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I am trying to classify all groups of order 36. I can first consider abelian ones, and use the Fundamental Theorem of Finite Abelian Groups to obain four of them: $$\mathbb{Z}_{2^2}\oplus \mathbb{Z}_{3^2},\, \mathbb{Z}_{2}\oplus \mathbb{Z}_{2}\oplus \mathbb{Z}_{3^2}, \,\oplus \mathbb{Z}_{2^2}\oplus \mathbb{Z}_{3}\oplus \mathbb{Z}_{3}, \,\oplus \mathbb{Z}_{2}\oplus \mathbb{Z}_{2}\oplus \mathbb{Z}_{3}\oplus \mathbb{Z}_{3}$$

Now I am unsure how to proceed with the non-abelian ones. A few things that I know: $36=2^2\cdot 3^3$, so by Burnside's Theorem all groups of order $36$ are solvable. How that might help I don't know. Another thing, using the Sylow's Theorems, there exists a Sylow 2-subgroups of order 4, and there exists a Sylow 3-Subgroup of order 9. Furthermore, if we denote $n_p$ the number of Sylow $p$-subgroups, then $n_2|9$, and $n_3|4$, and $n_p \equiv 1 \pmod{p}$. Therefore $n_2=1$ or $3$ or $9$, and $n_3=1$ or $4$. How can I proceed?

edit: also I looked at classify groups of order $36$, but I wasn't able to decipher much from the only answer, also the link in the answer is not working so I could not read the source material.

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    $\begingroup$ First, show that either the Sylow 2-subgroup or Sylow 3-subgroup is normal. Then you can do the classification by considering semidirect products: (Sylow 2-subgroup)$\ltimes$(Sylow 3-subgroup) and (Sylow 3-subgroup)$\ltimes$(Sylow 2-subgroup). So now it's a matter of classifying (a) groups of order 4; (b) groups of order 9; (c) the automorphism groups of these groups. $\endgroup$ – Steve D Oct 5 '17 at 17:37
  • $\begingroup$ @SteveD How can one show that either the Sylow 2-group or Sylow 3- subgroup is normal? Is it obvious that $n_3 =4$ implies $n_2 =1$? $\endgroup$ – Bysshed Oct 6 '17 at 16:19
  • $\begingroup$ If there are 4 Sylow 3-subgroups, then the action (by conjugation) of $G$ on this set gives a homomorphism $G\rightarrow S_4$. The kernel $K$ is the intersection of all the normalizers of all the Sylow 3-subgroups, and by order considerations we have $|K|=3$. That should tell you the image is $A_4$. Can you continue? $\endgroup$ – Steve D Oct 6 '17 at 18:18
  • $\begingroup$ $G/K \cong A_4$ is of order 12 and so has a unique Sylow 2-subgroup. So if $P$ and $Q$ are Sylow 2-subgroups of $G$, then $PK/K = QK/K$ and hence $PK = QK$. But $PK=QK$ is of order 12, and so has a unique Sylow 2-subgroup, namely $P$ and $Q$. Therefore $G$ has a unique Sylow 2-subgroup. Thanks @SteveD. $\endgroup$ – Bysshed Oct 6 '17 at 18:54

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