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(2ītk/n)+isin(2tk/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^O=cos((211•0)/5)+isin((211•0)/5) =cos(O)+isin(0) =1+Oi n=5,k=0 z^1=cos((211•1)/5)+isin((211•1)/5) Z^1=cos(21t/5)+isin(27/5) n=5,k=1 z^2=cos((21•2)/5)+isin((21•2)/5) z^2=cos(41/5)+isin(41/5) n=5,k=2 z^3=cos((211•3)/5)+isin((211•3)/5) z^3=cos(61/5)+isin(67/5) n=5,k=3 Z^4=cos((211•4)/5)+isin((211•4)75) Z^4=cos(81/5)+isin(81/5) n=5,k=4 Thus the group is a set with operation of multiplication as follows: G={{z^0,z^1,212,213,214 },x).

The following is the proof to show that z^n«z^m=z^(n+m). Work Justification Given m,nez and the radius is equal to one, we assume by Euler’s identity that z^n=cos(no)+isin(no) and zam=cos(mo) +isin(mo). z^n«z^m=(cos(no)+isin(no))•(cos(mo)+isin(mo)) Evaluate the expression with the given information. =cos(no)cos(mo) +icos(no)sin(mo)+isin(no)cos(mo)+i^2 sin(no)sin(mo) Used FOIL =cos(no)cos(mo)+icos(no)sin(mo)+isin(no)cos(mo)-sin(ne)sin(mo) i^2=-1 =cos(no)cos(mo)-sin(no)sin(mo)+i(cos(no)sin(mo)+sin(no)cos(mo)) Rewrite the equation. =cos(no+mo) +i(cos(no)sin(mo)+sin(no)cos(mo)) The angle sum identity states that cos(u+v)=cos(u) cos(v)-sin(u)sin(v) =cos(no+mo)+isin(n8+mo) The angle sum identity states that sin(u+v)=sin(u) cos(V)+cos(u)sin(v) =cos((n+m)0)+isin((n+m)0) Factor out the theta to show that when you multiply with exponents you can add the exponents. =z^(n+m) In order to show that set G is a group under complex multiplication, several things need to be proven.

Those things that need to be proven are closure, multiplicative identity, multiplicative inverse, and associative under multiplication (Nicodemi, 2006). If these are shown, then it proves that G is a group under complex multiplication. Cayley table for set G by complex multiplication is below. x Z10 z11 212 213 214 z^O z^0 z^1 z^2 z^3 z14 z^1 z11 Z^2 z13 214 z10 z^2 z^2 z^3 z14 z10 z11 z^3 z^3 214 z10 211 212 z14 z14 z^0 z11 z^2 z13 Closure under multiplication: This table to the left shows all of the elements of this set. This is an exhaustive list and all products of the elements are also in the set. Therefore, G is closed under multiplication. Multiplicative Identity: z^0xz^0=z\(0+0)=z^0 Z^0xz^1=z1(0+1)=z11 z^Oxz^2=z\(0+2)=z^2 Z^0xz^3=z^(0+3)=z13 z^O^z^4=z^(0+4)=z14 z^a EG, where a E{0,1,2,3,4}, we have z^Oz^a=z^(O+a)=z^a. Therefore, G has a multiplicative identity, namely z^0.

Multiplicative Inverse: The inverse element z^bEG is an element such that z^axz^b=zAbxz^a=z^0 for all {z^a,z] Aber where a,be{0,1,2,3,4}. Work Justification z^Oxz^O=z^(0+0)=z^0 z^O is the inverse of z^0. z^1xz^4=z^(1+4)=z^0 z^4 is the inverse of z^1. z^2xz^3=z^(2+3)=z^0 z^3 is the inverse of z^2. z^3xz^2=z^(3+2)=z^0 z^2 is the inverse of z^3. z^4xz^1=z^(4+1)=z^0 z^1 is the inverse of z^4. We see that for every element z^ae there exists an element Z^bEG, such that z^axz^b=z^0. Therefore, all multiplicative inverses are contained in G. Associative under multiplication: Let a,b,ce{0,1,2,3,4} and show (z^axz^b )<z^c=z^ax(z^bxz^c).

Work Justification (z^axz^b )xz^c=z^(a+b)xz^c Exponential Multiplication =z^(a+b+c)Exponential Multiplication =z^(a+(b+c)) Integer Associative under addition =z^axz^(b+c) Exponential Multiplication =z^ax [(z^bxz] ^c) This shows that G is associative under complex multiplication. All properties of a group are shown; therefore, G is a group under complex multiplication. Z_5 is the set ({0,1,2,3,4},+) and G is the set ({z^0,z^1,z^2, 213,z^4 },*). $(x)=(z^x )mod5 p(0)=(z^0 mod5=zA0 $(1)=(z^1 )mod5=z11 $(2)=(z^2 )mod5=z12 $(3)=(z^3 )mod5=z13 $(4)=(z14 )mod5=z14 Prove that G is isomorphic to Z_5 under addition. In order to prove that G is isomorphic, three things have to be proven: that is 1 to 1, onto, and operation preserving (homomorphism) (Nicodemi, 2006).

