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Jun 28, 2013 at 18:17 comment added YCor For some reason I didn't notice the comments earlier. You're right, I misread $S_i$ normal in $G$ instead of normal in $S_{i+1}$. Anyway as noticed by user35603, subnormal is enough and I edited accordingly.
Jun 28, 2013 at 18:15 history edited YCor CC BY-SA 3.0
little correction to fit with the hypotheses
Jun 26, 2013 at 11:24 vote accept user35603
Jun 26, 2013 at 11:24 comment added user35603 Thanks to all. I think you are right.
Jun 26, 2013 at 6:29 comment added j.p. @user35603: As there are only finitely many possibilities for $M_n$ ($\le S_i$), at least one of them shows up infinitely times often.
Jun 25, 2013 at 22:57 comment added user35603 But why $M$ does not depend on $n$?
Jun 25, 2013 at 22:40 comment added user35603 @Neil Strickland. Yes, $S_i$ is normal olny in $S_{i+1}$. In other words, $S_i$ is subnormal in $S_j$ for $j>i$. But it's O.K., the Lemma is valid in that case too. And I proved something similar too. But...
Jun 25, 2013 at 21:49 comment added Neil Strickland You seem to be assuming that $S_i\triangleleft S_j$ for all $j>i$, which does not automatically follow from the case $j=i+1$ as in the question. However, I suspect that the OP meant to assume the stronger version.
Jun 25, 2013 at 18:37 history answered YCor CC BY-SA 3.0