Need isomorphism theorem intuition

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).

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!