Consider $f: \mathbb Z_8 \rightarrow \mathbb Z_4$. List the cosets of the kernel of $f$ Consider $f: \mathbb Z_8 \rightarrow \mathbb Z_4$ given by 
$$\begin{pmatrix} 0&1&2&3&4&5&6&7 \\ 0&1&2&3&0&1&2&3 \end{pmatrix}$$
Verify the $f$ is a homomorphism, find its kernel $K$, and list the cosets of $K$.
I have verified that this is a homomorphism by a Cayley Table, and the kernel $K=\{0,4\}$.  
Now it says list the cosets of $K$. I am not sure of this.
Is the coset of $K$ the elements that $f$ does not map into the identity, i.e.: $\{1,2,3,5,6,7\}$?
Or, since the kernel of $f$ is a normal subgroup of the domain, in this case $\mathbb Z_8$ (i.e.: $K$ is subgroup of $\mathbb Z_8$), are the cosets of the kernel equal to $Ka$, where $a$ is any element of $\mathbb Z_8$?
 A: Hint: Note that the permutation corresponds to the map
$f:a\to a(\mod 4)$
Clearly, Kernel of the map is elements of $\mathbb{Z}_8$ which map to $0$ which are $0$ and $4$, which you have already written. $K=\{ 0,4\}$
What are cosets? Suppose $N$ is a normal subgroup of $G$
Then $G/N=\{ a\oplus_8N : a\in G\}$. Note $\oplus_8$ is addition modulo 8.
The distinct cosets are
$\mathbb{Z}_8/K=\{ K,1\oplus_8K,2\oplus_8K,3\oplus_8K\}$
The number of distinct cosets is given by $|\mathbb{Z}_8|/|K|$
Note 
$K=\{ 0,4\}$
$1\oplus_8K=\{ 1,5\}$
$2\oplus_8K=\{ 2,6\}$
$3\oplus_8K=\{ 3,7\}$
A: Since $|ker|=2$, so there will be $4$ elements in the coset. Now follow the usual rule for forming cosets of a subgroup, i.e., $aH=\{ah\mid a\in G\}$. Thus for this case the coset corresponding to the element $0,4$ is $K$ itself. For example, for $1$ its $1+K=\{1,5\}$. In a similar way cosets for other elements of $\mathbb{Z}_8$ can be formed, keeping in mind the operation on $\mathbb{Z}_8$. See that the cosets corresponding to $1$ and $5$ and $3$ and $7$ are the same.  
