# A first graduate course in abstract algebra by W.J. Wickless

By W.J. Wickless

Graduate textbooks usually have a slightly daunting heft. So it really is friendly for a textual content meant for first-year graduate scholars to be concise, and short adequate that on the finish of a direction approximately the total textual content could have been coated. This publication manages that feat, fullyyt with out sacrificing any fabric assurance. the normal issues of workforce thought, vector areas, modules, jewelry, box and Galois thought are coated, in addition to issues in noncommutative jewelry, workforce extensions, and chosen subject matters in abelian teams. a few of the brevity is end result of the the truth that rather than delivering huge chunks of workouts of middling hassle, Wickless (math, collage of Connecticut) has opted to supply fewer workouts of larger trouble.

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Extra resources for A first graduate course in abstract algebra

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Let X < G/N. Put H = {y & G : Ny e X}. For n&N,Nn = Ne e X. Hence, N C H. If y, z € H, then Ny, Nz e X. Thus, Nyz~l = (Ny)(Nz)~1 e X so that yz~1 e H. Therefore, by the subgroup test theorem, H < G. By definition, H/N = X. If H < G, let TT : G/N -> G/H be the factor map, n(Ng) = Hg. Since N C H it is easy to check that TT is well defined and, if well defined, TT is plainly a homomorphism. We have Ker TT = H/N = X, so X < G/N . Suppose X = H/N < G/N and let g e G. Then (Ng)-l(H/N)(Ng) C F/AT. Thus, (g-1Hg)/N C H/N.

Thus, s~, = 1 or ,ss = 6. If 85 = 1, then G has a unique-hence normal-Sylow ^-subgroup, and we're done. Suppose that 55 = 6. Since each Sylow ^-subgroup is isomorphic to Z(5), the pairwise intersection of any two distinct ones is < e > . Thus, the Sylow 5-subgroups account for 25 = 6 • 4 + 1 distinct elements. Next, look at 33. If 33 = 1, we're done. So, suppose that 53 = 10, the only other possibility. In this case, reasoning as above, we account for 20 = 10-2 additional elements. But we've already used more elements than are in the group!

If we number the elements of 5(3), we obtain a corresponding homomorphism c : G —> 84. The homomorphism c sends each g 6 G to the permutation induced by conjugation by g. Since G = 108 < 24 =| 84 |, the kernel of the map c will be a proper normal subgroup of G. Note that c cannot map each element g & G to the identity of 84, since any two 3-Sylow subgroups are conjugate. 4 Our final example has a different flavor. We construct a nonabelian group of G of order 21. Consider a group G with \ G \~ 3 • 7.