## Group Theory XII.4 (Category Theory: Functors)

In this section, we will explore further concepts in category theory. First, we shall talk about “maps” between categories.

Let C, D be categories. A (covariant) functor (written as F : C → D) is a map F : Ob(C) → Ob(D), and for any objects X, Y in C, a map F : Mor(X,Y) → Mor(F(X), F(Y)), which respects the identity morphism and composition of morphisms, i.e. if X, Y, Z are objects in C, then:

• F(1X) = 1F(X);
• F(gf) = F(g)F(f) for any f:X→Y and g:Y→Z.

Clearly, functors bring isomorphisms and inverses to isomorphisms and inverses respectively. Also, if F : CD and G : DE are both functors of categories, then so is GF : CE. The functor is an isomorphism if F is bijective on Ob(C) → Ob(D) and Mor(X, Y) → Mor(F(X), F(Y)) for any objects X, Y in C.

One common-occurring functor is that of inclusion:

Let C be a category. A subcategory D is given by subset Ob(D) of Ob(C), and for any objects X, Y in D, a subset MorD(X, Y) of MorC(X, Y), such that

• for any object X in D, 1X lies in MorD(X, X);
• for any f:X→Y and g:Y→Z in D, the composition gf:X→Z is also in D.

D is a full subcategory if MorD(X, Y) = MorC(X, Y) for any objects X, Y of D (i.e. D inherits the full morphism set from C).

Examples

1. In Set, we have the full subcategory FinSet, whose objects are finite sets and morphisms are functions f : XY.
2. In FinSet, we have the (non-full) subcategory FinSet0 whose objects are finite sets and morphisms are bijections fX → Y. Thus, Mor(X, Y) = Ø unless |X|=|Y|.
3. Ab is a full subcategory of Grp.
4. We have the forgetful functor GrpSet which takes a group to its underlying set and a group homomorphism to the underlying set function: thus, the functor forgets the group structure.
5. We have a functor GrpGrp which takes G to G × G since any homomorphism GH induces G × GH × H.
6. We have a functor SetAb which takes a set X to $P_X = \prod_{x\in X} \mathbf{Z}$ (|X| copies of Z), since a function XY gives a group homomorphism PXPY.
7. The free-group functor F : SetGrp takes X to F(X).
8. Abelianisation also gives a functor F : GrpAb since any homomorphism of groups GH induces a homomorphism GabHab.
9. We do not have a functor GrpGrp which takes G to its automorphism group Aut(G). The reason is that a group homomorphism GH does not induce a homomorphism Aut(G) → Aut(H).
10. If G and H are abelian (denoted additively), then the set of group homomorphisms G → H, denoted by Hom(G, H), is an abelian group under pointwise addition: (f + f’)(g) := f(g) + f’(g). Upon fixing G, Hom(G, -) is the functor AbAb which takes H to Hom(G, H). Any group homomorphism φHH’ will induce a map Hom(G, H) → Hom(G, H’) by composing with φ (i.e. f:G→H maps to φf:GH’). You can check that this is a group homomorphism also. We call this the Hom-functor.

Note that if we fix H, then the group homomorphism ψ : GG’ gives us a group homomorphism in the reverse direction:

Hom(G’, H) → Hom(G, H),  f maps to fψ.

This inspires us to consider another class of functors.

Let C, D be categories. A contravariant functor F is a map F : Ob(C) → Ob(D), and for any objects X, Y in C, a map F : Mor(X,Y) → Mor(F(Y), F(X)), which respects the identity morphism and composition of morphisms, i.e. if X, Y, Z are objects in C, then:

• F(1X) = 1F(X);
• F(gf) = F(f)F(g) for any f:X→Y and g:Y→Z.

Thus we see that the functor Hom(-, H) is a contravariant functor.

Conclusion: Hom(-, -) is contravariant in the first component and covariant in the second component!

The Hom-functor is extremely useful in higher algebra, in particular, cohomology theory – but we’re not going into that right now.

Exercise: prove that if X, Y are objects in a category C, we get contravariant functor Mor(X, -) and covariant functor Mor(-, Y), where:

• Mor(X, -) : CSet,  takes Y to the set Mor(X, Y);
• Mor(-, Y) : CSet (contravariant), takes X to the set Mor(X, Y).

