# hw6 - a 1 , half of the a 2 s make the dierence even and...

This preview shows page 1. Sign up to view the full content.

MATH 3360 HOMEWORK SOLUTION 6 (Some computational problems are omitted.) p188, E2 If f is 1-1, of course only one element can be mapped to e . If ker ( f ) = { e } , then for any two elements x,y , if f ( x ) = f ( y ) then f ( x - 1 y ) = e , then x - 1 y = e , i.e., x = y . This proves that f is 1-1. p188, E3 (i) { [0] , [5] } (ii) { [0] } (iii) { [0] , [3] , [6] , [9] } p193, E3 You can think of picking two linearly independent vectors in 2-dimensional F 3 × F 3 . The ﬁrst one has 3 2 - 1 = 8 choices (the only vector you cannot pick is (0 , 0)) and the second one has 3 3 - 3 = 6 choices (now you cannot pick any element which is linearly dependent to the ﬁrst one). So ﬁnally there are 8 × 6 = 48 elements in GL 2 ( F 3 ). p200, E12 There are 24 pairs, and for each pair ( a 1 ,a 2 ), there is a solution if and only if a 1 - a 2 is even. It is not diﬃcult to see that for each
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: a 1 , half of the a 2 s make the dierence even and half make the dierence odd. So nally half of the pairs have solutions. p201 E15 For (i) and (iii) you can uniformly prove the following: if r is an odd prime number &lt; q , then p 1( mod r ) cannot happen. Suppose not, i.e., p 1( mod r ). Then by plugging in p = 2 q + 1 we get q 0( mod r ), which for sure doesnt hold, as q is a prime number dierent from r . For (ii) we use the similar argument. The only dierence is that you get 2 q 0( mod 4), therefore q 0( mod 2), which also doesnt hold. p205, E5 The kernel is { [0] , [12] } and for the image, see p200 E12. 1...
View Full Document

Ask a homework question - tutors are online