Given:$$\Bbb Z_8 =\{0,1,2,3,4,5,6,7\}$$ and $$H=\{0,2,4,6\} = \left<[2]\right>$$, construct the factor group $\frac{\Bbb Z_8}{H}.$
Definition:
For a normal subgroup $H$ of $G$, the group $\frac{G}{H}$ is called the factor-/ or quotient group of $G$ modulo $H$. The set $\frac{G}{H}$ = {$aH | a \in G$} (left-coset) is a group under the operation $(aH)(bH)=abh$.
\begin{align}\\
\frac{\Bbb Z_8}{H}
& =\{gH|g\in \Bbb Z_8\} \\
& =\{g+H|g\in \Bbb Z_8\}
&\text{with }
& g+H=\{g+h|h\in H\}\\
\frac{\Bbb Z_8}{H}
& =\{H, 1+H, 2+H, …, 7+H\}\\
\end{align}
So we have:
\begin{align}\\
& 0+H=\{0,2,4,6\}&& 4+H=\{4,6,8,10 \}\\
& 1+H=\{1,3,5,7\}&& 5+H=\{5,7,9,11 \}\\
& 2+H=\{2,4,6,8\}&& 6+H=\{6,8,10,12\}\\
& 3+H=\{3,5,7,9\}&& 7+H=\{7,9,11,13\}\\
\end{align}
Taking modulo 8, we get:
\begin{align}\\
&0+H\mod 8=\{0,2,4,6\}&& 4+H\mod 8=\{4,6,0,2\}\\
&1+H\mod 8=\{1,3,5,7\}&& 5+H\mod 8=\{5,7,1,3\}\\
&2+H\mod 8=\{2,4,6,0\}&& 6+H\mod 8=\{6,0,3,4\}\\
&3+H\mod 8=\{3,5,7,1\}&& 7+H\mod 8=\{7,1,3,5\}\\
\end{align}
We see that:
\begin{align}\\
(1+H)\mod 8 =(3+H)\mod 8 =(5+H)\mod 8 =(7+H)\mod 8 &&=\{1,3,5,7\}
\\
(2+H)\mod 8 =(4+H)\mod 8 =(6+H)\mod 8 &&=\{0,2,4,6\}
\end{align}
I calculated the cosets of subgroup $H=\{0,2,4,6\}=\left<[2]\right>$ of $\Bbb Z_8$:
\begin{align}\\
& 0+\left<[2]\right> & =\{0,2,4,6\} \\
& 1+\left<[2]\right> & =\{1,3,5,7\} \\
\end{align}
I don't know how to proceed from here, or even if I'm on the right track. I thougt I would end up with a set of pairs, but I keep ending up calculating the same two sets.
Best Answer
Those two cosets do indeed form $\Bbb Z_8/H$ (under the operation $(a+H)+(b+H):=(a+b)+H$). Well done! You could improve your answer by noting that $0+H=H$.