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I'm studying the basics of quotient groups. I understand that if you build a quotient set from cosets of a group and the subgroup you are using to build them is normal then you end up with a group.

I fail to see why the fact that we can define operations in the quotient set and make it into a group is meaningful.

What is the motivation to do it? Either historical perspective, practical perspective or any other to understand why we care about this is welcome.

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    $\begingroup$ The group $\Bbb Z / n \Bbb Z$ is of fundamental importance to number theory and modern cryptograhy $\endgroup$
    – eepperly16
    Dec 13, 2017 at 17:10
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    $\begingroup$ Its beautiful, that's why its important. $\endgroup$ Dec 13, 2017 at 20:55
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    $\begingroup$ ^ Applied mathematicians vs. pure mathematicians, kids. $\endgroup$
    – imallett
    Dec 14, 2017 at 5:30
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    $\begingroup$ @amWhy I do not understand why you closed this question as a duplicate - they questions are completely different, as can be seen by the huge differences in the types of answers. One question asks "why is the idea of a quotient group like division?", while the other asks "why do I care that a quotient group is a group?"... $\endgroup$
    – user1729
    Dec 18, 2017 at 15:35
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    $\begingroup$ I'm kinda late to the party, but a surjective homomorphism of groups $G \to H$ is the same as a quotient group $G/N$. So if you care about homomorphisms, you care about quotient groups. For example, the surjection $SU(2) \to SO(3)$ (i.e. the fact that $SU(2)/ \mathbb{Z}/2\mathbb{Z} \cong SO(3)$) tells us a lot about the Hopf fibration $S^1 \to S^3 \to S^2$, an important object in both math and physics. $\endgroup$
    – leibnewtz
    Dec 22, 2017 at 0:20

7 Answers 7

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There are several reasons:

  • Sometimes, a group is too complicated to study as a whole. So we look at the quotient group, which is smaller. This can give us information about the original group structure.

An example to illustrate this:

If $Z(G)$ is the center of a group $G$, and the quotient group $G/Z(G)$ is cyclic, then the group $G$ itself is abelian. This fact can be used to prove that every group of order $p²$ (p prime) is abelian.

  • Some quotient groups are extremely important in mathematics. Think about the group $\mathbb{Z}/n\mathbb{Z}$, which is strongly related with modular arithmetic, number theory and cryptography.

Even when you were a kid, you were very familiar with the group $\mathbb{Z}/12 \mathbb{Z}$, without even realising: when you look at the clock, you start counting at $0$ again when it is $12$ o' clock.

Also, every cyclic group of order $n$ is isomorphic to this group, so knowing this group will allow you to fully understand a cyclic group, but there is more:

We can write every finite abelian group (up to isomorphism) as the direct product of quotient groups of the form $\mathbb{Z}/n\mathbb{Z}$. So understanding this one particular quotient group allows us to understand every abelian group!

  • In group theory, we are interested in building new groups using existing groups. Quotient groups are one way to build new (smaller) groups from an existing group. Other manners are direct products, semidirect products, etc.

  • Linking finite groups with quotient groups yields interesting methods to count the order of a group. For example, it is well known that $$sgn: (S_n, \circ) \to (\{-1,1\},.)$$ is a group homomorphism with kernel $A_n$

By the first isomorphism theorem, it follows that:

$$S_n/A_n \cong \{-1,1\} $$

Hence, $$|A_n| = \frac{|S_n|}{|\{-1,1\}|} = \frac{n!}{2}$$

In the same way, combinatorical identities can be proven.

  • Suppose you have an abstract group $G$ which you are not familiar with, but you manage to find an isomorphism $G \cong H/N$ where $H$ is a group you are familiar with. Then, because you know $H$ well, you also know the quotient group well (the operation on the quotient is the one induced by the operation on the group), and hence you have translated the information of this abstract group $G$ to something you can easily work with.

  • Quotient groups provide a way to show that all normal subgroups of a group $G$ are exactly the kernels of group homomorphisms $G \to H$.

Indeed, it is well known that the kernel of a group morphism is always a normal subgroup, and if $N$ is a normal subgroup of $G$, it is the kernel of the canonical epimorphism $G \to G/N: g \mapsto \overline{g}$

