how to find elements of a factor group, the index and if it is cyclic 

*We know that a subgroup H of an abelian group G is normal because for any a∈G,  aH={ah:h∈H}={ha:h∈H}=Ha. Thus, by Corollary 14.5 of the textbook, the set of all cosets (no matter whether left or right) of a normal subgroup under the coset multiplication is a group G/H, called a factor group of G by H.Based on the these arguments, answer the following:


(a) Find all elements of the factor group ℤ/4ℤ.
(b) Find the index of ℤ by 4ℤ, that is, (ℤ:4ℤ).
(c) Is ℤ/4ℤ cyclic and why?
I really am lost on this one and don't know where to start. Can someone point me in the right direction please?
 A: First of all, if $H \trianglelefteq G$, and $x+H,y+H \in G/H$, then $x+H = y+H \Leftrightarrow x-y \in H$ (using additive notation).
Two elements $n+\mathbb{Z}, m+\mathbb{Z} \in \mathbb{Z} / 4\mathbb{Z}$ are equal iff $n-m \in 4\mathbb{Z}$, that is, $4 \mid n-m$ or equivalenty $n \equiv m\ \text{mod}\ 4$, thus, the elements of $\mathbb{Z} / 4\mathbb{Z}$ are the congruence classes $\text{mod}\ 4$. This also tells you the index.
What can you say about the structure of $\mathbb{Z} / 4\mathbb{Z}$? Hint: $1+1+1+1 \equiv 0 \ \text{mod}\ 4$.
A: $\mathbb{Z}/4\mathbb{Z}$ is the quotient of $\mathbb{Z}$ by $4\mathbb{Z}$, this is the factor group they are referring to.
So $\mathbb{Z}=\{...,-3,-2,-1,0,1,2,3,...\}$ under addition, and
$4\mathbb{Z}=\{...,-12,-8,-4,0,4,8,12,...\}$ also under addition.
Can you deduce the elements, index, and whether or not the factor group is cyclic from here?
By way of example, to find the cosets, begin with the group $4\mathbb{Z}$ above, and then choose an element not in it from $\mathbb{Z}$ (I choose $1$) and then add it into every element of $4\mathbb{Z}$.
$1+\{...,-8,-4,0,4,8,...\}=\{...,-7,-3,1,5,9,...\}=[1]$
where $[1]$ denotes the equivalence class, modulo $4$, containing $1$. Do the same with $2$ and $3$ and then check that every element of $\mathbb{Z}$ is in either $[0]=4\mathbb{Z}$, $[1]$, $[2]$, or $[3]$.
