Question is to classify all groups of order $36$
I do not even know if it is of my level. Let me try this.
Sylow theorem says that there are sylow $2$ subgroups of order $4$ and sylow $3$ subgroups of order $9$.
Possible number of sylow $2$ subgroups are $1,3,9$
Possible number of sylow $3$ subgroups are $1,4$
Suppose $G$ has $4$ sylow $3$ subgroups, each subgroup has $8$ non identity elements, total $32$ non identity elements from $4$ sylow subgroups. Remaining $4$ elements would become one sylow $2$ subgroups.
This says that if $G$ has $4$ sylow $3$ subgroups then $G$ has $1$ sylow $2$ subgroup and thus, this sylow $2$ subgroup is normal and so $G$ is not simple.
Suppose $G$ has $1$ sylow $3$ subgroup then it is a normal subgroup and so $G$ is simple. In this case we can not decide on the number of sylow $2$ subgroups.. It can be any one of $1,3,9$..
I can classify abelian groups
- $\mathbb{Z}_4\times \mathbb{Z}_9$
- $\mathbb{Z}_4\times \mathbb{Z}_3\times \mathbb{Z}_3$
- $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_9$
- $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3\times \mathbb{Z}_3\cong \mathbb{Z}_6\times\mathbb{Z}_6$
One more information is.. If there are $4$ sylow $3$subgroups then there is exactly one sylow $2$ subgroup so normal.. If there is only one sylow $3$ subgroups it is normal... So, any such group has either a normal sylow $2$ subgroup or a sylow $3$ subgroup..
Could not go beyond this.