Timeline for Is there a precise notion of "almost all" such that almost all finite groups are Galois groups of extensions of the rationals?
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| Apr 30, 2014 at 21:58 | comment | added | Yassine Guerboussa | has an orbit in which all the members are $p$-nilpotent, then $|B_d|/|A_d| \rightarrow 1$ when $d \rightarrow \infty$. It is interesting to extend this definition to include all the primes. | |
| Apr 30, 2014 at 21:52 | comment | added | Yassine Guerboussa | For a finite $p$-group $P$, one can consider the class of all finite groups (up to isomorphism) containing $P$ as a Sylow $p$-subgroup; let us call this the orbit of $P$. It is shown by Henn and Priddy that almost all finite groups are $p$-nilpotent in the sense " for almost all $p$-groups $P$ of Frattini class $n$, any group in the orbit of $P$ is $p$-nilpotent". More precisely, If one denotes by $A_d$ the set of $p$-groups of Frattini class $n$that can be generated by $d$ elements, and one denotes by $B_d$ the subclass of $A_d$ having the property that any member $P$ of $B_d$, | |
| Apr 29, 2014 at 20:28 | vote | accept | Sylvain JULIEN | ||
| Apr 29, 2014 at 18:15 | history | answered | Nick Gill | CC BY-SA 3.0 |