Subgroup of the semidirect product of two subgroups with coprime orders [closed]

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 $$H\rtimes K$$.

closed as off-topic by YCor, Neil Hoffman, Derek Holt, Jeremy Rickard, მამუკა ჯიბლაძეOct 28 '18 at 5:12

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• Hint: try with $K$ of order 2 in a particular example: what are the subgroups of order 2? – YCor Oct 25 '18 at 17:39
• 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 $K$ . – Geoff Robinson Oct 25 '18 at 18:08
• 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. – Derek Holt Oct 26 '18 at 20:41
• OK thank you very much. But what we can say about this answer math.stackexchange.com/a/1330267/550778. Is it false. – Nourddine Snanou Oct 28 '18 at 15:05

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.