## Group Theory XII.3 (More Universal Properties)

In this section, we shall get more practice with universal properties for various algebraic constructions. First, take the following categories:

• Set = category of sets, with morphisms = set functions;
• Grp = category of groups, with morphisms = group homomorphisms;
• Ab = category of abelian groups, with morphisms = group homomorphisms.

Given any set X, we get the free group F(X) generated by X; we saw earlier that the injection i : XF(X) gives rise to the following correspondence:

MorSet(X, H) = MorGrp(F(X), H) for any group H,

where the LHS function fXH is obtained from g : F(X) → H by f = gi. This can be seen heuristically as follows: if we pick the word aba-2c3 in F(X) (where a, b, c are elements of X), then g must naturally map it to f(a)f(b)f(a)-2f(c)3 in H.

What if we wish to find an abelian group F’(X) such that:

MorSet(XH) = MorAb(F’(X), H) for any abelian group H?

Let’s see what goes wrong if we put F’(X) = free group F(X). If f : X → H is any function, we can forget for now that H is abelian and reason as before that it must induce a group homomorphism g : F(X) → H. The problem is that F’(X) is not abelian, so the RHS is not well-defined since we’re working in the category of abelian groups there. To rectify, we let F’(X) be the abelianisation of F(X):

$F'(X) = \oplus_{x \in X} \mathbf{Z},$

where we have |X| copies of Z, but we only take those elements with finitely many non-zero terms. This subgroup is called the direct sum of |X| copies of Z. E.g. the element 2a + 3b – 4c in F’(X) (where a, b, c in X) gets mapped to the element f(a)2f(b)3f(c)-4 in H.

Exercise : why do we only take the elements with finitely many non-zero terms? Thus, suppose we were greedy and take the set-theoretic product (i.e. the direct product):

$F'(X) = \prod_{x\in X} \mathbf{Z},$

what will go wrong? Hint: every f : X → H will still give F'(X) → H, but it is possible to find to different F'(X) → H which restricts to the same f.

Finally, we consider the third possibility: given a group G, we wish to find an abelian group F”(G) and group homomorphism πGF”(G) such that:

MorGrp(GH) = MorAb(F”(G), H) for any abelian group H,

where g : F”(G) → H corresponds to gπGH. We’ll give you the answer straight and let you do the verification by yourself: the abelianisation F”(G) = Gab.

Exercise : given a set X, we need a group E(X) and set-theoretic function σ : E(X) → X, such that

MorSet(G, X) = MorGrp(G, E(X)),

where g : G → E(X) corresponds to σg : G → X. Does such an E(X) exist?

Fibred Products. Next, we consider another case of universal properties. Let ρ1 : GK and ρ2HK be two fixed group homomorphisms. We wish to find the group P, together with homomorphisms π1 : PG and π2 : PH such that ρ1π1ρ2π2, and

• for any group Q and σ1Q → G and σ2Q → H such that ρ1σ1ρ2σ2, there is a unique f : QP for which π1fσ1 and π2fσ2.

We denote this group by G ×K H.

Exercise : prove that the group

$P := \{ (g,h) \in G \times H : \rho_1(g) = \rho_2(h) \}$

together with the projection maps to G and H, satisfies the universal property of the fibred product.

## Group Theory XII.2 (Category Theory)

Let’s take a closer look at the proofs and definition in the previous section. What concepts have we used? We have considered groups, homomorphisms between them, composition of homomorphisms, identity homomorphisms, isomorphisms and inverse homomorphisms. But the last two items can be expressed in terms of the other concepts: indeed f : GH is an isomorphism if and only if there exists gH → G such that gf = 1G and fg = 1H, in which case g is the inverse.

So we extract the essence of the remaining concepts.

A category C comprises of the following data:

• a collection of objects Ob(C),
• for any X, Y in Ob(C), a set Mor(X, Y) (whose elements are called morphisms), and
• for any X, Y, Z in Ob(C), a function Mor(YZ) × Mor(XY) → Mor(XZ) (called composition of morphisms).

