# Quotients and Homomorphisms for Beginners: Part 2, Introducing the Quotient Group

Consider the group $G= (\mathcal{G}, \cdot)$ where $\mathcal{G}$ is the set structure of $G$ and $\cdot$ is the group operation. Now, suppose we partition $\mathcal{G}$ into $\mathcal{G}_1 , \ldots \mathcal{G}_n$. Can we make a group with $\mathcal{G}_1 , \ldots \mathcal{G}_n$ as elements? In other words, can we divide $G$ into parts and make a group out of those parts?

Recall the definition of the (left) coset from the last post. We claim that the cosets of a group represent a partition of the group into disjoint subsets. This shall be proved presently with little more than the introduction of an alternative definition. But, first we must review the basic theory of equivalence classes.

Definition. An equivalence relation is a binary operation $\sim$ on a set $X$ such that for all $x \in X$ $x \sim x$ (reflexive), for any $x , y \in X$ if $x \sim y$, then $y \sim x$ (symmetric), and for any $x,y,z \in X$ if $x \sim y$ and $y \sim z$, then $x \sim z$ (transitive).

The equivalence class of an element $x \in X$ is defined as $[x] := \{ y \in X : x \sim y\}$. The set of all equivalence classes in $X$ (for a given relation $R$) is generally denoted $X/R$ (read: $X$ modulo $R$).

A more intuitive way of thinking about equivalence relations is to think of them as different ways of partitioning a set, i.e. separating a set into disjoint subsets.

Theorem. An equivalence relation $\sim$ on a set $S$ partitions $S$.

Proof. Clearly, every $x \in X$ belongs to the class $[x]$. Suppose $x \sim y$. If $x' \in [x]$, then $x' \sim x$, so $x' \sim y$ by transitivity. Hence, $x' \in [y]$. The same argument can be applied for $y' \in [y]$. Therefore, $[x]=[y]$. Moreover, suppose we have that $x$ and $y$ are not equivalent under $\sim$, then if $z \in [x] \cap [y]$, then $z$ is in both equivalence classes, hence $z \sim x$ and $x \sim y$. By symmetry, $x \sim z$, but similarly $x \sim y$. Thus, we have a contradiction, and so $[x] \cap [y] = \varnothing$. We have therefore shown that distinct equivalence classes are disjoint subsets. To finally prove this is a partition, we must show that the union of all subsets is equal to the original set. This is clear because $\displaystyle \cup_{x \in X} [x] = S$$\blacksquare$

It turns out that the converse is also true: every partition of a set is due to an equivalence relation! A formal proof of this fact is left to the reader as an exercise.

Now, let us define the coset using different language.

Definition. The left cosets of $H$ in $G$ are equivalence classes under the relation on $G$ given by $x \sim y \Leftrightarrow x^{-1}y \in H$, or equivalently if $xh=y$ for some $h \in H$.

To ensure the reader is involved, we leave the verification that the given relations above are equivalence relations and are indeed equivalent. It therefore follows that the cosets of $H$ in $G$ partition $\mathcal{G}$.

Now, can we use cosets to make a group? Let us try. Suppose we have a group $G$ and a subgroup $H$, then $G/H := \{ gH : g \in G \}$. Naturally, the operation should be $gH \cdot g' H := (gH)(g'H) := (gg')H$. The issue with this is that it is not consistent for coset representatives! If $aH=bH$, then we cannot guarentee that $(ab)H = (cb)H$! We need to fix coset representatives for the operation to be well-defined.

It turns out that normality is a necessary and sufficient condition for this to occur. For, if left and right cosets are equivalent, then

$\displaystyle (aH)(bH)=a(Hb)H=a(bH)H=(ab)HH=(ab)H$

Associativity is easily verified, the identity is simply $H$, and the inverse is $a^{-1} H$.

We call this important group the quotient group, and it is one of the most important groups in mathematics.