Classify all groups of order 66, up to isomorphism 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? 
 A: 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:


*

*The group $P$ has no automorphisms of order three because $Aut(C_{11})\simeq \Bbb{Z}_{11}^*$ is cyclic of order ten.

*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 $P$.

*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 $Q\unlhd G$.

*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.


*

*If $\epsilon_1=\epsilon_2=+1$ the group is abelian and hence cyclic.

*If $\epsilon_1=\epsilon_2=-1$ we get the dihedral group $D_{33}$, the symmetries of a regular $33$-gon.

*If $\epsilon_1=+1,\epsilon_2=-1$ we get $C_{11}\times D_3$.

*If $\epsilon_1=-1,\epsilon_2=+1$ we get $D_{11}\times C_3$.