Notation: elements of Mor(X, Y) are denoted by f : XY. Since the definition is so abstract, f may not even be a function, although it’s certainly helpful to keep that example in mind. For brevity, composition of fX → Y and gY → Z is denoted by gfX → Z. We also require the conditions:

• composing f : WX, g : XY, h : YZ is associative: h(gf) = (hg)f.
• for any object Y there is an identity morphism 1Y in Mor(Y, Y) such that 1Yf = f and g 1Y = g for any f : XY and g : YZ.

(Exercise : take a minute to prove that the identity morphism is unique.)

Following our observation above regarding inverses, we say f : XY is an isomorphism if there exists g : YX such that gf = 1X and fg = 1Y. When that happens, we say g is the inverse of f.  (Exercise : take another minute to prove that the inverse is unique if it exists.)

Following this definition, we can define the product of objects G and H to be a triplet (Pπ1, π2), where π1 : PG and π2 : PH are morphisms such that:

• for any such triplet (Qσ1, σ2) where Q is an object and σ1 : QX and σ2 : QY are morphisms, there is a unique morphism f : QP such that π1f = σ1 and π2fσ2.

The earlier proof that the product is unique can be copied verbatim to apply to this case. One very economical way of writing the universal property is:

Mor(Q, G × H) = Mor(Q, G) × Mor(Q, H),

via the correspondence f → (π1fπ2f). Note that the × means different things on both sides: on LHS, it refers to the categorical product, while on RHS it is the set product.

Mor(Q, (G × H) × K) = Mor(QG × H) × Mor(QK) = Mor(QG) × Mor(QH) × Mor(QK),

since set product is associative. The correspondence is likewise given by f → (π1fπ2f, π3f) for some π1, π2, π3 which maps (G × H) × K to G, H and K respectively.

Working through the same steps for G × (H × K), we find an identical universal property. Since objects which satisfy the same universal property must be isomorphic, we have:

$(G \times H)\times K \cong G \times (H \times K).$

Exercise: write out the categorical argument in full detail.

## Group Theory XII.1 (Introduction to Diagrams)

Given groups G and H, recall we have the product P = G × H and projection maps π1:P → G and π2:PH. There’s nothing mysterious about the projection maps: these just take (g, h) to g and h respectively. But what really makes (P, π1, π2) different from all other such triplets is that in a sense, it’s the “boss” of them all.

Let G, H be groups. A product of G and H is a triplet (Pπ1π2), where P is a group, π1:P → G and π2:P → H are group homomorphisms such that:

• for any triplet (Qσ1:Q → G and σ2:Q → H) there is a unique map f : QP such that π1f = σ1 and π2fσ2.

We call this the universal property of the product.

Indeed, our (G × Hπ1π2) satisfies this property since f must take x in Q to (σ1(x), σ2(x)) in G × H, no choice about it. So our G × H does satisfy the above universal property. And as the Chinese saying goes, you can’t have two tigers on the same mountain:

Any two products (Pπ1π2) and (P’π’1π’2) are isomorphic, in the sense that there is a unique isomorphism f : P’ → P such that π1fπ’1 and π2fπ’2.

Before we begin the proof, note that existence of f is guaranteed by the universal property; we only need to show it’s an isomorphism.

Proof:  First, apply the universal property to (Pπ1π2) itself, and we see that identity map h = 1P is the only P → P map for which π1hπ1 and π2hπ2.

Next, take product (Pπ1π2) and apply universal property to the triplet (Qσ1σ2) = (P’π’1π’2). This induces a unique map fP’ → P such that π1fπ’1 and π2fπ’2.

Do the same with product (P’π’1π’2) and apply universal property to (Pπ1π2); we get g : P → P’ such that π’1gπ1 and π’2gπ2.

Thus, π1fgπ1 and π2fgπ2; the first statement tells us fg = 1P. Likewise, since π’1gfπ’1 and π’2gfπ2, we conclude similarly that gf = 1P’. Thus f and g are isomorphisms and mutual inverses.

Ok, time out! The reader may have a foreboding that there’s much more to the above argument than mere cute abstraction. Yes, everyone knows what a group product looks like, and there’s little need to raise it to such an abstract level if that’s all we care about. But the above reasoning is powerful in the sense that we never took any elements from the groups. Instead, we’re merely tracing arrows around, composing them, and arguing based on sheer logic. There’re two advantages to this.

