4
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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. $\endgroup$ – Dietrich Burde Mar 25 '16 at 19:36
  • $\begingroup$ "all $14$ groups"?? I am not aware of them.. I know abelian groups.. I will write down. @DietrichBurde $\endgroup$ – user312648 Mar 25 '16 at 19:37
  • $\begingroup$ Every such group has either a normal $2$-Sylow or $3$-Sylow subgroup. $\endgroup$ – Dietrich Burde Mar 25 '16 at 19:38
  • 1
    $\begingroup$ 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. $\endgroup$ – Derek Holt Mar 25 '16 at 19:47
  • 1
    $\begingroup$ $A_4 \times C_3$. $\endgroup$ – Derek Holt Mar 25 '16 at 20:50
2
$\begingroup$

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 $D_{18}$.

Reference: These Notes.

$\endgroup$
1
  • $\begingroup$ Out of these groups does any group have Sylow subgroup Z/4 ? Except the trivial cases of direct product ? $\endgroup$ – Alexander Chervov Aug 5 '17 at 11:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy