Determine whether or not φ is a homomorphism: Let φ : Z6 -> Z2
be given by φ (x) = the remainder of x when divided by 2, as in the
division algorithm.
I know that this is the answer:
Solution: 
It is a homomorphism: φ (a + b) = φ (a) + φ (b) for all
a, b in Z6.

Now, I just don't get it.  First, why this solution use a + b instead of (1 + 2) when we do have numbers? such as Z6 and Z2?  
Also, how do I know when to use (a + b) or (a*b)? What could be another way to do it?( I am asking this because I saw other examples using (a * b) ) 
What about if the question is Let φ : Z12 -> Z2 or Let φ : Z21 -> Z2 should I get the same answer with (a + b) because it is given by φ (x) = the remainder of x when divided by 2
I hope you can understand my question because everything look so confuse to me.
Thanks,
 A: Okay, $Z_6$ and $Z_2$ are groups.  We can use notation $*$ or $+$.  They are just abstract notations so it doesn't matter which one we use.  A group only has one binary operation no matter what we call it.  For convenience and familiarity we use "+" because... well, we are doing modulo arithmetic.
You use a, b because they are variables and can stand for any numbers.  If we used specific number we'd have to list every possible combination.  That's 36 specific calculations.  Why do that when we can use variable place holders to show all at once?
To show that $\phi$ is a homomorphism  we have to show $\phi(a + b) = \phi(a) + \phi(b)$.
I'll leave it to you to figure out if $\phi:Z_{12} -> Z_2$ is a homomorphism.  But I will show you that $\phi: Z_{21} -> Z_2$ is not because $\phi(17 + 4) = \phi(0) = 0$ (because 21 = 0 mod(21)) but $\phi(17) + \phi(4) = 1 + 0 = 1 \ne 0$.  So it isn't a homomorphism.
A: A group is a set, S, associated with a binary operation, *, such that one of the elements, e, acts as an identity element.
$Z_6$ is a group.  It is the set {0, 1, 2, 3, 4, 5}.  There is an operation, +, that is defined a:
a + b = a + b if a + b < 6; or a + b = a + b - 6 if a + b >= 6.
$Z_6$ has an identity element of 0 (that means 0 + a = a + 0 = a for all a in $Z_6$.
=====
$Z_2$ is a group.  $Z_2$ = {0, 1}.  $Z_2$ has a binary operator, +, that is defined as 0 + 0 = 0;  0 + 1 = 1; 1 + 0 = 1; 1+ 1 = 0;
+++++++++++++++
Now... we can rewrite each element of $Z_6$ as the following:
0 = 2*0 + 0;  0 has a remainder of 0 when divided by 2.
1 = 2*0 + 1;  1 has a remainder of 1 when divided by 2.
2 = 2*1 + 0;  0 has a remainder of 0 when divided by 2.
3 = 2*1 + 1;  1 has a remainder of 0 when divided by 2.
4 = 2*2 + 0;  0 has a remainder of 0 when divided by 2.
5 = 2*1 + 1;  1 has a remainder of 0 when divided by 2.
We can generalize this as each element can be written as a = 2*c + i where c = { 0, 1, 2} and i = {0, 1}.  The remainder when a is divided by 2 is i.
So we define $\phi(a)$ = the remainder when divided by 2.  So $\phi(2*c + i) = i$. 
And we note that $i \in Z_2$ and so $\phi: Z_6 \rightarrow Z_2$.
But that's not enough.  We need to show that the group operation is preserved.  We need to show that $\phi(a + b) = \phi(a) + \phi(b)$.
So let $a = 2*c + i$ (c = 0, 1, 2 and i = 0, 1) and let $b = 2*d + j$ (ditto).
So 
Case 1:  $i = 0; j = 0$.
$a = 2*c; b= 2*d$ so $a + b = 2*c + 2*d = 2(c + d)$ if $2c + 2d < 6$; $a+b = 2*c + 2*d -6 = 2(c + d - 3)$ if $2c + 2d \ge 6$  The remainder when divided by 2 is 0.
So $\phi(a + b) = \phi(a) + \phi(b)$.
Case 2: $i = 1; j= 0;
$a = 2*c + 1; b = 2*d;  $a + b = 2(c+d) + 1$  $\phi(a) = 1; \phi(b) = 0$, $\phi(a + b) = 1. 
Case 3: $i = 0; j = 1;
$a = 2c; b = 2*d + 1;$ $\phi(a + b) = \phi (2(c + d) + 1) = 1 = 0 + 1 = \phi(a) + \phi(b)$.
Case 4: $i = 1; j = 1;
$a = 2c + 1; b= 2d + 1$  $a + b = 2(c + d) + 2 = 2(c + d + 1)$ remainder is 0.
So $\phi(a + b) = 0$  $\phi(a) + \phi(b) = 0 + 0 = 0$.
So $\phi$ is a homomorphism.
