## No simple group of order 132 exists

Prove that if $|G| = 132$ then $G$ is not simple.

Note that $132 = 2^2 \cdot 3 \cdot 11$. Suppose $G$ is simple.

The divisibility and congruence criteria of Sylow’s Theorem force the following.

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

Since $G$ is simple, we have $n_{11} = 12$, so that $G$ has $12 \cdot 10 = 120$ elements of order 11. If $n_3 = 22$, then $G$ has at least $120 + 22 \cdot 2 = 164$ elements, a contradiction. So $n_3 = 4$. 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.

• Martin  On October 15, 2011 at 10:49 pm

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

• martin  On October 16, 2011 at 11:29 pm

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?

• nbloomf  On October 19, 2011 at 9:51 am

You’re right. Thanks!

• Steven  On December 6, 2011 at 10:16 pm

Alternatively, you could argue that if neither $n_{11}, n_{3}, n_{2}$ are 1 then $n_{11} = 12$ and one particular option for the remaining n’s is $n_{3} =4, n_{2} =3$ (you can check that this is yields 132 elements). Then, if $P_{2} \in Syl_{2}(G)$ and $N = N_{G}(P_2)$, then $G$ acting on the cosets of $N$ induces an injective homomorphism from $G$ into $S_{3}$. However, this is impossible, since $|G| > 6$.

Is there anything wrong with this proof?

• nbloomf  On December 9, 2011 at 1:40 am

My main hangup is in seeing how any argument that eliminates the cases $n_2 = 11$ and $n_2 = 33$ does not also rule out $n_2 = 3$.

But also, the permutation representation induced by a subgroup is not necessarily injective. (For example, anytime $|G| > |G/H|!$.)