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Unformatted text preview: Math 113 Homework 4 – Solutions 1. Fraleigh, section 8, problem 21; section 9, problem 23. # 8.21: a. Verify that the six matrices 1 0 0 0 1 0 0 0 1 , 0 1 0 0 0 1 1 0 0 , 0 0 1 1 0 0 0 1 0 , 1 0 0 0 0 1 0 1 0 , 0 0 1 0 1 0 1 0 0 , 0 1 0 1 0 0 0 0 1 form a group under matrix multiplication. These matrices correspond to linear transformations which act on elements of R 3 by per mutation of the coordinates. Accordingly, for σ ∈ S 3 a permutation of { 1 , 2 , 3 } we denote by A σ ∈ GL (3 , R ) the matrix corresponding to the linear transformation ( x 1 , x 2 , x 3 ) mapsto→ ( x σ (1) , x σ (2) , x σ (3) ). We now check that the set H = { A σ , σ ∈ S 3 } forms a subgroup of GL (3 , R ): indeed, the composition of any two elements of H acts again on R 3 by permutation of the coordinates (namely, A σ A τ = A στ ), so H is closed under multiplication; H contains the identity element ( A id = I 3 ); and since A σ 1 = A σ , the inverse of an element of H is again in H . Therefore H is a subgroup of GL (3 , R ). b. What group discussed in this section is isomorphic to this group of six matrices? H is isomorphic to the symmetric group S 3 . Indeed, the map from S 3 to H defined by σ mapsto→ A σ (which is clearly a bijection) is a homomorphism since A σ A τ = A στ for all σ, τ ∈ S 3 . # 9.23: Mark each of the following true or false. a. Every permutation is a cycle. False: e.g., parenleftbigg 1 2 3 4 2 1 4 3 parenrightbigg ∈ S 4 has two nontrivial orbits { 1 , 2 } and { 3 , 4 } , so it is not a cycle. b. Every cycle is a permutation. True, by definition of a cycle. c. The definition of even and odd permutations could have been given equally well before Theorem 9.15. Without Theorem 9.15, we don’t know that a same permu tation can’t be both odd and even at the same time. Depending on your understanding of whether Definition 9.18 could allow for the possibility of a permutation being both odd and even, the statement is either true or false. d. Every nontrivial subgroup H of S 9 containing some odd permutation con tains a transposition. False: the 4cycle μ = (1 2 3 4) = (1 4) (1 3) (1 2) is odd, and the cyclic subgroup H = ( μ ) = { e, μ, μ 2 , μ 3 } does not contain any transposition. e. A 5 has 120 elements. False:  A 5  = 5! / 2 = 60. f. S n is not cyclic for any n ≥ 1 . False: S 1 = { e } and S 2 = { e, (1 2) } are cyclic. (On the other hand, for n ≥ 3 S n is not cyclic – for example, because it is not abelian.) g. A 3 is a commutative group. True: A 3 is in fact the cyclic group generated by the 3cycle (1 2 3), and isomorphic to Z 3 , hence abelian. (In fact  A 3  = 3! / 2 = 3, and we’ve seen that every group of order 3 is isomorphic to Z 3 .) h. S 7 is isomorphic to the subgroup of all those elements of S 8 that leave the number 8 fixed. True. Define a map f : S 7 → S 8 as follows: for σ ∈ S 7 , define...
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This note was uploaded on 03/14/2012 for the course MATH 113 taught by Professor Ogus during the Spring '08 term at Berkeley.
 Spring '08
 OGUS
 Matrices

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