About how to understand the third isomorphism theorem What is a good way to understand the intuition behind the third isomorphism theorem? Is it something looking like zooming out of the group structure?(i.e. discard the detailed info by modding out a normal subgroup).
 A: The following diagram is one classical way that the theorem is frequently  "illustrated" (excerpted from Burris and Sankappanavar's A Course in Universal Algebra). Essentially the diagrams illustrate the set-theoretic view of the matter, in terms of equivalence classes or, equivalently, partitions. Before studying the algebraic versions of these theorems, for motivation, it helps to have a good grasp on their much simpler set-theoretic versions, i.e. "forget" the algebraic structure (operations), so reducing from algebras and their maps (homs), to simply sets and functions. Then a congruence reduces to an equivalence relation (no operations to preserve), i.e. a partition, which leads to the vivid depiction below, in terms of finer and coarser partitions.


A: A very silly (and not "mathematically" rigorous) way I remember the theorem is by examining the analogue in rational numbers. Everyone who knows anything gets that $\dfrac{a/b}{c/b} = \dfrac{a}{b}\cdot\dfrac{b}{c} =\dfrac{a}{c}$. 
We can intuitively think of the Third Isomorphism theorem in the same fashion.
$$\dfrac{A/B}{C/B} ``=" \dfrac{A}{B}\cdot\dfrac{B}{C} \cong A/C$$
Again, not mathematically valid, but a good mnemonic.  
A: The best motivation I have comes from the classical groups:  $\mathbb{Z}$ and $\mathbb{Z}/n\mathbb{Z}$. In this way when you start with $m|n$ so that we have a subgroup relation to talk about
$$\mathbb{Z}/n\mathbb{Z}\bigg/n\mathbb{Z}/m\mathbb{Z}\cong \mathbb{Z}/(n/m)\mathbb{Z}$$
so that things really work like fractions, which is how the notation is setup, and which motivates the idea for more general groups.
This also works if you start with $A=\mathbb{Z}/n\mathbb{Z}$. Then taking $k|m|n$ we have
$$A/kA\bigg/mA/kA\cong A/(m/k)A$$
This is comparable to actual fractions since it looks like
$${n\over k}\bigg/{n\over m}={m\over k}$$
A: Your suggestion is fine.
Another intuition I think is important is that it's something to understand in the context of higher algebra: you're doing arithmetic with groups. Not with elements of groups, but with the groups themselves. The third isomorphism theorem is a basic algebraic identity used either for simplifying quotients of quotients, or conversely to write a quotient as a quotient of quotients as needed.
I often use it in the context of expressing a quotient $G/(HK)$ as first modding out by $H$ and then modding out by $K$, which can be justified as
$$ G / (HK) \cong (G / H) / (HK/H) \cong (G/H) / \pi(K) $$
where $\pi$ is the projection map $G \to G/H$.
I find the third isomorphism particularly appealing if you think in terms of setoids -- that is, you think in terms of sets equipped with equivalence relations, and modding out by a normal subgroup means equipping your group with the corresponding equivalence relation rather than constructing the set of equivalence classes. The third isomorphism theorem then says if you equip things with an equivalence relation, and then a coarser equivalence relation, you get the same result as if you just skipped the intermediate step.
