0
$\begingroup$

Suppose we have a semidirect product $G = N \rtimes K$, (here $N$ is the normal subgroup). I know that in general, questions about subgroups of $G$ are hard to answer, and that we don't have any nice theorem like Goursat's lemma for direct products.

However I was wondering: what happens if we ask some more specific questions? In particular, I would like to know: what are the subgroups of $G$ which are isomorphic to $K$ and intersect $N$ trivially? Are there any besides the conjugates of $K$? An answer with any single one of those two conditions is very welcome as well. By the way, I am working with finite groups.

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ Yes there are certainly semidirect products in which not all complements of $N$ are conjugate. There are small easy examples when $G$ is a direct product. When $N$ is abelian, the number of conjugacy classes of complements is equal to the order of the first cohomology group $H^1(G,N)$. $\endgroup$ – Derek Holt Oct 14 '17 at 8:11
2
$\begingroup$

Your question is equivalent to say that what are all complements of $N$ in $G$?

Answer of this question is still difficult. But there is a theorem when $(|K|,|N|)=1$.

In that case, it is known that if $H$ is also a complement of $N$ then $H=K^n$ for $n\in N$. This theorem is called as Schur-Zassenhaus theorem.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.