i have to prove group of order $36$ is not simple by using Sylow's theorems. please read whole solution as i have difficulties and i indicate them by " ? " now as $|G|= 36 = 3^2 . 2^2$ and $3^2 | 36$ then by Sylow I theorem $G$ must have Sylow $3$- subgroups of order $3^2 = 9$. now let number of Sylow $3$- subgroup be $n_3$ then according to Sylow 2nd and 3rd theorem $n_3 = 1 + 3k$ and $n_3 | O(G)$ this imply $1+3k | 36$ which is hold if $k= 0$ or $k =1$. if $k= 0$ then $n_3 = 1$ and hence we get unique Sylow $3$-subgroup of order " ? " and hence it is normal and hence $G$ can not be simple and hence we are done.
But if $k= 1$ then $n_3 = 4$ hence we get $4$ Sylow $3$- subgroups each of order $9$ and hence there are $8$ elements of order $9$, in each of these sylow-$3$-subgroups, is it true? hence there $4 . 8 = 32$ elrments of order $9$ is it true? and as $|G| = 36$ and hence remaining $4$ elements forms unique sylow $2$-subgroups of order $4$ is it true? i have confusion because number of sylow $2$- subgroup i.e. $n_2 = 1+ 2k$ must divides $|G|$ i.e $1+ 2k | 36$ it holds if $k = 0$ or $k= 1$ or $k=4$ and hence we get $n_2 = 1$ or $n_2 = 3$ or $n_2 = 9$ hence i have doubt .