Order 132

Let \(G\) be a group with order 132. We show that \(G\) is not simple.
Note that \(|G| = 2^2\cdot3\cdot11\).
Let us assume that \(G\) is simple and arrive at a contradiction. By simplicity, we know that \(n_{11} > 1, n_3 > 1,\) and \(n_2 > 1.\)
By Sylow Theorem (3), it is forced that \(n_{11} = 12.\) Also, \(n_3 \ge 4\) and \(n_2 \ge 3.\)

Note the following:

1. Intersection of any two Sylow-\(11\) subgroups is trivial.
2. Intersection of any two Sylow-\(3\) subgroups is trivial.
3. Intersection of any Sylow-\(p\) subgroup with any Sylow-\(q\) subgroup is trivial.

The above facts follow by considering the fact that the intersection would be a subgroup of the two bigger subgroups and would have to divide their orders.

Now, if we consider the union of all the Sylow-\(11\) and Sylow-\(3\) subgroups, it contains at least \(12(11-1) + 4(3-1) + 1 = 129\) elements. Thus, the remaining elements are at most \(3.\) However, we do need at least \(3\) more elements to form a Sylow-\(2\) subgroup.
These \(3\) elements, along with the identity must form the unique Sylow-\(2\) subgroup of order \(4.\)
However, we have reached a contradiction as we get that \(n_2 = 1.\)
Thus, we are done!