  • In abstract algebra, terms like quotient rings and quotient modules pop up all the same for several reasons analoguous to what was already written earlier. These things are in the first place quotient groups.
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  • $\begingroup$ Thanks!, are there any more concrete examples in the realm of geometry or daily life that you can think of?. $\endgroup$ Dec 13, 2017 at 18:47
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    $\begingroup$ Addition with even vs. odd integers (e.g., $\rm even + odd = odd$) and multiplication with positive vs. negative real numbers (e.g. $\rm positive \cdot negative = negative$) are examples of working within quotient groups you've probably been familiar with since you were 10 years old (although of course you probably wouldn't think of them that way until... much later $\ddot \smile$). As a bonus, these groups are isomorphic. $\endgroup$
    – pjs36
    Dec 13, 2017 at 19:07
  • $\begingroup$ Thanks pjs36, the thing is that subgroups of integers always seem to be normal subgroups, that's why I'm trying to get the concept of quotient groups kind of outside of those groups where you cant see it fail. Groups themselves always seem to pop up in nature, that's why i find them awesome, but quotient groups seem to be confined in more abstract groups like $\mathbb{Z} $ or $\mathbb{R} $, making them feel a lot more artificial and harder to grasp $\endgroup$ Dec 13, 2017 at 19:14
  • $\begingroup$ They'll always be normal as they're subgroups of an abelian group. Note that the conjugation condition for normality $nHn^{-1} = H$ is trivially satisfied if $H$ commutes with everything, because $nHn^{-1} = Hnn^{-1} = H$. $\endgroup$
    – Mark
    Dec 13, 2017 at 19:54
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Some of the first groups you're introduced to are $\Bbb Z/n \Bbb Z$. This is a quotient group. By the theorem about finitely generated abelian groups all finitely generated abelian groups are built up from these quotient groups.

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  • $\begingroup$ How does this fit in with OP question about normality, as every subgroup in an abelian group is normal ? $\endgroup$
    – Olórin
    Dec 13, 2017 at 17:21
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    $\begingroup$ @ujsgeyrr1f0d0d0r0h1h0j0j_juj (what a screenname) I was just responding to "why are quotient groups relevant" $\endgroup$
    – user223391
    Dec 13, 2017 at 17:24
  • $\begingroup$ But the question is "why the fact that a quotient group is a group is relevant?" Anynway. $\endgroup$
    – Olórin
    Dec 13, 2017 at 17:25
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    $\begingroup$ @ujsgeyrr1f0d0d0r0h1h0j0j_juj and I answered one case where it's relevant $\endgroup$
    – user223391
    Dec 13, 2017 at 17:25
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    $\begingroup$ How do you remember your username? $\endgroup$
    – D_S
    Dec 22, 2017 at 2:58
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The single biggest reason is that studying the quotients of a particular group can actually unlock information about the group itself. Concrete examples

1.) The proof of the 1st Sylow theorem (at least the ones I've seen) rely on looking at a quotient of a particular group. The third one as well.

2.) If $N$ is normal and solvable and $G/N$ is solvable, then $G$ is solvable.

3.) If $H$ is normal in $G$ and $gcd(|x|,|G/H|) = 1$ then $x \in H$

4.) If $G/Z(G)$ is cyclic, then $G$ is abelian

As you progress through algebra you will see it pop up again and again, and their usefulness will become clearer. The concept of a quotient group is one of the single most important developments in the history of mathematics.

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The fact that all groups are quotient groups of free groups probably doesn't hurt...

Furthermore, you have probably seen groups described as: "$G$ is generated from $x$ and $y$ subject to the equations $x^2=1$ and $y^3=1$" or things like $G=\langle x,y\mid x^2, y^3\rangle$. This is called a presentation of a group and is exactly the quotient group of the free group generated by (in this case) $\{x,y\}$. Finding presentations (ideally finite presentations) or knowing that a group is (finitely) presentable is often one of the goals of group theory. In many other cases, such presentations are the definitions of groups. For example, the case of $\mathbb{Z}/n\mathbb{Z}$ mentioned several times already can be straightforwardly presented as $\langle x\mid x^n\rangle$.

One of the benefits of having a presentation of a group is that a group homomorphism from it can be defined by giving a function of the generators and verifying that the function takes equivalent objects to the same object. So the $G$ from before has an infinite number of elements, but to define a group homomorphism, $f$, from it just requires specifying what $f(x)$ and $f(y)$ are and verifying that $f(x^2) = f(1)=1$ and $f(y^3)=f(1)=1$.

Finally, this pattern of freely generating a structure and then quotienting it is representative of algebraic objects generally. A good exercise is to show that given a congruence, $\sim$, on a group, the equivalence class of $1$ (the identity element), i.e. $\{g\in G\mid g\sim 1\}$, is a normal subgroup. Conversely, show $g\sim h \iff gh^{-1}\in N$ where $N$ is a normal subgroup of $G$ is a congruence. This is why normal subgroups are interesting. When you study rings and ideals, this is exactly what's happening there too. An ideal is the equivalence class of $0$ with respect to a ring congruence.

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The existing answers are about groups, but here is a more general perspective.


Suppose we have some kind of structure $S$ with a substructure $T$, meaning that $T$ inherits the operations and predicates from $S$ and is closed under those operations. We could ask the general question of whether we can partition $S$ into parts according to some equivalence relation $\sim$ such that $T$ is one of the parts, and all the operations and predicates commute with $\sim$. We can think of the partition as a projection from elements to parts.

