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Jun 28, 2016 at 22:25 comment added yakov Obviously, $G$ is supersolvable iff all indices of a chief series of $G$ containing $F(G)$ and lying below $F(G)$ are primes (i.e., $F(G)$ is supersolvable immersed in $G$, in terminology of R. Baer),
Apr 6, 2016 at 3:22 comment added majid arezoomand Yes, but you focused on my question.However, I changed the acceptance of answer again. Many thanks again.
Apr 6, 2016 at 3:11 comment added Geoff Robinson Stefan Kohl gave a much more comprehensive answer than I did, so I do not think you should have changed the acceptance of his answer.
Apr 6, 2016 at 3:11 comment added majid arezoomand Dear Geoff, Thank you for your answer.
Apr 6, 2016 at 3:09 vote accept majid arezoomand
Apr 6, 2016 at 3:19
Apr 5, 2016 at 20:22 history answered Geoff Robinson CC BY-SA 3.0