Homework4

# Homework4 - Homework 4 Solutions 1 e 2 = e thus e ∈ H and...

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Homework 4 Solutions 1. e 2 = e thus e ∈ H and H 6 = Ø. Now let x,y ∈ H , hence ∃ a,b ∈ G such that x = a 2 , y = b 2 . xy- 1 = a 2 ( b 2 )- 1 = ( ab- 1 ) 2 since G is abelian. Thus xy- 1 ∈ H and H ≤ G by the one step subgroup test. 2. (1) = (12)(12) thus (1) ∈ A n and A n 6 = Ø. Let a,b ∈ A n thus a and b can be written as a = α 1 .. .α r , b = β 1 .. .β s where α i , β i are transpositions and r and s are even. Since the α i ’s and β i ’s are transpositions: ab- 1 = ( α 1 .. .α r )( β 1 .. .β s )- 1 = α 1 .. .α r β- 1 s .. .β- 1 1 = α 1 .. .α r β s .. .β 1 Since r + s is even, ab- 1 is an even permutation and hence ab- 1 ∈ A n . Thus A n ≤ S n by the one step subgroup test. 3. [ G : H ] = | G | / | H | = ( | G | / | K | )( | K | / | H | ) = [ G : K ][ K : H ] 4. BWOC, suppose ∃ a ∈ H ∩ K such that a 6 = e . Let o ( a ) = l 6 = 1. a ∈ H and a ∈ K implies that l || H | and l || K | by the corollary to Lagrange’s Theorem. l 6 = 1...
View Full Document

## This note was uploaded on 04/19/2011 for the course MATH 21373 taught by Professor Johnson during the Spring '11 term at Carnegie Mellon.

Ask a homework question - tutors are online