I solved many questions about factor groups on Gallain's book. One question was
Let $G=\dfrac{\mathbb{Z}}{\langle 20\rangle}$ and $H=\dfrac{\langle 4\rangle}{\langle 20 \rangle}$. List the elements of $H$ and $\dfrac{G}{H}$.
I solved this question either. But then some question come to my mind, here is my question: What if let $G=\dfrac{\mathbb{Z}}{\langle 10\rangle}$ and $H=\dfrac{\langle 10\rangle}{\langle 20 \rangle}$. List the elements of $H$ and $\dfrac{G}{H}$
$$G=\dfrac{\mathbb{Z}}{\langle 10\rangle}=\lbrace a\in\langle 10\rangle:a\in \mathbb{Z} \rbrace=\lbrace 0+\langle10\rangle, 1+\langle10\rangle,...,9+\langle10\rangle \rbrace$$ and $$H=\dfrac{\langle 4\rangle}{\langle 20\rangle}=\lbrace a\in\langle20\rangle:a\in \langle 20\rangle \rbrace=\lbrace 0+\langle20\rangle, 4+\langle20\rangle,...,16+\langle20\rangle\rbrace$$
However, $$\dfrac{G}{H}=\lbrace a + H: a\in G \rbrace = \lbrace a+ \dfrac{\langle4\rangle}{\langle20\rangle}:a\in\ \dfrac{\mathbb{Z}}{\langle10\rangle} \rbrace = \lbrace (0+\langle 10\rangle)+\dfrac{\langle 4\rangle}{\langle 20\rangle}, (1+\langle 10\rangle)+\dfrac{\langle 4\rangle}{\langle 20\rangle},...,(9+\langle 10\rangle)+\dfrac{\langle 4\rangle}{\langle 20\rangle} \rbrace$$
But I couldn't reduce anymore. By the way I know that this must be isomorphic to $\mathbb{Z}_5$ since: $$\dfrac{\mathbb{Z}}{\langle 10 \rangle} \cong \mathbb{Z}_{10}, \;\; \dfrac{\langle 10 \rangle}{\langle 20\rangle}\cong \mathbb{Z}_2$$ Hence $$\dfrac{\mathbb{Z}_{10}}{\mathbb{Z}_2} \cong \mathbb{Z}_5$$
However, I couldn't show with definition where is the mistake or the part that I should do. Thanks in advance! Please avoid using first, second and third isomorphism theorems.
The factor group $\frac{G}{H}$ only exists if $H$ is a subgroup of $G$ (a normal subgroup, if $G$ is non-commutative). This is not the case in your example. The element $0+\left\langle20\right\rangle$, for example, is in $H$ but not in $G$.