I am having a hard time seeing the intuition behind quotient groups or rings. Intuitively, for a group, say $Z/nZ$ would the quotient groups be the different sub groups of order $0$ to $n-1$?

Or how would you be able to tell if there exists a subfield of R that is isomorphic to the ring $Q[x]/<x^2 - 2x - 1>$ ?

In the last example, I assume the part you are dividing by is similar to "modding out" but am having a hard time explicitly seeing it.

Thanks in advance for the help, and insights or general help would be appreciated.

  • $\begingroup$ You may find this question and its answers helpful; it’s in the language of homomorphisms rather than quotient groups, but these are just two ways of looking at the same thing. $\endgroup$ Apr 26, 2015 at 2:33
  • 1
    $\begingroup$ Quotient groups are what you get by collapsing parts of a group together. A geometric example: the unit circle is a quotient group of $\mathbf R$ since the circle looks like $\mathbf R/(2\pi \mathbf Z)$. You spent years in school saying things like $\pi = -\pi$ "as angles" even though $\pi \not= -\pi$ as numbers. Working with angles on a circle where $0 = 2\pi = \ldots$ is working in a quotient group. No subgroup of $\mathbf R$ looks like the quotient group $\mathbf R/(2\pi\mathbf Z)$, e.g., $\pi$ has order $2$ in $\mathbf R/(2\pi\mathbf Z)$ but no element has order $2$ in $\mathbf R$. $\endgroup$
    – KCd
    Apr 26, 2015 at 2:38
  • $\begingroup$ Close to being a duplicate of this. As I answered that one I should refrain from using my dupehammer. $\endgroup$ Aug 10, 2019 at 15:33

2 Answers 2


One of the key concepts behind quotient objects in an algebraic structure is the notion of congruence. But first, behind this there's an even more primitive notion, of equivalence. Now an equivalence relation on a set is really nothing more than forming a partition of "clumps" (subsets) of our original set. A very "useful" such partition is the "pre-image" partition of the domain of a function $f: A \to B$, in which our equivalence relation (on $A$) is formally given by:

$a_1 \sim a_2 \iff f(a_1) = f(a_2)$ so that $[a] = f^{-1}(f(a))$.

This lumps all pre-images of an element in $B$ together in the same equivalence class.

Equivalence is like a "relaxation" of equality, which nevertheless enjoys the properties of reflexiveness, symmetry, and transitivity that we use almost without thinking with the equals sign.

With a congruence, we have an equivalence relation that "respects" the algebraic operations of our structure. For example, an ADDITIVE congruence would be an equivalence relation on a set $S$ with an addition, +, such that:

if $[a] = [a']$ and $[b] = [b']$, then we also have: $[a + b] = [a' + b']$.

Let's see how this plays out in the context of groups. Now, any group $G$ has a "special" element, $e$, its group identity. So if we have a group congruence, perhaps the equivalence class of the identity is a special one, as well. Since we have a congruence, we know that:

$[g] = [e]$, and $[g'] = [e]$ together mean that $[gg'] = [ee] = [e]$, which means that if $g,g' \in [e]$, so is $gg'$, the equivalence (congruence) class of the identity is closed under the group multiplication. Now let's look at inverses:

Suppose that $g \in [e]$, that is, $[g] = [e]$. We would like to know what $[g^{-1}]$ is. Since we don't know, yet, let's just call it $[b]$. Since we have a congruence, we know that:

$[gg^{-1}] = [eb] = [b]$. But the (far) LHS is just $[e] = [gg^{-1}]$, so putting the two facts together, we have:

$[g^{-1}] = [b] = [e]$, which shows if $g \in [e]$, so is $g^{-1}$. Hence the equivalence class of $e$ is indeed special, it's a SUBGROUP of $G$, which is an unexpected bonus. So now we know there's more "structure" to our "clumps" of $G$ (the equivalence/congruence classes) than it might appear, at first.

Naturally, we might ask: what does a "typical" congruence class "look like"? I claim that:

$[g] = \{gx: x \in [e]\} = L_g$ (you probably know $L_g$ better as a left coset of $[e]$).

To show that $L_g \subseteq [g]$, we'll demonstrate that for each and every $x \in [e]$, that $[gx] = [g]$. From $x \in [e]$, we have $[x] = [e]$, and certainly $[g] = [g]$, so because we have a congruence:

$[gx] = [ge] = [g]$.

On the other hand, suppose that $y \in [g]$, so that $[y] = [g]$. Now $y = g(g^{-1}y)$, so we wish to show that $x = g^{-1}y \in [e]$. Again, since we have a congruence, we have from $[y] = [g]$ and $[g^{-1}] = [g^{-1}]$ that:

$[x] = [g^{-1}y] = [g^{-1}g] = [e]$, which shows $[g] \subseteq L_g$.

