Prove that if then is not simple.
Note that . Suppose is simple.
The divisibility and congruence criteria of Sylow’s Theorem force the following.
Since is simple, we have , so that has elements of order 11. If , then has at least elements, a contradiction. So . There are only 4 remaining elements, which must comprise a Sylow 2-subgroup, which is unique and thus normal- a contradiction.
Thus no group of order 132 is simple.