Subgroups

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• Sylow theorem

  Since G does not have normal Sylow 3-subgroups, then m is not equal to 1. Thus the possible numbers are m = 4 or m = 16. Then G may have 4 or 16 Sylow 3-subgroups. Determine the number of Sylow 2-subgroups and Sylow 3-subgroups that group G can have.

• Proof that There Are No Simple Groups of Order 56 or 148

  Now G can only be simple if it has 7 Sylow-2 subgroups and 8 Sylow-7 subgroups. We know from Sylow's second theorem that all Sylow-p subgroups are conjugate. In particular, all eight Sylow-7 subgroups must be conjugate.

• Normal Subgroups of a Group

  Subgroups of a Group Normal Subgroups of a Group Suppose that N and M are two normal subgroups of G and that N intersection M = (e).

• Prove that a cyclic group of prime order p has no non-trivial subgroups.

  Then the subgroups of G are in bijective correspondence with the positive divisors of p, namely, 1 and p; so the only subgroups of G are G itself and the trivial subgroup - in other words, G has no nontrivial *proper* subgroups.

• Intersection of two normal subgroups is a normal subgroup.

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• Conjugated Classes of Subgroups of A_5

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• Epimorphism proof

  So, assume that G can be written as a union of 3 proper subgroups A, B, C. Notice that any one of these subgroups cannot be contained in the union of the other two, for otherwise G is a union of two proper subgroups, impossible.

• Groups

  Show that G is not the internal direct product of any two of its proper subgroups. If G is the internal direct product of subgroups G_1 and G_2 show that G/G_1 is isomorphic to G_2 and G/G_2 is isomorphic to G_1.