Before Midterm 1

# Before Midterm 1 - Answer key No.1 to selected homework...

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Answer key No.1 to selected homework problems: Math. 110B 7.1 2. A permutation f either fxes one number 1 j 3 or two numbers or moves all. There are 3 elements fxing only one number (those interchang- ing the two numbers diﬀerent From j ). ±or such f , f f = I . There is only one which fxes two numbers; so, it fxes actually all three numbers 1 , 2 , 3, that is, I .T a k e f which moves all. Then f (1) = i and i ± = 1. IF f ( i ) = 1, then f interchanges 1 and i ; so, f fxes the rest j with j ± =1 and j ± = i . This is impossible because f moves all. Thus we conclude that f ( i )= j and j ± = i and j ± = 1. In other words, f = ± 1 ij 1 ² . The choice oF i and j are either ( i,j )=(2 , 3) or ( )=(3 , 2); so, there are two element with f f f = I (which can be verifed by the above shape oF f ). In conclusion, there are 3 elements g S 3 with g g = I , and there are two elements f with f f f = I ; the rest is I itselF. 2. (a) | Z 18 | = 18; (b) | D 4 | = 8, Four rotations (including the identity), one ²ip aFter a rotation (see Example in page 164-167). (c) 4!: number oF 4–permutations; (d) 5!: number oF 5–permutations; (e) By Theorem 2.11, non-units oF Z n are zero-divisors, which are mul- tiples oF proper divisors oF n = 0 (here “proper divisor d | n ” means 1 <d ). When n = 18, proper divisors are: 2 , 3 , 6 , 9 , 18. Remove all multiples oF the above numbers out oF integers 1 to 18 = 0, remaining ones are 1 , 5 , 7 , 11 , 13 , 17; so, | U 18 | =6 . 4. (a) Yes. Consider π : Z 10 Z 5 given by π ( x x mod 5. Check this is a homomorphism oF rings. Then π : { 2 , 4 , 6 , 8 }→ U 5 is one-to-one and onto. The product is just a product oF U 5 . Since we know that U 5 is a group; so, { 2 , 4 , 6 , 8 } is also a group. (b) Yes, because 2 x + y =2 x 2 y . Consider a Function f : Q G given by f ( x )=2 x . Then f is one-to-one and onto. Moreover f ( x + y f ( x ) * f ( y ). Since Q with addition + is a group, ( G, * ) is a group (the structure oF ( G, * ) is identical with ( Q , +) via f ). (c) No. 6 G does not have inverse in G . (d) Yes. Check associative law: ( a * b ) * c =( a + b - ab ) * c a + b - ab + c ) - ( a + b - ab ) c a + b + c - bc ) - a ( b + c - bc a * ( b + c - bc a * ( b * c ) . The identity exists: a * b = b a + b - ab = b a (1 - b )=0 b ± =1 = a =0 . Thus a = 0 is the identity. Inverse exists: a * b a + b - ab b = - a 1 - a G, 1

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because a ± = 1. Actually defning f : Q Q by f ( x )=1 - x , then f ( a * b - ( a + b - ab )=(1 - a )(1 - b )= f ( a ) f ( b ). f (0) = 1. This shows the group structure oF G and ( R × , × ) are identical by one-to-one onto Function f . Here R × is the number line 0 removed (the unit group oF the ring R ). (e) Yes. Consider f : R R f ( x x + 1. Then f ( a * b f ( a + b + ab a + b + ab +1 = ( a +1)( b +1) = f ( a ) f ( b ) and f ( - 1) = 0. Thus the group structure oF ( G, * ) and the group structure oF ( R × , × ) are identical via the one-to-one onto Function f . (F) Yes. By the condition cd = 0, the complex numbers in G are either real or purely imaginary (that is, r - 1 For real numbers r ). Since c + d ± =0 ,0 C is removed. Since product oF real or purely imaginary numbers stay in real or purely imaginary, G is a group with identity 1.
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Before Midterm 1 - Answer key No.1 to selected homework...

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