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Unformatted text preview: Lenny T. Evans MATH 534 Homework 3 February 11, 2010 7.7.4 If G is the group of nonzero elements of Z /p Z under for a prime p, then for a G, a 6 0 (mod p ) implies a p 1 1 (mod p ) . Proof: We see that G = { 1 , 2 ,..., p 1 } , which is p 1 elements. Clearly G does not include , so for any element a G, a 6 0 (mod p ) . The identity here is clearly 1 , since 1 a = a 1 = a. From problem 7 . 7 . 2 , we see a # G = e, so a p 1 = 1 . From this, we see that a p 1 is in the equivalence class of 1, under modulo p. Or, in other terms, a p 1 1 (mod p ) . 7.8.1 The subgroup H = { 1 ,f } of D n is not normal for all n 3 . Proof: A subgroup is normal if for all g D n , it is true that gH = Hg. Since r denotes a rotation by 2 n , clearly r 1 is a backward rotation by 2 n , or equivalently a rotation of 2  2 n = 2 ( n 1) n = r n 1 . So r 1 = r n 1 . As stated in Stillwell in section 7.4, for D n ,fr = r n 1 f Then, by the definition of cosets, we see that rH = { r,rf } , but Hr = { r,fr } = { r,r n 1 f } 6 = rH. Thus, since there is some g D n such that gH 6 = Hg, we know that H is not normal in D n . Look at rHrH. If this were well defined, this should be r 2 H = { r 2 ,r 2 f } . However, rH = { r,rf } , so rH could be written with rf as the coset representative, as rfH. Now look at rfHrH. If this were well defined this should be ( rf ) rH = r ( fr ) H = rr n 1 fH = fH. (Exploiting here the associative nature of elements of D n since D n is a group) However, since f H, rfHrH = H, so we see that the product is dependent on the chosen coset represen tative. 7.8.2 The subgroup of A 4 called H with the identity and three permutations that simultaneously transpose two pairs is normal. Proof: We see that H = { e, (1 2)(3 4) , (1 3)(2 4) , (1 4)(2 3) } . Since for any h H, hH = H = Hh, we see that for elements of H, the normality condition holds. As stated in Stillwell, A 4 has 12 elements, hence the index of H is 3. We know 3cycles are in A 4 from the previous homework, so we can try multiplying H by a 3cycle from the left to get a new coset. Look at the left cosets (1 2 3) H = { (1 2 3) , (1 3 4) , (2 4 3) , (1 4 2) } And (1 3 2) H = { (1 3 2) , (2 3 4) , (1 2 4) , (1 4 3) } From this, we see H, (1 2 3) H, and (1 3 2) H are disjoint, so these are all of the cosets of H, and all of these cosets contain all the elements of A 4 (since A 4 has 12 elements, and each coset has 4 distinct elements). 1 Look at the right cosets H (1 2 3) = { (1 2 3) , (2 4 3) , (1 4 2) , (1 3 4) } And H (1 3 2) = { (1 3 2) , (1 4 3) , (2 3 4) , (1 2 4) } ....
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 Spring '11
 KUMAR
 Math, Algebra

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