This is where the old homework is listed.
The text is "Contemporary Abstract Algebra" by J. Gallian, 5th edition.
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read pp. 3-9, skip from before example 6 (bottom of p. 9) to middle p. 14 read middle p. 14 through example 6, p. 16 skip after example 6, p. 16, to middle p. 17 read rest of chapter pp. 23--26: do problems #2,4,7,8,11,14,17--20,26,28,46,50. skip pp. 31--41 (we'll come back to some of it later) read pp. 42--52 pp. 53--56: do problems #1,4,7a,10-12,14,15,17,18,22--26,33--35.
Here is a short proof that p,p+2,p+4 all prime implies that p=3; this was our last quiz.
Consider the residue classes mod 3 of p,p+2,p+4; at least one of them must be 0 (that is, at least one of these three integers is divisible by 3). The reason is that adding +2 is the same as adding -1 (mod 3); exactly one of t,t-1,t-2 is zero (mod 3). The only way a number can be divisible by 3 and still be prime is if it is equal to 3. The least prime number is 2 so p+2 and p+4 can't be 3; hence, p must be 3.
Here are two exercises involving groups. Both of these should be written up for collection on Monday 1/23 as I asked in class on Friday.
#A1 (Mac Lane): Show that the set S consisting of all functions from the (x,y)-plane to itself with either of the two forms below s_b: (x, y) |----> (x + b, y) t_b: (x, y) |----> (-x + b, y) constitute a group with respect to composition, where b is any integer. For a,b any two integers, find s_a * s_b, s_b * s_a, s_a * t_b, t_b * s_a, and t_a * t_b, t_b * t_a. Also what is the identity of the group (S,*) and what are the inverses of s_a and t_a? Is the group commutative? Do you need to check associativity for the operation? #A2 (minimal conditions for a group) Let (G, *) be a non-empty set with an associative binary operation. If there exists e in G such that for all g in G, e*g = g (i.e., a left identity) and if, for every g in G, there exists an element g^(-1) such that g^(-1) * g = e (i.e., a left inverse), then G is actually a group; that is, e is also a right identity and g^(-1) is also a right inverse. Hint (Mac Lane): First show that (G, *) satisfies left cancellation (i.e., if g*h = g*k, then h = k).
Here's a sketch of the argument using Mac Lane's hint.
Assume existence of left identity and of left inverse. Then left cancellation is easy: ax = ay implies x = a^(-1)ax = a^(-1)ay = y. Now if ex = x for all x, then to show xe = x, by left cancellation it suffices to note that x^(-1)xe = e = x^(-1)x. Finally, to see that xx^(-1) = e, x^(-1)x x^(-1) = e x^(-1) = x^(-1) e.
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