# Download Course 311: Michaelmas Term 2001. Topics in Group Theory by Wilkins D.R. PDF

By Wilkins D.R.

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Example text

This equation is referred to as the class equation of the group G. Definition Let g be an element of a group G. The centralizer C(g) of g is the subgroup of G defined by C(g) = {h ∈ G : hg = gh}. 41 Let G be a finite group, and let p be a prime number. Suppose that pk divides the order of G for some positive integer k. Then either pk divides the order of some proper subgroup of G, or else p divides the order of the centre of G. Proof Choose elements g1 , g2 , . . , gr of G\Z(G), where Z(G) is the centre of G, such that each conjugacy class included in G \ Z(G) contains exactly one of these elements.

54 Let G be a group, let H1 and H2 be subgroups of G, where H1 H2 , and let J1 = H1 ∩ N , J2 = H2 ∩ N , K1 = H1 N/N and K2 = H2 N/N , where N is some normal subgroup of G. Then J1 J2 and K1 K2 . Moreover there exists a normal subgroup of H2 /H1 isomorphic to J2 /J1 , and the quotient of H2 /H1 by this normal subgroup is isomorphic to K2 /K1 . Proof It is a straightforward exercise to verify that J1 J2 and K1 K2 . Let θ: H2 → K2 be the surjective homomorphism sending h ∈ H2 to the coset hN . Now θ induces a well-defined surjective homomorphism ψ: H2 /H1 → K2 /K1 , since θ(H1 ) ⊂ K1 .

Using the Second Sylow Theorem, we see that any group of order 18 has just one Sylow 3-subgroup. This Sylow 3-subgroup is then a normal group of order 9, and therefore no group of order 18 is simple. Similarly a group of order 50 has just one Sylow 5-subgroup, which is then a normal subgroup of order 25, and therefore no group of order 50 is simple. Also a group of order 54 has just one Sylow 3-subgroup, which is then a normal subgroup of order 27, and therefore no group of order 54 is simple. On applying the Second Sylow Theorem, we see the number of Sylow 7-subgroups of any group of order 42 must divide 42 and be congruent to 1 modulo 7.