Show that no group of order 48 is simple
I was wondering if I was allowed to do something along this line of thinking:
Let $n_2$ be the number of $2$-Sylow groups.
$n_2$ is limited to $1$ and $3$ since these are the only divisors of 48 that are equivalent to $1 \mod 2$.
$n_2=3$ (since if $n_2=1$ the group is definitely not simple)
Each $n_2$ subgroup contains 1 distinct element and there are 3 of these subgroups hence there are 3 distinct elements.
We have $48-3=45$ elements to account for.
At this point, can I assume that these 45 elements form a subgroup and then solve this proof by proving a group of 45 elements form a p-Sylow normal subgroup?