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how to classify all groups of order 66, up to isomorphism?

Firstly, we have 66=$3\times 11\times 2$, suppose the number of the Sylow-11 group of this group is $n_k$, since 11 divides $n_k-1$ we can know that $n_k$=1 since $n_k$ must divide 6. If so, we can know any group with order 66 can only have one Sylow-11 subgroup,but now I am not sure how to continue analyze this question. Can someone tell me how to determine all the groups of order 66?

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  • $\begingroup$ According to Maple, there are four of them: with(GroupTheory); for n from 1 to NumGroups(66) do SmallGroup(66, n, form = fpgroup); end; $\endgroup$ Commented Mar 28, 2017 at 1:14
  • $\begingroup$ You can also ask WolframAlpha. One of the groups is cyclic, another is dihedral. $\endgroup$ Commented Mar 28, 2017 at 1:16
  • $\begingroup$ You can also ask NrSmallGroups(66); in GAP. $\endgroup$ Commented Mar 28, 2017 at 11:18

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You correctly deduced that such a group $G$ always has a normal cyclic subgroup $P$ of order $11$. Let us fix a generator $x\in P$. From that point on:

  1. The group $P$ has no automorphisms of order three because $Aut(C_{11})\simeq \Bbb{Z}_{11}^*$ is cyclic of order ten.
  2. So if $y\in G$ is of order three then the conjugation action of $y$ on $P$ must be trivial. In other words, $y$ centralizes $P$.
  3. The subgroup $N$ generated by $P$ and $y$ is therefore cyclic of order $33$. As $[G:N]=2$, $N\unlhd G$. As $Q=\langle y\rangle$ is a characteristic subgroup of $N$, we also have $Q\unlhd G$.
  4. Let $z\in G$ be an element of order two. It normalizes both $P$ and $Q$. The only automorphism of order two of either of those groups is taking the inverse. Therefore $zxz^{-1}=x^{\epsilon_1}$ and $zyz^{-1}=y^{\epsilon_2}$ with $\epsilon_1,\epsilon_2\in\{+1,-1\}$.

The choices for the two epsilons leave four possible combinations. All of them occur.

  1. If $\epsilon_1=\epsilon_2=+1$ the group is abelian and hence cyclic.
  2. If $\epsilon_1=\epsilon_2=-1$ we get the dihedral group $D_{33}$, the symmetries of a regular $33$-gon.
  3. If $\epsilon_1=+1,\epsilon_2=-1$ we get $C_{11}\times D_3$.
  4. If $\epsilon_1=-1,\epsilon_2=+1$ we get $D_{11}\times C_3$.
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