11 reputation
2
bio website
location Salerno, Italy
age 25
visits member for 2 years, 7 months
seen Jul 21 '12 at 10:01

May
7
awarded  Scholar
May
7
accepted Bounding the order of $\mathrm{Aut}(G)$
May
5
answered Bounding the order of $\mathrm{Aut}(G)$
May
3
comment Bounding the order of $\mathrm{Aut}(G)$
m<=log2(n) not m<log2(n)... so I'm going out of my mind!However ⌊log2(n−1)⌋ it's Right.
May
2
comment Bounding the order of $\mathrm{Aut}(G)$
Ok, but at the end I find that |Aut(G)| is less than (n-1)...(n-2^m) where m is less than log2(n)! Do you think that there is some problem in the proposition??? anyway thank you!
May
2
comment Bounding the order of $\mathrm{Aut}(G)$
of course... Sorry!
May
2
comment Bounding the order of $\mathrm{Aut}(G)$
"less" stands for <=
May
2
awarded  Student
May
2
asked Bounding the order of $\mathrm{Aut}(G)$