Here : https://groupprops.subwiki.org/wiki/A5_is_the_unique_simple_non-abelian_group_of_smallest_order
it is shown that the only simple subgroup of order $\ 60\ $ upto isomorphy is $A_5$
I do not understand this step :
The number of 2-Sylow subgroups is either 5 or 15: By fact (2) (the congruence and divisibility conditions on Sylow numbers), we have n_2 = 1,3,5,15. By fact (4), n_2 cannot be 1 or 3. Thus, n_2 = 5 or n_2 = 15.
How can we conclude with fact 4 (If $2^p-1$ is a Mersenne prime, then a group with order $(2^p-1)\cdot 2^p$ has a normal subgroup) that $n_2=1$ or $n_2=3$ is impossible ?
The only order for which fact 4 can play a role is $12$.
$n_2 = 1$ is of course impossible, since we are looking for simple groups.
For the rest, I guess that it should read "fact (5.4)" instead of "fact (4)", i.e., that $60$ divides $n_2!/2$. Then, obviously, $n_2 = 3$ is not possible.