Intuition behind quotient groups (2)

This follows the last article. At last article we discussed the intuition behind the cosets. At this article we discuss about the general intuition behind the quotient group.

Follow the ideas of On structural property of free groups.. We still consider first the quotient structure on underlying groups and then we try to complete it w.r.t. the group structure.

The quotient structure on set is simple, it’s just the set of equivalence classes under the equivalence relation $\sim$. Which gives

$$𝐺/\sim =\left\{\left[𝑎\right],\left[𝑏\right],\left[𝑐\right],\dots \right\}$$

Now we must let it respect the group structure, that is, we must define an group operation on $𝐺/\sim$. A natural choice is

$$\left[𝑎\right]\left[𝑏\right]=\left[𝑎𝑏\right]$$

Here we have a genuine question: Why do we have to define the group operation on the quotient structure this way? Why can’t we define it otherwise as long as it makes the set a group? The reason is that inside $\text{𝐒𝐞𝐭}$, the projection

$${𝜋}_{\text{𝐒𝐞𝐭}}:𝐺\to 𝐺/\sim$$

can be whatever set function as long as it’s surjective. But inside $\text{𝐆𝐫𝐩}$, the map

$${𝜋}_{\text{𝐆𝐫𝐩}}:𝐺\to 𝐺/\sim$$

must be a homomorphism, that homomorphism property forces us to have this definition.

However, notice that, since $\text{𝐆𝐫𝐩}$ carries more structure than sets, therefore we are expecting there are more requirements. Indeed, notice the following equivalence relation on $\mathbb{Z}$

$$𝑎\sim 𝑏\iff \left|𝑎\right|=|𝑏|$$

but if we define

$$\left[𝑎\right]+\left[𝑏\right]=\left[𝑎+𝑏\right]$$

that does not necessarily hold, since $1\sim -1$ but

$$\left[2\right]+\left[-1\right]=\left[1\right]$$

but

$$\left[2\right]+\left[1\right]=\left[3\right]$$

however clearly we have

$$\left[3\right]\ne \left[1\right]$$

But fundamentally since $\left[1\right]$ and $\left[-1\right]$ are the very same set, the $+$ here gives different outcome w.r.t. the same input, so it is not even a function.

The cause of this is all because we are required to define a function $+$ on the set, and intuitively, give any two expressions build from elements in $𝐺/\sim$ and $+$, we should be able to substitute equals for equals, therefore, since $\left[1\right]$ and $\left[-1\right]$ are equal, we should have $\left[2\right]+\left[-1\right]=\left[2\right]+\left[1\right]$. Furthermore, it should be no matter whether we are substituting the first factor or the second factor, which means that switching two components should yield the same result: $\left[-1\right]+\left[2\right]=\left[1\right]+\left[2\right]$. This is summarized by the direct consequence of the definition $\left[𝑎\right]\left[𝑏\right]=\left[𝑎𝑏\right]$, that is, this definition forces

$$𝑎\sim 𝑏\iff 𝑔𝑎\sim 𝑔𝑏\text{and}𝑎𝑔\sim 𝑏𝑔\forall 𝑔\in 𝐺$$

This of course becomes trivial once $𝐺$ is abelian.

Similarly. In last article we also explained the subgroup $𝐻$, notice that we said each subgroup $𝐻$ corresponds to the “null displacement” w.r.t. to an equivalence relation $\sim$, because we define $𝐻$ to be the equivalence class that contains $𝑒$, then we verified that it is a subgorup. But notice that in fact ANY subgroup of $𝐺$ must contains $𝑒$, this should have noticed you somehow that not only every equivalence relation on $𝐺$ gives such a subgroup $𝐻$, but every subgroup $𝐻$ also gives an equivalence relation on $𝐺$. Therefore combined with the Equation 10, this told us that $\sim$ is a relation that is compatible with group structure if and only if $𝑔𝐻=𝐻𝑔$ for all $𝑔\in 𝐻$, which means that $𝐻$ is normal.

Therefore, every normal subgroup $𝐻$ of $𝐺$ gives rise of a corresponding equivalence relation. We could equally say that, intuitively, each quotient group is determined completely by the equivalence class of the identity element $𝑒$. This match our previous explaination of the $𝐻$ subgroup, which is basically the “null displacements”. Therefore naturally once what displacements are null are determined, any two elements differed by such a displacement will be identical.

As a conlusion, we have finally the following intuition

1. The reason why we require the two condition on equivalence relation to ensure group structure is that the transitivity of the equivalence relation itself must be compatible with the group structure; that we must allow substitute equals for equals in a way that is compatible with group operation (this is just a fancy name for “substituting equal elements in the group operation yields same result”)

2. Every quotient group is completely determined by the elements that are identical to $𝑒$. And this intuition is justified by the “null displacement” interpretation of our previous article.