Adv Alegbra HW Solutions 55

Adv Alegbra HW Solutions 55 - S i has f nite index in G...

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

View Full Document Right Arrow Icon
55 2.60 (i) Prove that the stochastic group 6( 2 , R ) , the set of all nonsingular 2 × 2 matrices whose row sums are 1, is a subgroup of GL ( 2 , R ) . Solution. The proof is a straightforward calculation: note that the formula for the inverse is needed. The students should be encour- aged to try to show that 6( 3 , R ) is a group. (ii) De f ne 6 0 ( 2 , R ) to be the set of all nonsingular doubly stochastic matrices (all row sums are 1 and all column sums are 1). Prove that 6 0 ( 2 , R ) is a subgroup of GL ( 2 , R ) . Solution. The doubly stochastic group 6 0 is a subgroup because it is the intersection of the subgroups 6 and 6 0 . 2.61 Let G be a f nite group, and let S and T be (not necessarily distinct) nonempty subsets. Prove that either G = ST or | G |≥| S |+| T | . Solution. Absent. 2.62 (i) If { S i : i I } is a family of subgroups of a group G , prove that an intersection of cosets T i I x i S i is either empty or a coset of T i I S i . Solution. Absent. (ii) ( B. H. Neumann . ) If a group G is the set-theoretic union of f nitely many cosets, G = x 1 S 1 ∪···∪ x n S n , prove that at least one of the subgroups
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: S i has f nite index in G . Solution. Use induction on the number of distinct subgroups S i . 2.63 (i) Show that a left coset of h ( 1 2 ) i in S 3 may not be equal to a right coset of h ( 1 2 ) i in S 3 ; that is, there is α ∈ S 3 with α h ( 1 2 ) i ±= h ( 1 2 ) i α . Solution. Absent. (ii) Let G be a f nite group and let H ≤ G be a subgroup. Prove that the number of left cosets of H in G is equal to the number of right cosets of H in G . Solution. Consider aH 7→ Ha − 1 . 2.64 True or false with reasons. (i) If G and H are additive groups, then every homomorphism f : G → H satis f es f ( x + y ) = f ( x ) + f ( y ) for all x , y ∈ G . Solution. True. (ii) A function f : R → R × is a homomorphism if and only if f ( x + y ) = f ( x ) + f ( y ) for all x , y ∈ R . Solution. False....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online