Quotients of subgroups "inclusion reversing". Let $G$ be abelian group.
Let $H_1\subseteq H_2$ be subgroups of $G$. Is it true that $G/H_1 \supseteq G/H_2$?
For example $\mathbb{Z}/2\mathbb{Z}=\{0,1\}\subseteq\mathbb{Z}/4\mathbb{Z}=\{0,1,2,3\}$, and $4\mathbb{Z}\subseteq 2\mathbb{Z}$.
Thanks.
Thanks to 5xum who pointed out that strictly speaking as sets it is false. I am still interested in related theorems (up to isomorphism) or something like that. Thanks.
 A: No, it’s the other way around: There is a natural surjective map $f \colon G/H_1 → G/H_2$ given by enlarging cosets $gH_1 ↦ gH_2$. However, the axiom of choice asserts that there is a right inverse $r \colon G/H_2 → G/H_1$ to $f$ – that is, a map $r$ fulfilling $f∘r = \mathrm{id}_{G/H_2}$. Such a map has to be injective. The map $r$ is a choice of representatives for preimages under $f$.
So you can find some sort of embedding $G/H_2 → G/H_1$, but it highly relies on choice. The inclusion that you have given “$ℤ/2ℤ ⊆ ℤ/4ℤ$” is no literal inclusion (as already pointed out), and relies on choosing representatives of the preimages of the natural surjection
$$ℤ/4ℤ → ℤ/2ℤ,~0 ↦ 0,~1↦1,~2↦0,~3↦1.$$
Another remark. Note that if $H_1$ and $H_2$ are no normal subgroups, $G/H_1$ and $G/H_2$ have no group structure and may (for now) only be viewed as sets. If $H_1$ and $H_2$ are normal, then a choice of representatives $r \colon G/H_2 → G/H_1$ need not be a group homomorphism – in fact, there even might be no such choice $r$ that is also a group homomorphism!

This might be unnecessary, but here’s a bit more elaboration on the well-definedness of $f$.
Behind all these well-known isomorphism theorems lies a simple easy-to-show set-theoretic fact:

Let $M$, $\bar M$ and $N$ be sets, $π \colon M → \bar M$ a surjective map and $f \colon M → N$ be any map.  Then, if for all $x, y ∈ M$
  $$π(x) = π(y) \implies f(x) = f(y),$$
  there is a unique map $\bar f \colon \bar M → N$ such that $f = \bar f ∘ π$. (Furthermore, $\bar f$ is injective if and only if for all $x, y ∈ M$ you have $π(x) = π(y) ⇔ f(x) = f(y)$. And if $f$ is surjective, so is $\bar f$.)

Now consider $M = G$, $\bar M = G/H_2$ and $π \colon G → G/H_2,~g ↦ gH_2$ as well as $f \colon G → G/H_1,~g ↦ gH_1$. For any $g, g' ∈ G$,
$$π(g) = π(g') ⇒ gH_2 = gH_2 ⇒ gH_1 = gH_1 ⇒ f(g) = f(g'),$$
so the existence of $f$ as above is guaranteed.
