Assignment 3: Abstract Algebra

Task #3: Abstract Algebra
Competency 210.4.2: Groups & Competency 210.4.4: Fields
Jennifer Moore
Western Governor’s University

Part A: The image below is the fifth roots of unity. Using these fifth roots of unity and de Moivre’s formula to verify that the fifth roots of unity form a group under complex multiplication. de Moivre’s formula is z^k=cos(2πk/n)+isin(2πk/n),k=0,1,2,…,n-1 (Nicodemi, 2006) for the 5th roots unity of n=5. The following show the fifth roots of unity using de Moivre’s formula. de Moivre’s Formula z^0=cos((2π∙0)/5)+isin((2π∙0)/5) =cos⁡(0)+isin(0)
=1+0i n=5,k=0 z^1=cos((2π∙1)/5)+isin((2π∙1)/5) z^1=cos(2π/5)+isin(2π/5) n=5,k=1 z^2=cos((2π∙2)/5)+isin((2π∙2)/5) z^2=cos(4π/5)+isin(4π/5) n=5,k=2
Since S & T are subfields, each containing c, it follows that c^(-1)∈S and c^(-1)∈T where (c^(-1) )×c=c×(c^(-1) )=g. Hence, c^(-1)∈S∩T.
Commutative under addition:
If a,b ∈F, then a+b=b+a. That is the elements a,b ∈F are commutative under addition. If a,b ∈S∩T, then a,b ∈S and a,b ∈T. Since S & T are subfields, each containing elements a and b, it follows that a+b=b+a ∈S and a+b=b+a ∈T. Hence, a+b=b+a ∈S∩T and the elements in the intersection are commutative under addition.
Commutative under multiplication:
If a,b ∈F, then ab=ba. That is the elements a,b ∈F are commutative under multiplication. If a,b ∈S∩T, then a,b ∈S and a,b ∈T. Since S & T are subfields, each containing elements a and b, it follows that ab=ba ∈S and ab=ba ∈T. Hence, ab=ba ∈S∩T and the elements in the intersection are commutative under multiplication.
Associative under addition:
If a,b,c ∈F, then (a+b)+c=a+(b+c). That is the elements a,b,c ∈F are associative under addition. If a,b,c ∈S∩T, then a,b,c ∈S and a,b,c ∈T. Since S & T are subfields, each containing elements a, b, c it follows that (a+b)+c=a+(b+c) ∈S and (a+b)+c=a+(b+c) ∈T. Hence, (a+b)+c=a+(b+c)∈S∩T and the elements in the intersection are associative under