Need isomorphism theorem intuition I am currently trying to understand the isomorphism theorems. The issue I am having is that I am struggling to find a way to think about them. 
In Stillwell's Elements of Algebra, I found a way to understand the first theorem ($\frac{G}{\ker \phi} \simeq \operatorname{Im} \phi$ for any homomorphism $\phi:G\rightarrow G'$). It was proven in terms of set functions (since there is a one-to-one correspondence between the elements $e \in \operatorname{Im} \phi$ and $\phi^{-1}(e)$). However, the second and third theorems are not even shown. Also, my curriculum is taken from Fraleigh's A First Course in Abstract Algebra. There, the first isomorphism theorem is longer than just the isomorphism between $\frac{G}{\ker \phi}$ and $\operatorname{Im}\phi$; He adds that there is a unique isomorphism $\mu: G/\ker\phi \rightarrow \phi[G]$ such that $\phi(x) = \mu(\gamma(x))$ for each $x \in G$. Here, $\gamma$ is the canonical homomorphism from G to $\frac{G}{\ker\phi}$.
I still do not understand what the canonical homomorphism is since it is not in the index of the book. More importantly, I am really having a hard time just thinking about the definitions involved. I can see a proof, and understand why each step is accurate. Yet, it is too abstract for me to process.
This has happened once before when I was learning measure- and integration theory. I couldn't understand anything for months until one day I read a definition of the Lebesgue integral in Euclidean space, and suddenly everything just "clicked"; Within a week I could understand, rather than just know, the entire curriculum. I am hoping for another one of those moments before exam.
I have the same problem now with certain aspects of abstract algebra. They generally involve factor groups and/or mappings. So, I have a suspicion that if I "get" the isomorphism theorems, the rest will connect.
 A: Suppose $\phi:(G, \cdot) \to (G', *)$ is a surjective homomorphism, and put $K = \ker\phi$. The main ideas behind the first homomorphism theorem and the canonical homomorphism $\gamma:G \to G/\ker\phi$, defined by $\gamma(a) = aK$, are:


*

*The "level sets" of $\phi$, a.k.a., the primages of singleton sets, are precisely the (left) cosets of $K$, because if $a$ and $b$ are elements of $G$, then
\begin{align*}
\phi(a) = \phi(b) \quad&\text{if and only if} && \phi(a^{-1} \cdot b) = e', \\
&\text{if and only if} && a^{-1} \cdot b \in K, \\
&\text{if and only if} && a^{-1} \cdot bK = K, \\
&\text{if and only if} && aK = bK.
\end{align*}

*The formula $\mu(aK) = \phi(a)$ determines a (single-valued/well-defined) mapping $\mu:G/K \to G'$, and $\mu$ is bijective. (If $aK = bK$, then $\mu(aK) = \phi(a) = \phi(b) = \mu(bK)$, so $\mu$ is well-defined. Conversely, if $\mu(aK) = \mu(bK)$, then the preceding chain of logical equivalences gives $aK = bK$, i.e., $\mu$ is injective. Finally, $\mu$ is surjective because $\phi$ is surjective.)

*Coset multiplication on $G/K$ corresponds with the operation of $G'$:
$$
\mu(aK) * \mu(bK)
  = \phi(a) * \phi(b)
  = \phi(a \cdot b)
  = \mu\bigl((a \cdot b)K\bigr).
$$
Conceptually, it may help to think of $\phi$ as a pair of dark sunglasses; when you look at $G$ through $\phi$, you can't see individual elements $a$, only cosets $aK$. The mapping $\mu$ gives the coset $aK$ the "label" $\phi(a)$. That is, $\phi$ makes $G$ look like $G/K$, which after relabeling by $\mu$ looks identical to $G'$ (with regard to group structure).
A: If I'm not mistaken, the canonical homomorphism maps an element to its equivalence class. My intuition with quotient/factor groups is that it's very much like regular division. When you take, for example, 6/3 one way to think about this is that you're seeing what's left when you take every 3 that can be in 6 and collapse them to 1. You can express 6 as 3 and 3 which "collapses" to 1 and 1. If you did that the exact same thing with a group though you'd get just a bunch of identity elements. The homomorphism in the isomorphism theorems is a map that lets you distinguish each cluster you're collapsing so each piece isn't ambiguous. Hope that helps!
