Group inclusion- Quotiens Inclusion Given a group $G$ , and two subgroups $ G_1, G_2 $ such that $ G_1 \subseteq G_2 $ , 
is it true that $ G/G_2 \subseteq G/G_1 $ ? 
Thanks in advance ! 
 A: What you probably had in mind is the following:
$$G/G_2\cong\left(G/G_1\right)/\left(G_2/G_1\right)$$
by the 2nd or 3d. isomorphism theorem, so what you asked can't generally be as $\,G/G_2\,$ is a quotient of $\,G/G_1\,$ , i.e.: there exists a surjective group homomorphism
$$G/G_1\longrightarrow G/G_2\,\,,\,\,xG_1\to xG_2$$
A: The quotients may not make sense, if $G_1$ and $G_2$ are not normal.
However, it is true that every left-$G_2$ coset is completely contained in a $G_1$ coset. That is, the equivalence relation induced by the larger subgroup is "coarser" than that of the larger group. In other words, $x{}_{G_1}\equiv y$ implies $x{}_{G_2}\equiv y$, but not necessarily conversely.
To see this, simply remember that $x{}_{G_i}\equiv y$ if and only if $y^{-1}x\in G_i$. If $x{}_{G_1}\equiv y$, then $y^{-1}x\in G_1\subseteq G_2$, so $x{}_{G_2}\equiv y$. But if $G_2$ properly contains $G_1$, then let $x\in G_2-G_1$; then $x{}_{G_2}\equiv e$, but $x{}_{G_1}\notequiv e$. 
Similar results hold for congruent-on-the-right (equivalently, right cosets). 
If both $G_1$ and $G_2$ are normal, so that the set of equivalence classes can be given a group structure, then the coarser equivalence relation induces a smaller group, so that $G/G_2$ is a quotient of $G/G_1$. 
