Understanding the quotient map

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The notions of normal subgroup, quotient map and quotient group are somewhat tricky to understand the first time. These notions are also extremely crucial to the structure theory of groups. In this survey article, we study the differing approaches we can take to studying and understanding the quotient map. We begin with a few examples.

Quotient maps for sets

In arithmetic, we think of a quotient as a division of one number by another. While this description is somewhat relevant, it is not the most appropriate for quotient maps of groups. A better way is to first understand quotient maps of sets.

A quotient map, and surjective homomorphism, of sets

Suppose $A$ and $B$ are two sets, and $f:A\to B$ is a surjective function (or set map). In other words, for every $b\in B$, the set $f^{-1}(b)=\{a\in A\mid f(a)=b\}$ is nonempty. Then, we can associate, to each $b\in B$, the set $f^{-1}(b)$.

Observe that:

• The sets $f^{-1}(b)$ form a partition of $A$ into disjoint, nonempty subsets.
• Two elements $a_{1},a_{2}\in A$ are in the same part if and only if $f(a_{1})=f(a_{2})$
• So, the sets $f^{-1}(b)$ are the equivalence classes in $A$ under the equivalence relation $a_{1}\sim a_{2}\iff f(a_{1})=f(a_{2})$

Now, suppose that instead of a map $f:A\to B$, we started with an equivalence relation $\sim$ on $A$. Then, we could construct a set $A/(\sim )$ as the set of equivalence classes under $\sim$, and define a map $f':A\to A/\sim$ as sending each $a\in A$ to its equivalence class. This is termed the quotient map for the equivalence relation $\sim$.

Thus, we have:

• A way of using a surjective function to construct an equivalence relation (namely, the equivalence relation of having the same image)
• A way of using an equivalence relation to generate a surjective function (namely, the quotient map of that equivalence relation)

These two associations are inverses of each other in a weak sense. Namely, if we start with a function $f:A\to B$, obtain the equivalence relation, and then consider the quotient map $f':A\to A/\sim$, then we can construct a natural bijection $g:A/\sim \to B$ such that $f=g\circ f'$. The bijection simply sends an equivalence class of elements in $A$ to the image of any of those elements, in $B$.

Conversely, if we start with an equivalence relation $\sim$, construct the quotient map, and take the equivalence relation arising from that quotient map, we recover the same equivalence relation $\sim$.

Here's an example. Suppose $A=\{1,2,3\}$, $B=\{4,5\}$ and $f$ is given by:

$f(1)=f(3)=4,f(2)=5$

Then the equivalence relation generated has equivalence classes $\{1,3\}$ and $\{2\}$. The quotient is the set:

$\{\{1,3\},\{2\}\}$

with the map $f'$ being:

$f'(1)=f'(3)=\{1,3\},f'(2)=\{2\}$

and the natural bijection $g$ is given by:

$g(\{1,3\})=4,g(\{2\})=5$

When we impose the group structure

Quotient maps and congruences

We'd like to say something analogous to what was said for sets, but respecting the group structure. In other words, we want to relate surjective homomorphisms of groups, with quotient maps from certain equivalence relations.

One direction is easily done. Given any surjective homomorphism $\varphi :G\to H$, we obtain an equivalence relation on $G$. However, not every equivalence relation on $G$ can arise from a surjective homomorphism. The equivalence relation has some special properties. Let's look at this more closely.

The homomorphism condition tells us that if $a_{1},a_{2}\in G$, then $\varphi (a_{1}a_{2})=\varphi (a_{1})\varphi (a_{2})$. In particular, it tells us that the image of $a_{1}a_{2}$ depends only on the images of $a_{1}$ and $a_{2}$. Thus, we have:

$\varphi (a_{1})=\varphi (b_{1}),\varphi (a_{2})=\varphi (b_{2})\implies \varphi (a_{1}a_{2})=\varphi (b_{1}b_{2})$

This imposes the following condition on the equivalence relation:

$a_{1}\sim b_{1},a_{2}\sim b_{2}\implies a_{1}a_{2}\sim b_{1}b_{2}$

Similarly, we obtain that:

$a\sim b\implies a^{-1}\sim b^{-1}$

An equivalence relation $\sim$ satisfying the above conditions is termed a congruence on a group. We now see that, starting with any congruence on a group, we can take the corresponding quotient map, and give a group structure to the quotient, so that the map is a surjective homomorphism. Namely, given two equivalence classes, we multiply their representatives, and take the equivalence class of the product.

So, we have two associations:

• A surjective homomorphism gives rise to a congruence, namely, the equivalence relation of having the same image.
• A congruence gives rise to a surjective homomorphism, namely, its quotient map.

Further, the two associations are inverses of each other, in the following weak sense. Suppose $\varphi :G\to H$ is a surjective homomorphism of groups, and $\sim$ is the equivalence relation that this generates on $G$. This equivalence relation is a congruence. Consider the quotient map $\varphi ':G\to G/\sim$. Then, there is a natural isomorphism $\sigma :G/\sim \to H$ such that $\sigma \circ \varphi '=\varphi$.