Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to prove something regarding solvable groups, but I got stuck near the end with this problem: Let $K \unlhd H \leq G$ and $N \unlhd G$. I'm trying to prove that $NK \unlhd NH$. I've tried to do this through conjugation-invariance: $$(nh)^{-1}n'k(nh) = h^{-1}n^{-1}n'knh$$ but got stuck. I'd really appreciate a clue :)

share|improve this question

1 Answer 1

up vote 0 down vote accepted

Just continue: $$ \begin{align} (nh)^{-1}n'k(nh) = h^{-1}n^{-1}n'knh & = h^{-1}n^{-1}n'h\cdot h^{-1}kh\cdot h^{-1}nh \\ &\in NKN \end{align} $$ and as $N$ is normal, $KN=NK$ (because $kn=(knk^{-1})\cdot k$).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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