Skip to main content
6 events
when toggle format what by license comment
Nov 4, 2016 at 22:52 vote accept Igor Rivin
Nov 4, 2016 at 20:16 comment added YCor @FriederLadisch: I assumed explicitly that $(p-1)/2$ is prime. So there are exactly two maximal subgroups: the group of squares, and the group $\{\pm 1\}$. So OK I forgot $-1$. Then the probability that a pair of elements in the multiplicative group lies in a maximal subgroup is $1/4+3/(p-1)^2$, i.e. the probability it generates is $3/4-3/(p-1)^2$.
Nov 4, 2016 at 18:00 comment added Frieder Ladisch @YCor. That's probably the easiest explanation why the probability in question can be at most $3/4$. I think the second part of your comment is not correct: the probability of generating $H$ is exactly $3/4$ iff $|H|$ is a power of $2$ (thus $p$ is a Fermat prime). Otherwise, there are other maximal subgroups of $H$ than the group of squares, and we must exclude the possibility that a pair is in these other subgroups.
Nov 4, 2016 at 16:48 comment added YCor The group $H$ of units modulo $p$ is a cyclic group of order $p-1$. So for $p>2$ the group of squares $H^2$ has index 2. So the probability that a pair in $H$ lies in $H^2$ is 1/4. Thus the probability of generating is $\le 3/4$. For those $p$ such that $(p-1)/2$ is prime, the probability of generating $H$ is thus exactly $3/4$, and the probability of generating inside the affine group will tend to 3/4. (This is a particular case of Jan's answer)
Nov 4, 2016 at 12:39 answer added Jan-Christoph Schlage-Puchta timeline score: 5
Nov 4, 2016 at 11:10 history asked Igor Rivin CC BY-SA 3.0