How do I show that $\mathbb{Z}_4\times{}\mathbb{Z}_2/K$ is isomorphic to $\mathbb{Z}_2\times{}\mathbb{Z}_2$, where $K=\langle(2,0)\rangle$.
I'm getting confused with the details involved here, I will attempt using what I know. If you do answer, please could you please not assume I know why each step works? Thanks.
So $$K=\langle(2,0)\rangle=\{\dots,(-4,0),(-2,0),(0,0),(2,0),(4,0),\dots\}$$ $$=\{(\overline0,\overline0),(\overline2,\overline0)\},\text{ with equivalence classes w.r.t to mod 4 and then mod 2 in the pairs respectively.}$$
Can I say this is then isomorphic to $=\{(\overline0,\overline0),(\overline1,\overline0)\}$ w.r.t mod $2$ for both parts in the pair, hence is isomorphic to $\{0\}\times\mathbb{Z}_2$? which is isomorphic to $\mathbb{Z}_2.$
How would I continue / are my steps ok so far?
edit: (definition requested)
So your $K=\{\left([0]_4,[0]_2\right), \,\, \left([2]_4,[0]_2\right)\}$ (henceforth I will drop the subscripts). The quotient group is given by \begin{align*} (\mathbb{Z}_4 \times \mathbb{Z}_2)/K& =\left\{(a,b)+K \, | \, a \in \mathbb{Z}_4, b \in \mathbb{Z_2}\right\} \end{align*} But note that $(0,0)+K=K$, $(2,0)+K=K$, and $(3,0)+K=(1,0)+K$ etc. thus \begin{align*} (\mathbb{Z}_4 \times \mathbb{Z}_2)/K& =\left\{K,\,\, (0,1)+K,\,\, (1,0)+K,\,\, (1,1)+ K\right\} \end{align*} To show precisely that this is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$ we need to find an isomorphism $f:(\mathbb{Z}_4 \times \mathbb{Z}_2)/K \longrightarrow \mathbb{Z}_2 \times \mathbb{Z}_2$ (as @Shaun said you can directly use FIT for this from $\mathbb{Z}_4 \times \mathbb{Z}_2 \rightarrow \mathbb{Z}_2 \times \mathbb{Z_2}$ and show that the kernel is $K$).
Or you can proceed as follows (which is just rehashing of FIT in a sense).
Let $f((a,b)+K)=([a]_2,[b]_2)$. Now see if this $f$ satisfies all the characteristics of an isomorphism.