Adv Alegbra HW Solutions 54

Adv Alegbra HW Solutions 54 - 54 2.54 (i) Define the...

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Unformatted text preview: 54 2.54 (i) Define the special linear group by SL(2, R) = { A ∈ GL(2, R) : det( A) = 1}. Prove that SL(2, R) is a subgroup of GL(2, R). Solution. It suffices to show that if A , B ∈ SL(2, R), then so is AB −1 ; that is, if det( A) = 1 = det( B ), then det( AB −1 ) = 1. Since det(U V ) = det(U ) det(V ), it follows that 1 = det( E ) = det( B B −1 ) = det( B ) det( B −1 ). Hence, det( AB −1 ) = det( A) det( B −1 ) = 1. (ii) Prove that GL(2, Q) is a subgroup of GL(2, R). Solution. Both the product of two matrices with rational entries and the inverse of a matrix with rational entries have rational entries. 2.55 Give an example of two subgroups H and K of a group G whose union H ∪ K is not a subgroup of G . Solution. If G = S3 , H = (1 2) , and K = (1 3) , then H ∪ K is not a subgroup of G , for (1 2)(1 3) = (1 3 2) ∈ H ∪ K . / 2.56 Let G be a finite group with subgroups H and K . If H ≤ K , prove that [G : H ] = [G : K ][ K : H ]. Solution. If G is a finite group with subgroup H , then [G : H ] = |G |/| H |. Hence, if H ≤ K , then [G : K ][ K : H ] = (|G |/| K |) · (| K |/| H |) = |G |/| H | = [G : H ]. 2.57 If H and K are subgroups of a group G and if | H | and | K | are relatively prime, prove that H ∩ K = {1}. Solution. By Lagrange’s theorem, | H ∩ K | is a divisor of | H | and a divisor of | K |; that is, | H ∩ K | is a common divisor of | H | and | K |. But | H | and | K | are relatively prime, so that | H ∩ K | = 1 and H ∩ K = {1}. 2.58 Prove that every infinite group contains infinitely many subgroups. Solution. An infinite group have only finitely many cyclic subgroups. 2.59 Let G be a group of order 4. Prove that either G is cyclic or x 2 = 1 for every x ∈ G . Conclude, using Exercise 2.44, that G must be abelian. Solution. If G has order 4, then the only possible orders of elements in G are 1, 2, and 4. If there is an element of order 4, then G is cyclic with that element as generator. Otherwise, every element has order 1 or 2, so that x 2 = 1 for every x ∈ G . ...
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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