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.

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## Comments

I think that the argument can be shortened. You can change n_2\ge 11 by n_2\ge 3.

In the second paragraph you have:

“If n_2 \geq 11 and P_2 and Q_2 are distinct Sylow 2-subgroups of G….”

Why can’t you say: If n_2\geq 3 and….

Then you get the same conclusion you had with 11 and that would be the end of the proof?

You’re right. Thanks!

Alternatively, you could argue that if neither are 1 then and one particular option for the remaining n’s is (you can check that this is yields 132 elements). Then, if and , then $G$ acting on the cosets of induces an injective homomorphism from into . However, this is impossible, since .

Is there anything wrong with this proof?

My main hangup is in seeing how any argument that eliminates the cases and does not also rule out .

But also, the permutation representation induced by a subgroup is not necessarily injective. (For example, anytime .)