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 path approximately the whole textual content could have been lined. This ebook manages that feat, fullyyt with out sacrificing any fabric insurance. the normal subject matters of team concept, vector areas, modules, earrings, box and Galois thought are lined, in addition to themes in noncommutative earrings, team extensions, and chosen themes in abelian teams. many of the brevity is as a result of the truth that rather than offering huge chunks of workouts of middling trouble, Wickless (math, college of Connecticut) has opted to supply fewer routines of larger hassle.

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Another possible definition would be that G is a p-group if and only if g € G ==>• g — p fc ^, where k(g) is a nonnegative integer depending on g. Using what we've done so far, it's easy to prove that these two definitions are equivalent. 1 Let G be a finite group. Prove that G is a p-group if and only if g 6 G : 9\=P M9) We are ready for the first Sylow Theorem . In the statement, pn refers to an arbitrary prime power with n > I. 2 (First Sylow Theorem) Let pn divide G I . Then G has a subgroup of order pn.

Now consider the map L : G —>

A(kr)). 4 Let a, j3 6 Sn. Prove that f3 = aaa~l for some a G Sn if and only if a, /? have the same cycle structure, that is they factor into products of the same number of disjoint cycles of the same lengths. 5 Prove that the commutator subgroup of Sn is An. , (1, n) >= Sn. 8 Regard 83 < 64 in the natural way, 83 being identified as the subgroup {a £ 84 : cr(4) = 4}. Use the right coset decomposition of 84 mod 83 to get a list of the elements of 84. Is 83 < 84? 9. 9 Show, by filling in the details in the steps below, that the group A4, a group of order 12, has no subgroup of order 6.

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