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I'm reading Abstract Algebra of Dummit and I got stuck on this point. When we try to build the semi-direct product of 2 group predefined, such as $H$ and $K$, then all the semi-direct product groups is determined by the homomorphism from $K$ to $\operatorname{Aut}(H)$. The problem is to determine which couple of result groups are isomorphic to each other.

For example, we have 2 homomorphisms $\varphi_{1}, \varphi_{2}$. In some problems they reason that if $\operatorname{Ker}(\varphi_{1}) \cong \operatorname{Ker}(\varphi_{2})$, then 2 semidirect product groups of $H$ and $K$ defined by $\varphi_{1}$ and $\varphi_{2}$ are isomorphic, too. Can any one please explain for me whether this is true in general? If not, in which case this is true, and how to prove that?

One more thing, if two kernels are not isomorphic, how can we prove that two semidirect product groups are not isomorphic?

Thanks so much.

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    $\begingroup$ In what problems do they reason that? Can you give some link or quote? $\endgroup$ – DonAntonio Jun 15 '13 at 11:01
  • $\begingroup$ Certainly you'll need much stronger hypotheses than $\ker(\phi_1) \cong \ker(\phi_2)$. It is very common for $\ker(\phi_1) = \ker(\phi_2) = 1$ amongst non-isomorphic semi-direct products. $\endgroup$ – Jack Schmidt Jun 15 '13 at 14:34
  • $\begingroup$ Yes, for example , consider semidirect product of 2 groups: $T = Z_{2}$ and $V = Z_{2} \times Z_{2}$, they consider that there're 3 homomorphism from $V$ to $Aut(T)$ by specifying their kernels as one of three subgroups of order 2 in $V$, and since 3 kernels are isomorphic, then 3 semidirect product groups are isomorphic $\endgroup$ – le duc quang Jun 16 '13 at 5:34
  • $\begingroup$ I've already updated the question. If 2 kernels are not isomorphic, how can we prove that 2 semidirect product groups are not isomorphic, thanks. $\endgroup$ – le duc quang Jun 16 '13 at 10:44
  • $\begingroup$ Perhaps you misread “kernel” versus “image”? $\endgroup$ – Jack Schmidt Jun 16 '13 at 11:58

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