Are these groups isomorphic

Group G = {1,-1,i,-i} where i^2 = -1
Group a = {[0],[1],[2],[3]}
Group b = {([0].[0]),([0],[1]),([1],[0]),([1],[1])…

From my Cayley table, I came up with the answer that G and a are isomorphic and G and b are are isomorphic.

Is this correct?

no, but it is helpful to see WHY you are incorrect.

note that G is cyclic, with generator i:

= {i,i^2,i^3,i^4} = {i,-1,-i,1} because:

i^2 = -1, i^3 = i(i^2) = i(-1) = -i, i^4 = (i^2)^2 = (-1)^2 = 1.

note that A = {[0],[1],[2],[3]} is also cyclic, with generator [1].

let's PROVE that G and A are isomorphic. how do we do this?

we exhibit an isomorphism, that'll do it. remember an isomorphism must be 2 things:

1) a homomorphism
2) a bijective function.

here is my "proposed" isomorphism: φ:A-->G given by φ([k]) = i^k. it is clear we have a bijection:

[0]--->i^0 = 1
[1]--->i^1 = i
[2]--->i^2 = -1
[3]--->i^3 = -i

note that if m is in [k], then m = k + 4n, so i^m = i^(k+4n) = (i^k)(i^(4n)) = (i^k)(i^4)^n

= (i^k)(1^n) = (i^k)(1) = i^k, so this mapping is well-defined, it depends only on [k], not on k.

but...is φ a homomorphism? we must show that:

φ([k]+[m]) = (φ([k]))(φ([m])).

now φ([k] + [m]) = φ([k+m]) = i^(k+m) = (i^k)(i^m) = (φ([k]))(φ([m])), so yes, φ

is indeed an isomorphism.

now, how can we PROVE B = {([0],[0]),([0],[1]),([1],[0]),([1],[1])… is not isomorphic to G?

let's take a detour:

LEMMA: if G and H are isomorphic, and G is cyclic, then H is cyclic.

proof: suppose φ is our isomorphism, and that g generates G. i claim φ(g) generates H.

we can write ANY member of G as g^k for some integer k. now, φ is onto,

so every element h in H is the image of some element in G, that is h = φ(g^k).

but φ is a homomorphism, so φ(g^k) = [φ(g)]^k, so φ(g) generates H.

now...let's prove B isn't cyclic. that will settle the matter.