Question

Classify all groups of order 66 up to isomorphism.

Short answer

There are 4 groups of order $66,{\mathrm{\mathbb{Z}}}_{66},D_{33},D_{11}\times {\mathrm{\mathbb{Z}}}_3$and $D_3\times {\mathrm{\mathbb{Z}}}_{11}$.

Step-by-step solution

Referring to Corollary 9.16, and Theorem 9.17 (Third Sylow Theorem)

Corollary 9.16

Let G be a finite group and K be a Sylow p-subgroup for some prime p. Then, K is normal in G if and only if K is the only Sylow p-subgroup.

Theorem 9.17 (Third Sylow Theorem)

The number of Sylow p-subgroups of finite group G divides $\left|G\right|$ and is of the form 1+pk for some non-negative integer k.

Classifying all groups of order 66 up to isomorphism

Let’s consider any group of order 66.

Then, the order of G can be written as:

$\begin{array}{rcl}n& =& 66\\ & =& 2.3·11\end{array}$

Therefore, according to Theorem 9.17 and Corollary 9.16, any group of order 66 has a normal Sylow 11-subgroup.

Therefore, .

Let $b\in G$be any element of order 3.

Because it is normal, it must have $b^{-1}ab=a^k$for some $k\in \mathrm{\mathbb{Z}}$.

Therefore, the subgroup must be abelian of order 66, and since it is abelian, it must be cyclic.

This implies, it must be isomorphic to ${\mathrm{\mathbb{Z}}}_{66}$.

Now, suppose $c\in G$to be an element of order 2 in the group.

So, the only isomorphism of order 2 of two subgroups is possible by considering their inverse.

By applying this, we get 4 distinct combinations representing group G which are written as:

Hence, there are 4 groups of order $66,{\mathrm{\mathbb{Z}}}_{66},D_{33},D_{11}\times {\mathrm{\mathbb{Z}}}_3$and $D_3\times {\mathrm{\mathbb{Z}}}_{11}$.