• The same concept may be repeated for various algebraic structures, e.g. finite products exist for sets, rings, vector spaces, modules, topological spaces, Hilbert spaces, Banach spaces, homological sequences, sheaves, schemes, etc. It’s nice to do the above work once and stick to it.
• For some algebraic structures like sheaves and schemes, the definition is so involved that directly manipulating the “elements” (if there were any to speak of) is like swimming through butter. The above approach is so much neater.

Trust us: once you start learning schemes, you’ll be thanking your lucky stars for such abstraction.

Let’s continue our abstract nonsense: denoting the (abstract) product by G × H now, prove the following as an exercise.

For any groups G, H, K, L,

• $G \times H \cong H \times G$;
• $(G \times H) \times K \cong G \times (H \times K)$;
• if g1 : GK and g2 H → L are maps, then we get a unique g : G×HK×L such that πKg = g1πG and πLgg2πH where the projection maps are:

$\pi_G : G\times H \to G, \pi_H : G\times H \to H,$

$\pi_K : K\times L \to K, \pi_L : K\times L \to L.$

[ Hint for the first two problems: prove that the RHS satisfies the universal property for the LHS, or vice versa ]

## Group Theory XI.4 (Nilpotent Groups: Examples etc)

We will talk more about the properties of nilpotent groups, before narrowing down to the concrete examples (as well as non-examples). The first thing we shall prove is:

All p-groups are nilpotent.

Proof. Let G be a p-group; its centre Z(G) is non-trivial. Assume, by induction hypothesis, that G/Z(G) is nilpotent. We get a sequence of normal subgroups $G_n \triangleleft G$ containing Z(G) such that G/Gn+1 commutes with Gn/Gn+1 and finally Gm = Z(G). Now we just append the term Gm+1 = 1 and everything’s ok since G/Gm+1 = G commutes with Gm/Gm+1 = Z(G) by definition of centre. q.e.d.

We already know that a product of two nilpotent groups is nilpotent. Hence:

A product of finitely many p-groups is a nilpotent group.

The nice thing is that the converse is true, i.e. a finite nilpotent group must be a product of p-groups. The whole of next part is devoted to the proof.

Proof. Let G be finite nilpotent; let P be a Sylow p-subgroup of G. We shall prove that P is unique, and hence normal in G; then since the intersection of a p-subgroup and a q-subgroup must be trivial, we see that G is the product of all the Sylow p-subgroups (see here for the background theory).

To prove P is unique, recall that the number of Sylow p-subgroup P is [G : N(P)]. Now we need two facts:

• N(N(P)) = N(P) for a Sylow p-subgroup P;
• if HG is a proper subgroup of nilpotent G, then HN(H) is a proper subgroup.

Proof of first claim: let N=N(P); since P is normal in N, it is the unique Sylow p-subgroup of N. So

$x \in N(N) \implies xNx^{-1} = N \implies xPx^{-1} \subseteq N \implies xPx^{-1} = P \implies x \in N.$

Proof of second claim: pick index i such that $G_i \not\subseteq H, G_{i+1} \subseteq H$. Hence, let x be in Gi but not H. Then:

$H \supseteq G_{i+1} = [G, G_i] \supseteq [G, x] \supseteq [H, x]$

from which it follows that xHx-1 is contained in H. Replacing x by its inverse, we get the reverse containment. Thus, xHx-1 = H. q.e.d.

Let’s summarise what we’ve learnt.

A finite group is nilpotent if and only if it is a product of p-groups.

Hence by the criteria, the nonabelian group of order pq for p|(q-1) is not nilpotent, although it is solvable. Likewise, the group of upper-triangular invertible real matrices is solvable but not nilpotent, although the subgroup:

$U = \left\{ \begin{pmatrix} 1 & * & * \\ 0 & 1 & * \\ 0 & 0 & 1 \end{pmatrix}\right\}$

is nilpotent – we shan’t prove this, but classically, this was the example which inspired the definition for nilpotent groups.

