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 semidirect product $H\rtimes K$.
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closed as offtopic by YCor, Neil Hoffman, Derek Holt, Jeremy Rickard, მამუკა ჯიბლაძე Oct 28 '18 at 5:12
This question appears to be offtopic. The users who voted to close gave these specific reasons:
 "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – YCor, Neil Hoffman, მამუკა ჯიბლაძე
 "This question does not appear to be about research level mathematics within the scope defined in the help center." – Derek Holt, Jeremy Rickard

3$\begingroup$ Hint: try with $K$ of order 2 in a particular example: what are the subgroups of order 2? $\endgroup$ – YCor Oct 25 '18 at 17:39

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 Robinson Oct 25 '18 at 18:08

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 Holt Oct 26 '18 at 20:41

$\begingroup$ OK thank you very much. But what we can say about this answer math.stackexchange.com/a/1330267/550778. Is it false. $\endgroup$ – Nourddine Snanou Oct 28 '18 at 15:05
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As YCor said, something as easy as $S_3=C_3\rtimes C_2$ provides a counterexample. The most important result dealing with the situation you are interested in is the SchurZassenhaus Theorem.

$\begingroup$ OK thank you very much. But what we can say about this answer math.math.stackexchange.com/questions/1330088/…. Is it false. $\endgroup$ – Nourddine Snanou Oct 28 '18 at 17:39