Now normally, the idea of congruence modulo a subgroup is introduced, for quotient groups. This is the same idea as $[e]$, which we already saw was in fact a subgroup, which we might as well call $H$. As we saw, the equivalence classes of a congruence, are of the form $gH = \{gh: h \in H\}$. The only "missing link" at this point, is showing that the equivalence class of the identity, $[e] = H$ is a special KIND of subgroup, called a NORMAL subgroup. This entails showing that $[g] = gH = Hg$, which is clear from the fact that we have a congruence (we show that $[xg] = [g]$ for $x \in H$, the same way we showed $[gx] = [g]$).

Let's go "one step further" to give the equivalence relation for a congruence as it is usually defined: for a normal subgroup $H$ of $G$, the equivalence relation given by: $x \sim y \iff x^{-1}y \in H$ is, in fact, a congruence. Note that this is really the same as saying $x \sim y \iff xH = yH$. We show this is a congruence, the usual way:

Let $[x] = [x']$ (that is, $xH = x'H$), and $[y] = [y']$ (so $yH = y'H$). We need to demonstrate $[xy] = [x'y']$, that is: $(xy)^{-1}x'y' \in H$. Now:

$(xy)^{-1}x'y' = y^{-1}x^{-1}x'y'$. Since $[x] = [x']$, we know $x^{-1}x' \in H$. Thus we need to show $y^{-1}Hy' = H$, or equivalently, $Hy' = yH$. Here is where the normality of $H$ is crucial:

BECAUSE $H$ IS NORMAL, $Hy' = y'H$, and since $[y] = [y']$, we have $Hy' = y'H = yH$. Working backwards, we then know that $y^{-1}Hy' = H$, and since $x^{-1}x' \in H$, so is $y^{-1}x^{-1}x'y'$, and we have a congruence.

The idea of a quotient group, or quotient ring, does seem unnatural, at first. The key idea is that it allows us to "clump" together "equivalent members" of a group, in such a way that we get a smaller group. The same idea applies to congruences of rings, which have to "respect" BOTH addition and multiplication, and the type of "thing" the congruence class of the identity (zero) becomes, is called an IDEAL.

  • $\begingroup$ I love this answer, looking at quotient groups from the perspective of equivalence classes and then working backwards to get the definition helped my intuition a lot, thankyou for this contribution! $\endgroup$
    – Szmagpie
    Apr 2, 2016 at 19:31

Intuitively, for a group, say $\mathbb Z/n\mathbb Z$ would the quotient groups be the different sub groups of order $0$ to $n-1$?

This question is ill-formed. When $n$ is some fixed number, say $n=7$, we can speak of $\mathbb Z/7\mathbb Z$, which is one quotient group, namely the quotient of $\mathbb Z$ by its subgroup $7\mathbb Z$ (the additive group of all integer multiples of $7$).

The elements of $\mathbb Z/7\mathbb Z$ are the residue classes (or more generally, cosets) that contain $0$ through $6$: $$\begin{array}{l}[0] = 7\mathbb Z+0 = \{\ldots,-14,-7,0,7,\ldots\}\\ [1] = 7\mathbb Z+1 = \{\ldots,-13,-6,1,8,\ldots\}\\ [2] = 7\mathbb Z+2 = \{\ldots,-12,-5,2,9,\ldots\}\\ [3] = 7\mathbb Z+3 = \{\ldots,-11,-4,3,10,\ldots\}\\ [4] = 7\mathbb Z+4 = \{\ldots,-10,-3,4,11,\ldots\}\\ [5] = 7\mathbb Z+5 = \{\ldots,-9,-2,5,12,\ldots\}\\ [6] = 7\mathbb Z+6 = \{\ldots,-8,-1,6,13,\ldots\}\end{array} $$

But these elements are not "quotient groups" or "subgroups" (except that $[0]$ is a subgroup, namely $7\mathbb Z$ -- but that is not the property that's important about it when we view it as an element of $\mathbb Z/7\mathbb Z$).

Or how would you be able to tell if there exists a subfield of $\mathbb R$ that is isomorphic to the ring $\mathbb Q[x]/\langle x^2−2x−1\rangle$?

Well, I would be able to tell that, because I know some field theory, so I know that the ring is isomorphic to a subfield of $\mathbb R$ because $x^2-2x-1$ has real roots.

However, this fact does not follow "obviously" from the concept of a quotient ring. There's some additional work to be done here even after one understands the quotient ring concept in itself.


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