For example, for a binary operation $\circ$ on $S$, we want to have $x \circ y \sim x' \circ y'$ if $x \sim x'$ and $y \sim y'$. And for a predicate $P$ on $S$, we want to have $P(x) \equiv P(x')$ if $x \sim x'$.


First let us see what we get if we do have such a partition.

The commuting of the operations and predicates with the partition implies that we can define the same operations and predicates on the parts themselves, simply by going forward and backward along the projection. Symbolically, for every $A,B \in S/\sim$ we define $P(A) \overset{def}\equiv ( P(x) \text{ where } x \in A )$ and $A \circ B \overset{def}= ( ( x \circ y ) / \sim \text{ where } x \in A \land y \in B )$. These definitions are valid precisely because the choice of representatives does not matter. (Here $x/\sim$ is naturally defined as the part containing $x$.)

If $\circ$ is associative with identity $e$ and inverse function $i$, then each part $A$ is generated by the action of $T$ on any one member. Symbolically, for any $a \in A$ we have $A = a \circ T \overset{def}= \{ a \circ x : x \in T \}$ and by symmetry $A = T \circ a$. This is one crucial point. Let us see why this is true. Note that $e \in T$ because $T$ is a substructure. For any $x \in T$ it is clear that $a \circ x \sim a \circ e = a$ and hence $a \circ x \in A$. Also, for any $a' \in A$ we have $i(a) \circ a' \sim i(a) \circ a = e \in T$, and hence $i(a) \circ a' = y$ for some $y \in T$, giving $a' = a \circ y$.

If we have just one such operation $\circ$, we get the isomorphism theorems in terms of cosets. For the first isomorphism theorem, any structure homomorphism $φ$ on $S$ would induce such a partition via ( $x \sim x'$ iff $φ(x) = φ(x')$ ), where $T = Ker(φ) \overset{def}= \{ x : x \in S \land φ(x) = φ(e) \}$, and hence the typical definition of the quotient structure $S / T$ as the cosets $\{ a \circ T : a \in S \}$ with the appropriate structure inherited from $S$ makes total sense since it is just $S / \sim$, which is obviously isomorphic to $Im(φ)$. Similar kind of reasoning gives the other isomorphism theorems.


Next let us see what we need to get such a partition.

All the following structures have an associative invertible operation $\circ$, and hence the crucial point above applies to all the following structures. Thus we know that the desired partition must be given by $x \sim y \overset{def}\equiv x \circ T = y \circ T$. Therefore, if $x \circ T = T \circ x$ for every $x \in S$, then given any $x \sim x'$ and $y \sim y'$ we have $x \circ y \circ T = x \circ y \circ T \circ T = x \circ T \circ y \circ T$ $= x' \circ T \circ y' \circ T = \cdots = x' \circ y' \circ T$, and hence $x \circ y \sim x' \circ y'$. So the algebraic identity ( $x \circ T = T \circ x$ ) is equivalent to the commuting of $\circ$ with the desired $\sim$. This is the second crucial point.

  • For a group $(G,*)$ to be partitioned by subgroup $(H,*)$, it is necessary and sufficient to have $g*H = H*g$ for every $g \in G$, by the second crucial point. And equivalently, $g*H*g^{-1} \subseteq H$ for every $g \in G$. This is precisely why the definition of normal subgroups works!

  • For a ring $(R,+,*)$ to be partitioned by subring $(S,+,*)$, we need to have $r+S = S+r$ for every $r \in R$, but that is automatic since $R$ is a ring. We also need to have for every $x,y \in R$ that $(x+S)*(y+S) = x*y+S$, equivalently $x*y+x*S+S*y+S = x*y+S$. This would be satisfied if we have $x*S = y*S = S$. This is the reason for the definition of ideals!

  • For an $R$-module $(M,+)$ to be partitioned by submodule $(M,+)$, it is completely automatic by commutativity of addition.


The notion of structure-preserving partitions is a common one. Vector spaces provide a convenient visual example, since such partitions of them are literally projections onto vector subspaces. The first isomorphism theorem here is equivalent to the rank-nullity theorem.

Another important example is the real numbers. They can be constructed from the structure $S$ of Cauchy sequences of rationals with pointwise addition and multiplication, by partitioning them according to whether their difference tends to zero. The pointwise operations clearly commute with the partition, and hence the ring properties carry straight over from $S$ to the reals. We can also define $<$ on $S$ via ( $f < g$ iff $f(n) < g(n)$ as $n \to \infty$ ), which we can prove is well-defined and a total order, and that it commutes with the partition. Thus it also carries over to the reals. The only really special fact to be proven is the existence of multiplicative inverse for a nonzero real. Well, the zero real is the part of $S$ containing the zero sequence, and every nonzero real is a part of $S$ whose sequences are eventually bounded away from zero. So the partitioning has effectively collapsed all the 'non-invertible' sequences in $S$ to a single zero real.

