Is the function $f:\mathbb{Z}_{8}\rightarrow\mathbb{Z}_2$ where $f([x]_8)=[x]_2$ a group morphism? Let $[x]_n$ denote the equivalence class of $\mathbb{Z}_n$ that contains $x$. 
I am not certain where to begin with this question:
$f:\mathbb{Z}_8\rightarrow \mathbb{Z}_2$, where, $f([x]_8)=[x]_2$.
I am not sureif the following is true:  $$f([x]_8\cdot [y]_8)=[x]_2\cdot [y]_2 = f([x]_8)\cdot f([y]_8)$$. 
I am not certain what the binary operation is and as such I do not know if it is a group morphism. I am trying to understand group morphisms more fully and any help would be appreciation. If I am missing something fundamental, please let me know. 
 A: Since you're talking about $\Bbb Z_8$ as a group, I assume the operation is addition. We have:
$$f([x]_8 + [y]_8) = f([x + y]_8) = [x+y]_2 
$$
and 
$$
 f([x]_8) + f([y]_8) = [x]_2 + [y]_2 = [x + y]_2
$$
and we see that these two are equal. Thus we only need to see that inverses are preserved for it to be a homomorphism in full. This is easier to prove indirectly by observing that all odd numbers are sent to $[1]_2$ and all even numbers to $[0]_2$. Since all elements of $\Bbb Z_8$ have the same parity as their inverse, we are done.
A: The groups $\mathbb{Z}_n$ (for $n$ a positive integer) denotes the group on the set $\{0,1,2,\dots, n-1\}$ with binary operation given by addition modulo $n$. 
In order to determine if this is a morphism, you need to show that $$f([x]_8 +_8 [y]_8) = f([x]_8) +_2 f([y]_8)$$ where $+_n$ denotes addition modulo $n$. 
My suggestion is to divide your proof into cases. Consider what the results of taking an integer modulo 2 could yield. We are are going to explore three cases of numbers in $\mathbb{Z}_8$: two evens, two odds, or 1 even and 1 odd. With each case, see if you can prove the homomorphism property. 
