Proving isomorphism on additive group $(\Bbb{Z}_4,+)$ and multiplicative group $(\Bbb{Z}_5^*, \times)$ The question I'm having problems with involves proving the above groups are isometric. Therefore, I have to prove they are bijective (1-1, and onto) and homomorphic.  I have done the group operations table for each and come up with:
Set $A = (\mathbb{Z}_4,+) = {0,1,2,3}$
Set $B = (\Bbb{Z}_5^*, \times) = {1,2,3,4}$
So $B$, can be mapped from $A$ with the function:    $\alpha(x)=x+1$
We can see no element of $B$ is the image of more than one element in $A$, therefore we have proven a 1-1 correspondence, and onto.
Now I have to prove they are isomorphic by exhibiting a 1-1 corresponds α between their elements such that:
$a+b \equiv c\ (\text{mod}\ 4)$ if and only if $\alpha(a) \cdot \alpha(b)\equiv \alpha(c)(\text{mod}\ 5)$
I'm stuck here... do I just plug in all the possible values of a, b, and c?  I suppose there should be 4 ways to do this... 
As an example: 
$a = 1, b = 2, c = 3$
$1 + 2 = 3(\text{mod}\ 4)$
$3\equiv 3(\text{mod}\ 4)$
and
$\alpha(1) \cdot \alpha(2) \equiv \alpha(3)(\text{mod}\ 5)$
$2\cdot 3 \equiv 4(\text{mod}\ 5)$
$6\equiv 4(\text{mod}\ 5)??????$ 
I feel like I'm missing a key concept here. Been watching a ton of videos, but I'm missing something!
 A: Your error is right at the start, with the map $x \to x+1$. That is a bijection, but not a group isomorphism since it does not respect the operations in the group.
In your map $0$ must go to $1$ (can you see why?). Then try to find a place for $1$ to go. Once you decide on that, the group operations will force the rest of the map. So experiment.  
A: Define your function as $$0\to 1$$
$$1\to 2$$
$$2\to 4$$
$$3\to 3$$
and everything works fine.
Not every one-to-one function is an isomorphism.
A: Hint:
Any group of order $4$ is isomorphic to $(\Bbb Z_4,+)$ or to the Klein Vierergruppe.  In the latter, the square of any element is the identity.  
A: Let $\alpha(.)$ be the mapping from $(\mathbb{Z_4},+)$ to $(\mathbb{Z^*_5},\times)$
$\alpha(0+0) = \alpha(0) \times \alpha(0) \implies \alpha(0) = 1 $
Therefore $0$ maps to 1
Let $\alpha(1) = a, \alpha(2) = b \implies a^2 = b \implies a = 2 $ and $b = 4$
Therefore 1 maps to 2 and 2 maps to 4. Similar operations for 3 = 2 + 1 will show that 3 maps to 3.