Exercise : Prove that if G is finite and nilpotent and m | |G|, then G has a subgroup of order m. Thus, the “converse” of Lagrange’s theorem holds for nilpotent groups!

Find non-nilpotent groups for which this holds.

## Group Theory XI.3 (Nilpotent Groups)

Related to solvable groups is the concept of nilpotent groups.

Let G be a group. We say G is nilpotent if there is a sequence of decreasing normal subgroups of G:

$G = G_0 \ge G_1 \ge G_2 \ge \dots \ge G_m = 1, \,\, G_i \triangleleft G,$

such that G/Gn+1 commutes with Gn/Gn+1 for all n.

Once again, we apply our earlier trick to fix the sequence of groups Gn. Explicitly, to ensure that G/N commutes with Gn/N, all commutators [a, b] with $a \in G, b \in G_n$ must lie in N. Thus, if we let [GnG] be the subgroup of G generated by all these [a, b], then [GnG] ≤ N. Hence, we can replace the above Gn by the following series and obtain and equivalent definition.

Let G be a group. The lower central series of G is defined to be:

G0 = GGn+1 = [Gn, G].

If Gn = 1 for some n, then we say G is nilpotent.

( Exercise : make this argument rigourous! Also prove that each Gn in the lower central series is a normal subgroup of G. )

Next, some simple observations:

• If GnGn-1, then Gn+1 = [G, Gn] ≤ [G, Gn-1] = Gn. So the Gn are decreasing.
• Compared to the derived series, we clearly have GnG(n). Hence if Gn is trivial, then so is G(n). Thus, nilpotent implies solvable.
• If HG, then HnGn, so the subgroup of a nilpotent group is nilpotent.
• If N is a normal subgroup of G, then (G/N)n = image of Gn in G/N; hence if G is nilpotent so is G/N.
• If G and H are nilpotent, then so is G × H, since we just have (G × H)n = Gn × Hn.

Warning: unlike the case of solvable groups, it is not true that if N and G/N are nilpotent, then so is G. Thus, nilpotency is not preserved under extension. We will see some explicit examples in the next section.

## Group Theory 0 (Preface + Table of Contents)

Welcome. In this series of notes, I’ll be covering some basic materials in Group Theory. Ok, truth is, even though I said “basic”, there’ll probably be enough materials to cover two semesters of undergraduate algebra. The aim of this set of notes is to explain the various concepts in a more relaxed / informal setting, without sacrificing too much mathematical rigour. Needless to say, the approach is entirely subjective and the astute reader may soon learn to develop his own viewpoint.

This set of notes is not intended to completely replace a textbook. Many of the proofs are omitted or contracted, there’re not enough exercises, and it’s certainly not healthy to stick to a single point-of-view.

Chapter I presents a first look at the best example of groups: the set of permutations, before plunging the reader into the abstract notion of groups.

Chapter II is a first look at abstract groups, with some basic concepts like cancellation laws, order of group/group elements. Examples are also provided for both finite & infinite groups.

Chapter III introduces the concept of subgroups: subsets of a group which inherits the group structure.

Chapter IV talks about cosets. Unlike subsets of a set, a subgroup actually partitions a group nicely as a disjoint union of equal-sized pieces, called cosets.

Chapter V talks about normal subgroups, where the cosets themselves actually form a group – called the quotient group.

Chapter VI introduces homomorphisms:  a function from one group to another which respects the underlying group operation.

Chapter VII follows an abstract group back to its “roots” by looking at its elements as permutations on a certain set. The key result is the Class Formula! We can’t stress this enough.

Chapter VIII presents the crown jewel of finite group theory: the three Sylow theorems. The results are exceptionally beautiful and useful!

Chapter IX discusses the long-standing pipedream of group theorists: that of classifying groups up to isomorphism.

Chapter X introduces groups which are defined by generators and relations. Burnside’s problem presents an interesting application of such groups.

Chapter XI talks about solvable and nilpotent groups. Solvable groups are basically those which can be “built” from abelian groups by extension.

Chapter XII takes things to a higher level of abstraction, by taking a first look at category theory. In fact, you can skip here straight after Chapter VI if you’ve the stomach for it.

And finally, the epilogue.