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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jun 2, 2016 at 20:42 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Dec 19, 2015 at 15:36 answer added Yakov timeline score: -1
Dec 12, 2014 at 15:58 comment added Derek Holt I am really not sure about that - I would not like to bet either way on what happens for general finite solvable groups, for example.
Dec 12, 2014 at 14:45 comment added Pablo @DerekHolt this is great! Do you know what happens if we allow $G$ to be any finite group and not just a $p$-group?
Dec 12, 2014 at 13:40 comment added Derek Holt I think this is true with $a_k=1/k$ i.e. $|L| \le |H|$ in this situation. It can be proved by induction on $|G|$. Choose a normal subgroup $N$ of $G$ with $|N|=p$, and apply induction to $G/N$ distinguishing between the cases $N \le H$ and $N \cap H=1$ (note that $L \cap N=1$ in the second case).
S Dec 12, 2014 at 12:11 history suggested Sam Jones CC BY-SA 3.0
Formatted link correctly.
Dec 12, 2014 at 12:07 review Suggested edits
S Dec 12, 2014 at 12:11
Dec 12, 2014 at 10:37 history asked Pablo CC BY-SA 3.0