It is well known that if $\gcd (|H|,|K|)=1$ then all subgroups of $H\times K$ are of the form $H^{\prime }\times K^{\prime }$ such that $H^{\prime}$ is a subgroup of $H$ and $K^{\prime}$ is a subgroup of $K$.why it is not true for the subgroups of the semi-direct product $H\rtimes K$.
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4$\begingroup$ Hint: try with $K$ of order 2 in a particular example: what are the subgroups of order 2? $\endgroup$– YCorCommented Oct 25, 2018 at 17:39
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3$\begingroup$ It might be more realistic to expect that each subgroup of the semidirect product might be a semidirect product of subgroups $H_{1},K_{1},$ with $H_{1}$ isomorphic to a subgroup of $H$ and $K_{1}$ isomorphic to a subgroup of $K.$ In fact, in your situation, $H_{1}$ can be taken to be a subgroup of $H,$ and you might consider the possible relationship between $K_{1}$ and $K$ . $\endgroup$– Geoff RobinsonCommented Oct 25, 2018 at 18:08
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3$\begingroup$ The reason it is not true is because there are counterexamples, and in fact the smallest example of a nontrivial semidirect product with factors of coprime orders provides a counterexample. $\endgroup$– Derek HoltCommented Oct 26, 2018 at 20:41
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$\begingroup$ OK thank you very much. But what we can say about this answer math.stackexchange.com/a/1330267/550778. Is it false. $\endgroup$– Nourddine SnanouCommented Oct 28, 2018 at 15:05
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As YCor said, something as easy as $S_3=C_3\rtimes C_2$ provides a counter-example. The most important result dealing with the situation you are interested in is the Schur-Zassenhaus Theorem.
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$\begingroup$ OK thank you very much. But what we can say about this answer math.math.stackexchange.com/questions/1330088/…. Is it false. $\endgroup$ Commented Oct 28, 2018 at 17:39
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$\begingroup$ What we can say is that this answer uses the Schur-Zassenhaus theorem. Note also that a conjugate of the (subgroup of the) acting group is used there. $\endgroup$ Commented May 1, 2019 at 21:06