Originally Posted by

**Deveno** Z4 is cyclic of order 4. in Z4, 1 is an element of order 4 which generates the group (3 is also a generator).

if T and Z4 are to be isomorphic, T has to have an element of order 4.

well, it's not e, that just leaves 3 to check. what is the order of x?

x^2 = e, so x is of order 2. let's keep looking.....

y^2 = x. this looks promising....

y^3 = (y^2)y = xy = z. so y is not of order 1,2 or 3...what's left? just to be sure, let's check:

y^4 = (y^3)y = zy = e, so yep, y is of order 4.

a homomorphism has to map identity to identity:

so e--->0 is already chosen for us.

an isomorphism has to send elements to images of the same order, so we have two choices for y:

y-->1, or y-->3. let's pick y--->1. we can now write T as {e,y,y^2,y^3} (since we know y^2 = x, and y^3 = z)

now, convince yourself that sending y^k <---> k (mod 4) is an isomorphism.

the bijection part is easy...why is it a homomorphism (can you think of a way to prove it "in general" without verifying all 16 possible products in T)?