jul 15 '21 at 04:59
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    @kabenyuk. I think you are right. So what do you think about the following example.Let $H$ be nonabelian of order $6$. Let $\varphi_1(h)=id_{H} $ and $\varphi_2(h)(g) = hgh^{-1}$. Then $H\rtimes _{\varphi_1 }H\cong H\rtimes _{\varphi_2}H$ but $\ker(\varphi_1) = H$ and $\ker(\varphi_2) = 1$. – نورالدين سنانو Jul 15 '21 at 21:57
  • You are right. @Derek Holt gave a similar example in the first comment to the question https://math.stackexchange.com/questions/3453460/requested-hint-for-if-ker-varphi-1-not-cong-ker-varphi-2-then-h-rtim?rq=1 – kabenyuk Jul 16 '21 at 06:06

1 Answers1

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We have $\ker \varphi_1 \cong \ker \varphi_2$ whenever $|K|$ and $|H/Z(H)|$ are coprime.

To see that, note that $|H/Z(H)| = |{\rm Inn}(H)|$ so, if this condition is met, then $\varphi_1(K) \cap {\rm Inn}(H) = 1$ and no element of $K$ can induce a nontrivial inner automorphism of $H$.

Then $C_{H \rtimes_{\phi_1} K}(H) = H \rtimes \ker \varphi_1$ and similarly $C_{H \rtimes_{\phi_2} K}(H) = H \rtimes \ker \varphi_2$, so $$H \rtimes_{\phi_1} K \cong H \rtimes_{\phi_2} K \Rightarrow \ker \varphi_1 \cong \ker \varphi_2.$$

Derek Holt
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