# classify groups of order $36$

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

1. $\mathbb{Z}_4\times \mathbb{Z}_9$
2. $\mathbb{Z}_4\times \mathbb{Z}_3\times \mathbb{Z}_3$
3. $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_9$
4. $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3\times \mathbb{Z}_3\cong \mathbb{Z}_6\times\mathbb{Z}_6$

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.

• Well, all $14$ different groups of order $36$ are solvable by Burnsides $p^aq^b$-theorem. We have $4$ abelian groups and $4$ nilpotent non-abelian groups. – Dietrich Burde Mar 25 '16 at 19:36
• "all $14$ groups"?? I am not aware of them.. I know abelian groups.. I will write down. @DietrichBurde – user312648 Mar 25 '16 at 19:37
• Every such group has either a normal $2$-Sylow or $3$-Sylow subgroup. – Dietrich Burde Mar 25 '16 at 19:38
• Your argument that $4$ Sylow $3$-subgroup implies a unique Sylow $2$-subgroup does not work (although the conclusion is true). You are assuming that the $4$ Sylow $3$-subgroups are disjoint, but they might not be. – Derek Holt Mar 25 '16 at 19:47
• $A_4 \times C_3$. – Derek Holt Mar 25 '16 at 20:50

There are $$4$$ abelian groups of order $$36$$ and $$10$$ non-abelian solvable groups of order $$36$$. We can do the classification according to the following case distinction. Let $$G$$ be a group of order $$36$$.
1.) There are only $$2$$ groups of order $$36$$ having no normal subgroup of order $$9$$, namely $$C_3\times A_4$$ and $$C_9\ltimes (C_2\times C_2)$$.
2.) There are $$12$$ groups of order $$36$$ having a normal subgroup of order $$9$$, namely the $$4$$ abelian groups, and $$8$$ others, e.g., a nontrivial extension of the dihedral group $$D_9$$ by $$C_2$$, $$C_9\ltimes (C_2\times C_2)$$, $$(C_3\times C_3)\ltimes C_4$$, $$(C_3\times C_3)\ltimes (C_2\times C_2)$$, and $$D_{18}$$.