There are also many results in abstract algebra where normal subgroups are incredibly important. For example, for any finite normal field extension $K / F$ and any subgroup $G$ of $Aut(K/F)$ (the automorphisms of $K$ that fixes elements of $F$), we have that $G$ is a normal subgroup of $Aut(K/F)$ iff the elements of $K$ fixed by $G$ is a normal extension of $F$. And the unsolvability of the quintic turns out to be because the alternating group $A_5$ has no non-trivial normal subgroup.

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  • $\begingroup$ This is the first time we've both answered the same question :) I am not certain that the rank-nullity theorem would be equivalent to the first isomorphism theorem. $\endgroup$
    – David Reed
    Dec 27, 2017 at 4:53
  • $\begingroup$ @DavidReed: It's a vague notion of equivalence. Any two true sentences are logically equivalent, and so any two theorems are also equivalent. But the idea here is that the core of both theorems are the same. That's all. =) $\endgroup$
    – user21820
    Dec 27, 2017 at 5:06
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    $\begingroup$ Nice answer. Especially the central theme of structure preserving partitions is very illuminating. +1 $\endgroup$
    – Paramanand Singh
    Dec 27, 2017 at 16:26
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One reason is that properties of the quotient group can be "backtracked" to properties of the group. As an explicit example, in geometric and combinatorial group theory (that is, the study of finitely generated, infinite groups using funky techniques) there is an extremely important class of groups called hyperbolic groups. These groups are well-behaved; for example, they are all finitely presented. Think of hyperbolic groups as generalised free groups. It is a natural question to ask if their finitely generated subgroups are well-behaved. For example:

Question. Are all the finitely generated subgroups of a hyperbolic group also hyperbolic?

Rips* proved the following result.

Theorem. For every finitely presented group $Q$ there exists a hyperbolic group $H$ and a finitely generated, normal subgroup $N$ of $H$ such that $H/N\cong Q$.

Given a finite presentation of $Q$, Rips explicitly constructs the group $H$. This result is usually referred to as Rips' construction. Rips' takes the group $H$ to be a "torsion free small cancellation" group, which is a much nicer class of groups than the class of hyperbolic groups. Rips' paper can be read easily with only a basic understanding of small cancellation theory (it is 3 pages long!).

We can then backtrack properties of the group $Q$ to obtain results about the subgroup $N$. For example,

  • $Q$ may be taken to have insoluble word problem. Then $N$ has insoluble membership problem: it is not decidable whether or not a given element $h\in H$ is contained in $N$.
  • $Q$ may be taken to contain a finitely generated but not finitely presentable subgroup. Then $N$ contains a finitely generated but not finitely presentable group $K$. In particular, $K$ is finitely generated hut not hyperbolic. This gives a negative answer to the above question. Yay. [In fact, it turns out that in Rips' construction the subgroup $N$ is finitely presentable if and only if the image group $Q$ is finite.]

For more examples of applications of Rips' construction, see either Rips paper or this old blog post.

* E. Rips, Subgroups of small Cancellation Groups, Bulletin of the London Mathematical Society, Volume 14, Issue 1, 1 January 1982, pp45–47, doi link.

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It is relevant by the first isomorphism theorem.

If $\phi: G \rightarrow H$ is a surjective homomorphism of groups, and $N$ is the kernel of $\phi$, then $N$ is a normal subgroup of $G$, and the function $\bar{\phi}: G/N \rightarrow H$ defined by $\bar{\phi}(xN) = \phi(x)$ is an isomorphism of groups.

I would say the reason we care that $G/N$ has a group structure when $N$ is a normal subgroup, is that for surjective homomorphisms as above, we can forget about the object $H$ entirely, identifying it with the quotient group.

Example: let $S^1$ be the group of complex numbers with absolute value $1$ (the unit circle), and define $\phi: \mathbb{R} \rightarrow S^1$ by $\phi(x) = \cos 2 \pi x + i \sin 2\pi x$. Then $\phi$ is surjective with kernel $\mathbb{Z}$, so $\phi$ induces an isomorphism $\bar{\phi}: \mathbb{R}/\mathbb{Z} \rightarrow S^1$ by $\bar{\phi}(x + \mathbb{Z}) = \cos 2\pi x + i \sin 2\pi x$. The isomorphism $\bar{\phi}$ allows us to identify the circle with $\mathbb{R}/\mathbb{Z}$.

This techically applies to all normal subgroups of a group $G$, since a normal subgroup is exactly the same thing as the kernel of a surjective homomorphism with domain $G$.

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