Intuition behind quotient groups

Consider quotienting group $𝐺$ by an equivalence relation $\sim$, the proposition § Proposition 7.4 of Aluffi tells us that

$$𝑎\sim 𝑏⟺𝑎𝐻=𝑏𝐻$$

where $𝐻$ is a subgroup of $𝐺$ as well as the equivalence class of ${𝑒}_{𝐺}$.

The proof is rather easy but I’d like the carve out the intuitions behind this. Namely what does it mean for $𝑎\sim 𝑏$ makes the left coset of $𝐻$ identical?

First consider $\mathbb{Z}/3\mathbb{Z}$, the $𝐻$ group of this group is $3\mathbb{Z}$, so consider the element $\left[1\right]=\left\{\dots ,1,4,7,10,\dots \right\}$, clearly we have $1+3\mathbb{Z}=4+3\mathbb{Z}=10+3\mathbb{Z}$. This is a concrete example of the above principle. Notice that here we have $4-1=3\in 3\mathbb{Z}$, similarly $7-1=6\in 3\mathbb{Z}$. So intuitively we have some sense that $𝐻$ represents the relative displacement between to elements that are considered identical w.r.t. to the relation $\sim$.

What does this mean? First we need to know that multiply a coset by an element in the equivalence class permutes the equivalence class, as we can see from the example $1+3\mathbb{Z}=4+3\mathbb{Z}=7+3\mathbb{Z}$. Secondarily, we need to notice that if $𝑏\sim 𝑎$, then ${𝑎}^{-1}𝑏\sim 𝑒$, put in $\mathbb{Z}/3\mathbb{Z}$, this means that $4-1\sim 7-1\sim 0$. Which hints us that ${𝑎}^{-1}𝑏$ represents the relative displacement between two elements $𝑎$ and $𝑏$. And if that displacement is considered to be “equivalent” to $0$, then naturally $𝑎$ and $𝑏$ are considered equivalent.

So what does $𝐻$ do? $𝐻$ happens to record all “displacements” that are “equivalent” to $0$ (or in general, ${𝑒}_{𝐺}$). We call such displacements “null displacements” . So if $𝑎\sim 𝑏$, then we must have $𝑎$ and $𝑏$ differed by a null displacement, and since $𝐻$ records all such displacements, we can say that $𝑎$ and $𝑏$ differed by an element of $𝐻$. So multiplying $𝑎$ by any element in $𝐻$ yields an element ${𝑎}^{\prime }\in 𝑎𝐻$ congruent to $𝑎$, and similarly, multiplying $𝑏$ by any elements of $𝐻$ must yield an element ${𝑏}^{\prime }\in 𝑏𝐻$ congruent to $𝑏$, as $𝑎$ and $𝑏$ are congruent at very first, by transitivity of the equivalence relation we have<intro:null-displacement>

$${𝑎}^{\prime }\sim 𝑎\sim 𝑏\sim {𝑏}^{\prime }$$

therefore ${𝑎}^{\prime }\sim {𝑏}^{\prime }$, this means that we must have $𝑎𝐻=𝑏𝐻$.

From the intuition perspective, we need to use the previous fact that multiplying $𝐻$ by two elements in the same equivalence class only permutes the class. So since originally $𝑎\sim 𝑏$, which means $𝑎$ and $𝑏$ are in the same equivalence class, therefore naturally $𝑎𝐻$ and $𝑏𝐻$ are just the permutations of the same class.

From another perspective, as $𝐻$ contains all “null displacements”, multiplying it by two already equivalent elements $𝑎$ and $𝑏$ must feels like “staying still”, that is, if let $𝑎$ “displaced” by $ℎ\in 𝐻$, then there must be an element ${ℎ}^{\prime }\in 𝐻$ such that displacing $𝑏$ by ${ℎ}^{\prime }$ gives the same result as displacing $𝑎$ by $ℎ$. Because otherwise ${ℎ}^{\prime }$ wouldn’t be a “null displacement” as it causes actual differences that distinguishes $𝑎$ and $𝑏$ w.r.t. the relation $\sim$. You may think of this as elevators, different equivalence classes correspond to different elevators, and the floors in each elevator corresponds to different elements in the same equivalence class. The elevator mimics the equivalence relation in the sense that if you look from above, then no matter how a person move up or down in the same elevator, it looks exactly the same to you. And each person in an elevator can move to any floor in that elevator by moving up or down (this is the analogous of the “null displacement”, moving up or down here are null displacements because they don’t affect the person you looked from above.) And the only way a person cannot reach a specific floor is because it is at another elevator. Now, consider $𝑎$ a person, $𝑎𝐻$ is all the floor it can reach from in his elevator, now $𝑎𝐻=𝑏𝐻$ means that, no matter $𝑎$ reach which floor, $𝑏$ can reach that floor too by moving up or down for possibly different number of floors compared to $𝑎$, but never the less he must be able to reach it, because the only possible way he can’t reach it is because he’s inside another elevator, which is impossible because we assume $𝑎\sim 𝑏$ at very start, that they initially must be in the same elevator.