Math 103ASolutions11:30AM Monday 3/19/20071. True: (a) (d) (f)False: (b) (c) (e)2. (a)Supposex,y∈Z(G). Thenxg=gxandyg=gyfor allg∈Gby the definitionof the center. Hencegy-1=y-1gand so(xy-1)g=xy-1g=xgy-1=gxy-1=g(xy-1).By the definition ofZ(G),xy-1∈Z(G).(b) Sincegz=zgfor allz∈Z(G) and allg∈G, it follows thatgZ(G) =Z(G)gforallg∈G.3. Using the convention that 1-cycles are omitted, here are order: example.1: e2: (12)(34)3: (123)4: (12)(3456)5: (12345)6: (12)(34)(567)7: (1234567)4. (a)IfGis Abelian,φ(gh) = (gh)2=g2h2=φ(g)φ(h).Ifφis a homomorphism,φ(g)φ(h) =φ(gh) and sog2h2= (gh)2=ghgh. Multi-plying byg-1on the left andh-1on the right, we getgh=hg.(b) Letg∈Ghave order 2n-1. We haveφ(gn) =g2n=g2n-1g=g. (We did notactually use Abelian.)5. SinceKis non-cyclic, its order cannot be a prime. Since|N|=|G|/|K|, the possiblevalues of|N|are 53×7 divided by numbers which are not prime.Thus we have55275×7.6. (a)The identity is 0 and the polynomials with zero derivative are the constants.
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