Applications

# Applications - 69 III APPLICATIONS by A. N . M ilgram A....

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69 III APPLICATIONS by A. N. Milgram A. Solvable Groups. Before proceeding with the applications we must discuss certain questions in the theory of groups. We shall assume several simple propo- sitions: (a) If N is a normal subgroup of the group G, then the mapping f(x) = xN is a homomorphism of G on the factor group G/N. f is called the natural homomorphism. (b) The image and the inverse image of a normal subgroup under a homomorphism is a normal subgroup, (c) If f is a homomorphism of the group G on G 1 , then setting N f = f(N), and defining the mapping gasg(xN) = f(x)N',w e readily see that g is a homomorphism of the factor group G/N on the factor group G'/N 1 . Indeed, if N is the inverse image of N' then g is an isomorphism. We now prove THEOREM 1. (Zassenhaus). If U and V are subgroups of G, u and v normal subgroups of U and V, respectively, then the following three factor groups are isomorphic: u(UnV)/u(Unv), v(UnV)/v(unV), (UnV)/(unV)(vnU). It is obvious that U n v is a normal subgroup of U n V. Let f be the natural mapping of U on U/u. Call f(UnV) = H and f(Unv) = K. Then f'^H) = u(UnV) and f^K) = u(Unv) from which it follows that u(UnV)/u(Unv) is isomorphic to H/K. If, however, we view f as defined only over U n V, then f^K) = [un(UnV)](Unv) = (unV)(Unv) so that (UnV)/(unV)(Unv) is also isomorphic to H/K.

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70 Thus the first and third of the above factor groups are isomorphic to each other. Similarly, the second and third factor groups are isomorphic. Corollary 1. If H is a subgroup and N a normal subgroup of the group G, then H/HnN is isomorphic to HN/N, a subgroup of G/N. Proof: Set G = U, N = u, H = V and the identity 1 = v in Theorem 1. Corollary 2. Under the conditions of Corollary 1, if G/N is abelian, so also is H/HnN. Let us call a group G solvable if it contains a sequence of sub- groups G = G 0 D Gj D. . .D G g = 1, each a normal subgroup of the preceding, and with G^ /G i abelian. THEOREM 2. Any subgroup of a solvable group is solvable. For let H be a subgroup of G, and call H £ = HnG. . Then that H.^/H. is abelian follows from Corollary 2 above, where G M , G. and H. j play the role of G, N and H. THEOREM 3. The homomorph of a solvable group is solvable. Let f(G) = G 1 , and define G\ = fCG^ where G. belongs to a a sequence exhibiting the solvability of G. Then by (c) there exists a homomorphism mapping G il /G i on GJ^/GJ. But the homomorphic image of an abelian group is abelian so that the groups G'. exhibit the solvability of G' » B. Permutation Groups. Any one to one mapping of a set of n objects on itself is called a permutation. The iteration of two such mapping is called their product.
71 It may be readily verified that the set of all such mappings forms a group in which the unit is the identity map. The group is called the symmetric group on n letters. Let us for simplicity denote the set of n objects by the numbers 1,2, . .. , n. The mapping S such that S(i) = i + l mod n will be de- noted by (123. . .n) and more generally (i j. . . m ) will denote the map- ping T such that T(i) = j, . . . ,T(m) = i. If(ij. ..m ) has k numbers, then it will be called a k cycle. It is clear that ifT = (ij. ..s) then T- 1 =(s. ..ji>.

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## This note was uploaded on 11/14/2011 for the course MATH 367 taught by Professor Sdd during the Spring '11 term at Middle East Technical University.

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Applications - 69 III APPLICATIONS by A. N . M ilgram A....

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