Thread: Group ismorphism (Z2xZ4)

Hi,

Can you show that (Z2xZ4,+)/N isomorphic to (Z4,+)?
(N=<(1,2)>)

My solution is: N={(0,0),(1,2)} and f((a,b))->x (mod4) (a(mod2),b(mod4))

But f(0,0)=f(2,4)=f(1,2)+f(1,2) so f(1,2)=2(mod4). Hence N is not kerf so i cannot use first ismorphism theorem. Am I wrong?

it suffices to find a homomorphism from Z₂xZ₄ onto Z₄ whose kernel is N (do you understand why? this is "the hard part").

let's look at the cosets of Z₂xZ₄/N:

we have N = {(0,0),(1,2)}, the identity

we have (0,1) + N = {(0,1),(1,3)} (this is also (1,3) + N)

we have (0,2) + N = {(0,2), (1,0)} (which is (1,0) + N, as well)

we have (0,3) + N = {(0,3), (1,1)} (which is (1,1) + N).

so if we map (0,k) + N ---> k, it appears we will have an isomorphism.

so it looks like we want a mapping that sends (0,k) ---> k and (1,k) ---> k+2 (mod 4) (there are other possible homomorphisms).

can we combine these two into a single condition?

how about sending (k,m) ---> m + 2k (mod 4)?

now verify that f(k,m) = m + 2k (mod 4) is indeed an onto homomorphism, and that ker(f) = N.

Originally Posted by Deveno

it suffices to find a homomorphism from Z₂xZ₄ onto Z₄ whose kernel is N (do you understand why? this is "the hard part").

let's look at the cosets of Z₂xZ₄/N:

we have N = {(0,0),(1,2)}, the identity

we have (0,1) + N = {(0,1),(1,3)} (this is also (1,3) + N)

we have (0,2) + N = {(0,2), (1,0)} (which is (1,0) + N, as well)

we have (0,3) + N = {(0,3), (1,1)} (which is (1,1) + N).

so if we map (0,k) + N ---> k, it appears we will have an isomorphism.

so it looks like we want a mapping that sends (0,k) ---> k and (1,k) ---> k+2 (mod 4) (there are other possible homomorphisms).

can we combine these two into a single condition?

how about sending (k,m) ---> m + 2k (mod 4)?

now verify that f(k,m) = m + 2k (mod 4) is indeed an onto homomorphism, and that ker(f) = N.

Is my solution wrong?

you don't exactly say what your solution IS.

you write f((a,b)) = x (mod 4) <---what is x????