Show 1-1 (injective): One – to – one is “A function for which every element of the range of the function corresponds to exactly one element of the domain” (Simmons, 2016). $(x)=P(y) only happens when x=y. Φ(0)=zΛΟ →(1)=z^1+z^0 $(2)=z^2+z^0 03)=x^3+z^0 $(4)=z^4+z^0 This shows that if $(x)=z^0=°(y), then x=y=0 because all other mappings of $ don’t map to the value of z^0. Similarly, $(1)=z^1 &$(0)#z^1,$(2)#z^1,$(3)#z^1,$(4)#z^1 $(2)=z^2 & $(O#z^2,0(1)#z^2,0(3)#z^2,0(4)#z^2 °(3)=z^3 & $(0)#z13,0(1)#z13,0(2)#z^3,0(4)#z^3 0(4)=z^4 & (O)#z^4,$(1)#z^4,0(2)#z^4,0(3)#z^4 Therefore, is one – to – one. Show Onto: The function is said to be onto (surjective) if for every YE{z^0,211,212,213,214 } there is an xE{0,1,2,3,4}.

Part B: “A field, F, is a commutative ring in which every non-zero element has a multiplicative inverse. A subfield is a subset of a field such that with respect to the operations + &• of the field, the subset in itself is a field” (Chapter 1: Fields). Let F be a field. Let S and I be subfields of F. Prove that the intersection, SAT, must also be a field. The following are the axioms of a field: Additive Identity Multiplicative Identity Not empty set Closure under addition Closure under multiplication Additive Inverse Multiplicative Inverse Commutative under addition Commutative under multiplication Associative under addition Associative under multiplication Distributive We use these axioms to show that the intersection of S and I are also a field. Additive Identity: Fis a field.

Therefore, F contains an additive identity element, e. S is a subfield which is also a field. This means that S also has an additive identity element, e. Tis a subfield which is also a field. This means that I also has an additive identity element, e. Since EES and EET, we know EESNT. This also shows that SnT is not empty. Multiplicative Identity: Fis a field. Therefore, F contains a multiplicative identity element, h. S is a subfield which is also a field. This means that s also has a multiplicative identity element, h. Tis a subfield which is also a field. This means that T also has a multiplicative identity, h. Since hes and het, we know hESNT. Closure under addition: Sis a subfield of F. This means that S is closed under addition. Similarly, K is a subfield of Fand is also closed under addition. If a,besnt, then a,bes and a,bet. Since both S and I are closed under addition, we know that a+bES and a+bET. This means that a+bESNT. Therefore, the intersection of S and T is closed under addition.

Closure under Multiplication: Sis a subfield of F. This means that S is closed under multiplication. Similarly, K is a subfield of Fand is also closed under multiplication. If a,besnt, then abes and a,bet. Since both S and I are closed under multiplication, we know that axbes and axbET. This means that axbesnt. Therefore, the intersection of S and T is closed under multiplication. Additive Inverse: If CESNT, then CES and CET. Since &T are subfields, each containing c, it follows that -CES and-cET where (-c)+c=c+(-C)=g. Hence, -CESNT. Multiplicative Inverse: If CESNT, then CES and CET. Since S&T are subfields, each containing c, it follows that c^(-1)ES and c^(-1)ET where (C^(-1) )xc=cx(c^(-1) )=g. Hence, C^(-1)ESNT. Commutative under addition: If a,b EF, then a+b=b+a. That is the elements a,b EF are commutative under addition. If a,b esnt, then a,b ES and a,b ET. Since S & T are subfields, each containing elements a and b, it follows that a+b=b+a ES and a+b=b+a ET.

Hence, a+b=b+a ESNT and the elements in the intersection are commutative under addition. Commutative under multiplication: If a,b EF, then ab=ba. That is the elements a,b EF are commutative under multiplication. If a,b ESNT, then a,b es and a,b ET. Since S&T are subfields, each containing elements a and b, it follows that ab=ba ES and ab=ba ET. Hence, ab=ba ESOT and the elements in the intersection are commutative under multiplication. Associative under addition: If a,b,c EF, then (a+b)+c=a+(b+c). That is the elements a,b,c EF are associative under addition. If a,b,c Esnt, then a,b,ces and a,b,cET. Since S&T are subfields, each containing elements a, b, c it follows that (a+b)+c=a+b+c) ES and (a+b)+c=a+(b+c) ET. Hence, (a+b)+c=a+(b+c)ESNT and the elements in the intersection are associative under addition. Associative under multiplication: If a,b,c EF, then (axb)xc=ax(bxc). That is the elements a,b,c EF are associative under multiplication.

If a,b,c ESNT, then a,b,c es and a,b,c ET. Since S&T are subfields, each containing elements a, b, c it follows that (axb)xc=ax(bxc) ES and (axb)xc=ax(bxc) ET. Hence, (axb)xc=ax(bxC)ESNT and the elements in the intersection are associative under multiplication. Distributive: If a,b,c EF, then a(b+c)=ab+ac. That is the elements a,b,c EF are distributive. If a,b,c ESOT, then a,b,c ES and a,b,cet. Since S&T are subfields, each containing elements a, b, c it follows that alb +c)=ab+ac ES and a(b+c)=ab+ac ET. Hence, a(b+c)=ab+acesnT and the elements in the intersection are distributive. If a,b,c EF, then ab+acra(b+c). That is the elements a,b,c EF are distributive. If a,b,c ESOT, then a,b,ces and a,b,c ET. Since S&T are subfields, each containing elements a, b, c it follows that ab +ac=a(b+c) ES and ab+ac=a(b+c) ET. Hence, ab+ac=a(b+c)ESNT and the elements in the intersection are distributive. All these axioms above show that SOT is also a field.