## § 7.7 Simple Groups: solved problems

15. Prove that there are no simple groups of order 200.

Solution: Suppose that |G| = 200 = 23 · 52. The number of Sylow 5 subgroups must be a divisor of 8 and congruent to 1 modulo 5, so it can only be 1, and this gives us a proper nontrivial normal subgroup.

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16. Sharpen Exercise 7.7.3 (b) of the text by showing that if G is a simple group that contains a subgroup of index n, where n > 2, then G can be embedded in the alternating group An.

Solution: Assume that H is a subgroup with [G:H] = n, and let G act by multiplication on the left cosets of H. This action is nontrivial, so the corresponding homomorphism µ : G -> Sn is nontrivial. Therefore ker(µ) is trivial, since G is simple. Thus G can be embedded in Sn. Then An µ(G) is a normal subgroup of µ(G), so since G is simple, either µ(G) An, or An µ(G) = {e}. The second case implies |G| = 2, since the square of any odd permutation is even, and this cannot happen since n > 2.

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17. Prove that if G contains a nontrivial subgroup of index 3, then G is not simple.

Solution: If G is simple and contains a subgroup of index 3, then G can be embedded in A3. If the subgroup of index 3 is nontrivial, then |G| > 3 = | A3 |, a contradiction.

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18 Prove that there are no simple groups of order 96.

Solution: Suppose that |G| = 96 = 25 · 3. Then the Sylow 2-subgroup of G has index 3, and so the previous problem shows that G cannot be simple.

An alternate proof is to observe that |G| is not a divisor of 3!.

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19. Prove that there are no simple groups of order 132.

Solution: Let G be a group of order 132 = 22 · 3 · 11. For the number of Sylow subgroups we have n2(G) = 1, 3, 11, or 33; n3(G) = 1, 4, or 22; and n11(G) = 1 or 12. We will focus on n3(G) and n11(G). If n3(G) = 4 we can let the group act on the Sylow 3-subgroups to produce a homomorphism into S4. Because 132 is not a divisor of 24 = | S4 | this cannot be one-to-one and therefore has a nontrivial kernel. If n3(G) = 22 and n11(G) = 12 we get too many elements: 44 of order 3 and 120 of order 11. Thus either n3(G) = 1 or n11(G) = 1, and the group has a proper nontrivial normal subgroup.

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20. Prove that there are no simple groups of order 160.

Solution: Suppose that |G| = 160 = 25 · 5. Then the Sylow 2-subgroup of G has index 5, and 25 · 5 is not a divisor of 5! = 120, so G must have a proper nontrial normal subgroup.

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21. Prove that there are no simple groups of order 280.

Solution: Let G be a group of order 280 = 23 · 5 · 7. In this case for the number of Sylow subgroups we have n2(G) = 1, 5, 7, or 35; n5(G) = 1 or 56; and n7(G) = 1 or 8. Suppose that n5(G) = 56 and n7(G) = 8, since otherwise either there is 1 Sylow 5-subgroup or 1 Sylow 7-subgroup, showing that the group is not simple. Since the corresponding Sylow subgroups are cyclic of prime order, their intersections are always trivial. Thus we have 8 · 6 elements of order 7 and 56 · 4 elements of order 5, leaving a total of 8 elements to construct all of the Sylow 2-subgroups. It follows that there can be only one Sylow 2-subgroup, so it is normal, and the group is not simple in this case.

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22. Prove that there are no simple groups of order 1452.

Solution: Let G be a group of order 1452 = 22 · 3 · 112. We must have n11(G) = 1 or 12. In the second case, we can let the group act by conjugation on the set of Sylow 11-subgroups, producing a nontrivial homomorphism from the group into S12. But 1452 is not a divisor of | S12 | = 12! since it has 112 as a factor, while 12! does not. Therefore the kernel of the homomorphism is a proper nontrivial normal subgroup, so the group cannot be simple.

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