Prove that if G 132 then G is not simple Prove that if G 1

Solution

we have to prove lGl = 132

132 = 2*2*3*11 = 2^{2 *}3*11

Let us assume that Gis simple. Then by Syow\'s theorem we can say

n₂ = {1,3,11,33}

n₃ = {1, 4 ,22)

n₁₁={1,12}

Now since we have assumed that, G is simpe, so that G has 12*10 = 120 elements of the order 11.

If n₃ = 22 then G has 120 + (2*22) = 164 elements. But given that lGl = 132. Hence contradiction.

Then consider n_{3 =} 4, There are only 4 remaining elements which must comprise a sylow 2-subgroup which is unique & thus normal. But we have consider G as simple. Hence again contradiction.

Hence it is clear that lGl = 132 can not be simple.