Prove that if |G|=132 then G cannot be simple

Question

Okay so I have done this but I would like a heads up if it is enough to prove it.

$132=2\cdot 2\cdot 3\cdot11=2^2\cdot3\cdot11$ Let us assume that G is simple. Then from Sylow's theorem, we can say

$$n_2\in\{1,3,11,33\}$$ $$n_3\in \{1,4,22\}$$ $$n_{11}\in\{1,12\}$$

Since we have assumed that $G$ is simple we have $n_{11}=12$, such that $G$ has $12\cdot 10=120$ elements of order $11$.

If $n_3=22$ then $G$ has $120 +(2\times22)$ elements, but that's $164$ and $|G|=132$, hence contradiction.

So then $n_3=4$. There are only $4$ remaining elements, which must comprise a Sylow $2$-subgroup which is unique and thus normal but this is a contradiction.

Hence $|G|=132$ cannot be simple.

You need to be careful being precise. $n_2$ is a single number that must (by the Sylow theorems) be one of those in the given set. It is not equal to the set. — Tobias Kildetoft, Dec 10 '13 at 19:04
it should be the symbol like a backwards E? I just didn't know how to type that here? — ZZS14, Dec 10 '13 at 19:10
Yes, precisely. It is typed as \in (put between dollar-signs, yielding $\in$). — Tobias Kildetoft, Dec 10 '13 at 19:14
See also http://math.stackexchange.com/questions/601737/show-that-g-1000-is-not-simple and http://math.stackexchange.com/questions/13111/no-simple-group-of-order-300. — Dietrich Burde, Dec 10 '13 at 20:04

score 4 · Answer 1 · answered Dec 10 '13 at 19:02

4

That exactly works. Indeed, this is the standard approach.

answered Dec 10 '13 at 19:02

mathematics2x2life

13,578

Prove that if |G|=132 then G cannot be simple

1 Answers1

Linked