-2
$\begingroup$

Let $H$ and $K$ be a finite groups and $G'$ be a normal subgroup of the internal semidirect product $H.\,K$. Take $\,H'=\,H \cap \,G'$ and $K'= (G' \,H) \cap \,K$. We can see easely that $\mid G' \mid =\mid H'\mid \mid K'\mid$.

  1. Is it true that $G'=H'.\,K'$.
  2. In the case that the answer is "no": Are there any reasonable constraints to $H$ and $K$ such that the answer is "yes"?

Any help would be appreciated so much. Thank you all.

$\endgroup$

closed as off-topic by Chris Godsil, YCor, Derek Holt, Gabriel C. Drummond-Cole, Pace Nielsen Feb 11 at 17:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Chris Godsil, YCor, Derek Holt, Gabriel C. Drummond-Cole, Pace Nielsen
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ In general $K'$ is not a subgroup of $G'$. For example $G'$ could be a complement of $H$ that is disjoint from $K$. That can happen even with a direct product. $\endgroup$ – Derek Holt Feb 11 at 10:26