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103a_07w_fs

# 103a_07w_fs - Math 103A 1 True(a(d(f Solutions False(b(c(e...

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Math 103A Solutions 11:30AM Monday 3/19/2007 1. True: (a) (d) (f) False: (b) (c) (e) 2. (a) Suppose x,y Z ( G ). Then xg = gx and yg = gy for all g G by the definition of the center. Hence gy - 1 = y - 1 g and so ( xy - 1 ) g = xy - 1 g = xgy - 1 = gxy - 1 = g ( xy - 1 ) . By the definition of Z ( G ), xy - 1 Z ( G ). (b) Since gz = zg for all z Z ( G ) and all g G , it follows that gZ ( G ) = Z ( G ) g for all g G . 3. Using the convention that 1-cycles are omitted, here are order: example. 1: e 2: (12)(34) 3: (123) 4: (12)(3456) 5: (12345) 6: (12)(34)(567) 7: (1234567) 4. (a) If G is Abelian, φ ( gh ) = ( gh ) 2 = g 2 h 2 = φ ( g ) φ ( h ). If φ is a homomorphism, φ ( g ) φ ( h ) = φ ( gh ) and so g 2 h 2 = ( gh ) 2 = ghgh . Multi- plying by g - 1 on the left and h - 1 on the right, we get gh = hg . (b) Let g G have order 2 n - 1. We have φ ( g n ) = g 2 n = g 2 n - 1 g = g . (We did not actually use Abelian.) 5. Since K is non-cyclic, its order cannot be a prime. Since | N | = | G | / | K | , the possible values of | N | are 5 3 × 7 divided by numbers which are not prime. Thus we have 5 5 2 7 5 × 7. 6. (a) The identity is 0 and the polynomials with zero derivative are the constants.
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