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.

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  • 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|!.)

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