- #1
JackNicholson
- 1
- 1
0
Hint: Show that the isomorphism preserves the order of the element
My solution:
C4 = {e,r,r^2,r^3} where e-identity element and r is rotation by 90°
Z4 = {0,1,2,3}
LEMMA:
! Isomorphism preserves the order of the element !
(PROOF OF IT)Now we calcuate the order of the elements of both groups.
ord(e)=1 -------------- ord(0)=1
ord(r)=4 -------------- ord(1)=4
ord(r^2)=2 ----------- ord(2)=2
ord(r^3)=4 ----------- ord(3)=4
We see that there is 1 element in both groups with order equal 1, 2 elements with order equal 4 and 1 element with order equal 2.
So we can write 2 mappings:
ψ : C4 -> Z4
ψ(e)=0
ψ(r)=1
ψ(r^2)=2
ψ(r^3)=3
ϕ: C4 -> Z4
ϕ(e)=0
ϕ(r)=3
ϕ(r^2)=2
ϕ(r^3)=1
We can see it clearly that those mapping are bijective.And now how show that those 2 functions are also homomorphism? I know that homomorphism is when: ϕ(xy)=ϕ(x) + ϕ(y) but how show it in this certain case without writing every possible situation?
Hint: Show that the isomorphism preserves the order of the element
My solution:
C4 = {e,r,r^2,r^3} where e-identity element and r is rotation by 90°
Z4 = {0,1,2,3}
LEMMA:
! Isomorphism preserves the order of the element !
(PROOF OF IT)Now we calcuate the order of the elements of both groups.
ord(e)=1 -------------- ord(0)=1
ord(r)=4 -------------- ord(1)=4
ord(r^2)=2 ----------- ord(2)=2
ord(r^3)=4 ----------- ord(3)=4
We see that there is 1 element in both groups with order equal 1, 2 elements with order equal 4 and 1 element with order equal 2.
So we can write 2 mappings:
ψ : C4 -> Z4
ψ(e)=0
ψ(r)=1
ψ(r^2)=2
ψ(r^3)=3
ϕ: C4 -> Z4
ϕ(e)=0
ϕ(r)=3
ϕ(r^2)=2
ϕ(r^3)=1
We can see it clearly that those mapping are bijective.And now how show that those 2 functions are also homomorphism? I know that homomorphism is when: ϕ(xy)=ϕ(x) + ϕ(y) but how show it in this certain case without writing every possible situation?