Isomorphisms between C4 & Z4 Groups

[JackNicholson]

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?

 

[pasmith]

You can write your mappings more easily as \psi^{-1}(n) = r^{n} and \phi^{-1}(n) = r^{-n}.