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Let $G$ a group and $N$ a normal subgroup of $G$. If $G$ it have a subgroup $H$ s.t. $H \cap N $ is the trivial subgroup and $H$ is isomorphic to $G/ N$ then $G$ is isomorphic to $N\rtimes H$.

Could someone help me to understand the general idea and motivation of this construction sorry but at this moment seems totally bizarre, jaja ?

I have an idea but I'm not sure if we show that the composition of the canonical map from $G\to G/N$ and the isomorphism $G/N$ to $H$ which we call $f$ is the identity under the elements in $H$ and is kernel is $N$, so we can write the elements in $G$ as product on the form $G= NH$ and from here construct the homomorphism from $H\to Aut (N)$ and hope that this give us the isomorphism, my intuition is slightly correct?

I have to go, later on write down the details...

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  • $\begingroup$ Is $G$ a finite group? $\endgroup$ – Quang Hoang Aug 27 '14 at 14:14
  • $\begingroup$ @QuangHoang. No necessarily $\endgroup$ – Jose Antonio Aug 27 '14 at 14:18
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    $\begingroup$ In that case your second sentence is not necessarily true, because $G/N$ could have a proper subgroup isomorphic to itself, and then it is possible that $NH \ne G$. $\endgroup$ – Derek Holt Aug 27 '14 at 16:08
  • $\begingroup$ @DerekHolt I'm not completely sure but I think it suffice to show that $G_{\pi} \to G/N_f\to H$ the composition $g=\pi\circ f$ is the identity for elements in $H$ and $\ker g =N$ and using the fact that $N\cap H=\{1\}$ and this implies that $NH = G$ and from here is trivial the construction of the isomorphism to $N\rtimes H$. $\endgroup$ – Jose Antonio Aug 27 '14 at 20:16
  • $\begingroup$ Yes I agree with that. But, since you are assuming that $f$ is an isomorphism, ${\rm ker} g = {\rm ker \pi} = N$ holds anyway,so yopu don't need that as an extra assumption. $\endgroup$ – Derek Holt Aug 28 '14 at 7:52

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