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When we consider the classification of the group G by semidirect product, we need to consider all the homomorphisms from K to Aut(H), Where G=HK and H$\unlhd$G,H$\bigcap$K=1

But by the theorem: Suppose G is a group with subgroups H and K such that, (1) H$\unlhd$G (2) H$\bigcap$K=1. Let $\Phi$:K$\to$Aut(H) be the homomorphism defined by mapping k$\in$K to the automorphism of left conjugation by k on H. Then HK$\cong$H$\rtimes_{\Phi}$K.

we can get if G=HK and H$\unlhd$G,H$\bigcap$K=1, then G$\cong$H$\rtimes_{\Phi}$K with the $\Phi$ showed above.Thus it seems if G=HK and H$\unlhd$G,H$\bigcap$K=1, then G is defined only by H and K. In this sense, it seems when we consider the classification of the group G by semidirect product, we don't need to consider other homomorphisms as they are isomorphic to G and G$\cong$H$\rtimes_{\Phi}$K with the $\Phi$ showed above.

Let me show my analysis in an example, in the groups of order 12. Let V$\in Syl_2(G)$ and T$\in Syl_3(G)$. By the Lagrange theorem V$\bigcap$T=1. And we consider the case that V$\cong Z_2\times Z_2$, T$\cong Z_3$ and V$\unlhd$G. By semidirect product we can get G=$A_4$ or $ Z_2\times Z_2\times Z_3$. Now we apply the theorem, here $A_4$=VT and V$\unlhd$$A_4$,V$\bigcap$T=1, similarly we have $ Z_2\times Z_2\times Z_3$=VT and V$\unlhd$$ Z_2\times Z_2\times Z_3$, V$\bigcap$T=1. Thus apply the theorem to both $ Z_2\times Z_2\times Z_3$ and $A_4$ , we can get $ Z_2\times Z_2\times Z_3$$\cong$V$\rtimes_{\Phi}$T(with the $\Phi$ showed above), and $A_4$$\cong$V$\rtimes_{\Phi}$T, therefore we can get $ Z_2\times Z_2\times Z_3$ is isomorphic to $A_4$.A contradiction.

I know my analysis is wrong, but could you tell me where I made the mistake? Thank you!

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  • $\begingroup$ I am not sure if I understand exactly what you asking. For any specific group $G$ of this type there is a specific $\Phi:K \to {\rm Aut}(H)$ for which $G \cong H \rtimes_{\Phi} K$. But if you change $\Phi$ then you get a different group $G$. One of the difficulties in carrying out this classification is that different homomorphisms $\Phi$ sometimes give rise to isomorphic groups $G$, so you have to determine when this happens. $\endgroup$ – Derek Holt Aug 2 '15 at 9:35
  • $\begingroup$ I think the symbol $H\rtimes K$ is misleading - it should be $H\rtimes_{\Phi} K$, as the group so constructed depends on $\Phi$ $\endgroup$ – Prahlad Vaidyanathan Aug 2 '15 at 9:40
  • $\begingroup$ @DerekHolt, let me be more specific, to the groups with order n, in order to make the classification we need to find the possible normal subgroup H and subgroup K with |HK|=n, H$\bigcap$=1 and try to find out all the homomorphisms from K to Aut(H), then by a triple (H,K,$\phi$), we can define a group G with order n. However, by the theorem above, it seems if we have normal subgroup H and subgroup K with HK=G, then G is determined by H$\rtimes_{\Phi}$K with Φ:K→Aut(H) be the homomorphism defined by mapping k∈K to the automorphism of left conjugation k on H. it seems all the triples are same $\endgroup$ – 6666 Aug 2 '15 at 9:51
  • $\begingroup$ In the case of $G=HK$ with $H$ normal, $K$ acts on $H$ by conjugation giving you the "missing" homomorphism $\Phi$. $\endgroup$ – j.p. Aug 2 '15 at 10:01
  • $\begingroup$ @Joseph I am still unsure what you are asking. What you write all seems correct, so what is the problem? As I said, any give $G$ of this form detemrines $\Phi$, but different $\Phi$ may result in nonisomorphic groups $G$. $\endgroup$ – Derek Holt Aug 2 '15 at 10:35

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