## Basic Algebra I.5 (Operations on Ideals)

Given ideals I and J of ring R, we can perform the following operations to obtain new ideals:

• $I+J := \{ x + y : x \in I, y \in J\}$ is an ideal of R;
• $I\cap J$ is an ideal of R;
• $IJ := \{ x_1 y_1 + \dots + x_r y_r : x_i \in I, y_i \in J\}$ is an ideal of R. Thus, IJ is the set of all finite sums of xy where x, y belong to I, J respectively.

Verifying that these sets are ideals is a rather routine process so we’ll leave it to the reader. Instead, we’ll spend some time explaining what these operations mean. The intersection IJ needs no explanation: that’s the biggest ideal of R contained in both I and J. The sum I+J is the “dual” : it’s the smallest ideal of R which contains both I and J. Finally, IJ is the smallest ideal which contains xy for all x in I and y in J.

Note : $IJ \subset I \cap J$. Indeed, if $x \in I, y \in J$, then $xy \in I$ since I is an ideal; likewise, $xy \in J$ since J is an ideal. Hence $xy \in I \cap J$.

The case of R = Z illustrates this quite nicely. Note that all ideals of Z can be written as nZ for some non-negative integer n. Also, $a\mathbf{Z} \subseteq b\mathbf{Z}$ if and only if b | a. Thus:

• mZ + nZ is the smallest ideal containing mZ and nZ, i.e. dZ where d is the largest value dividing m and n; thus mZ + nZ = gcd(m, n)Z;
• mZnZ = lcm(m, n)Z;
• mZ · nZ = mnZ.

Generalisation. One can generalise the intersection and sum to arbitrary collections of ideals. Thus if {Iλ} is a collection of ideals, we can define ∩λ Iλ as the set of x which occurs in all Iλ and ∑λ Iλ as the set of all finite sums of elements from Iλ (we can only take finite sums since we can’t sum an infinite number of elements here without the concept of convergence).

Although we can’t take an infinite product of Iλ, we can take any finite product of I1, I2, …, In inductively: first take I1I2, then (I1I2)I3, then ((I1I2)I3)I4 … It turns out taking product of ideals is associative (proof: exercise), so we can drop the brackets and write I1I2In.

Another way of looking at the n-term product II1I2In is that I is the smallest ideal which contains all products of the form $x_1 x_2 \dots x_n$, for $x_1 \in I_1, x_2 \in I_2, \dots, x_n \in I_n$.

Another example.  Consider the ring R[x, y] of polynomials in x, y with real coefficients. Define the ideals $I = \left$ and $J = \left< y\right>$, i.e. I (resp. J) is the set of polynomials divisible by x (resp. y). Then:

• I + J = set of all polynomials in x, y without constant term;
• $I \cap J = \left< xy\right>$ = set of multiples of xy;
• $IJ = \left< xy \right>$ = set of multiples of xy.

Exercise. Let R and S be rings and take the ring product R × S. Prove that if I and J are ideals of R and S respectively, then I × J is an ideal of R × S. Note that the quotient is: $(R \times S) / (I \times J) \cong (R/I) \times (S/J).$

Prove also that conversely, any ideal of R × S must be of the form I × J for some ideals I and J of R and S respectively.

The ideals {0} × S and R × {0} give quotients which are isomorphic to R and S respectively. Hence, even though R and S are not subrings of R × S, they’re its quotient rings.

## Basic Algebra I.4 (Ideals and Ring Quotients)

Our first naïve attempt is to take a ring quotient R/S for a subring S of R.  First, (S, +) is a subgroup of (R, +) so we can denote every coset R/S by x+S for some x in R. Naturally we’ll want our 1+S to be the identity element in R/S, which immediately gives us a problem: since 1 is in S, 1+S = 0+S and we have 1=0 in R/S. This means r’ = r’×1 = r’×0 = 0 for any r’ in R/S which is ridiculous.

This forces us to abandon our approach. Instead we look at subrngs of R. Let I be any subrng. Once again, every coset R/I is represented by x+I for some x. In order for multiplication to be well-defined in R/I, we need the following condition: $x+I = x'+I,\,\, y+I = y'+I \implies xy+I = x'y'+I.$,

i.e. $x - x' \in I, \,\, y-y' \in I \implies xy - x'y' \in I.$

But xy – x’y’ = x(yy’) + (xx’)y’ and a moment of reflection tells us that our condition above is equivalent to the following:

if $x \in I$, $y \in R$, then $xy, yx \in R$.

Hence we define an ideal to be a subset I of R such that:

• I contains 0;
• if xy lies in I, then x+y lies in I;
• if x lies in I and y lies in R, then xy and yx lie in I.

Sometimes, one also says I is a two-sided ideal; the reason for this will become apparent when we think of ideals as a special case of modules.

Note: from the conditions above, it follows that I is a subgroup of (R, +); indeed to show that I is closed under subtraction, write x-y = x + (-1)y and apply the last property.

Examples of ideals

1. In Z, the only ideals are nZ where n is a non-negative integer.
2. In a field K, the only ideals are {0} and K. [ Indeed, if I is an ideal containing some non-zero element x, then x has an inverse y; since xy = 1, we see that I contains 1 and hence every element of K.  ]
3. In a commutative ring R, for any $x\in R$, we can take the set $\left := \{ xy : y\in R\}$  of all multiples of x. Clearly, this is non-empty, closed under addition, and closed under multiplication by any element of R.
4. Recall that the set of upper-triangular real n × n matrices forms a ring. The set of strictly upper-triangular matrices (whose diagonal entries are 0) forms an ideal.
5. The ring M(n, R) of n × n real matrices forms a ring with no non-trivial ideals. [ Sketch of proof : let x be a non-zero matrix in ideal I. Left- and right-multiply by appropriate matrices such that the resulting matrix has only 1 non-zero entry…. ]

In example 5, the ring in question has no ideals other than 0 and itself. We call such a ring a simple ring (just like a person who has no ideals is a simple person).

## Basic Algebra I.3 (Subrings and Ring Products)

The concept of subrings follow naturally.

Let R be a ring. A subring is a subset S of R such that:

• (S, +) is a subgroup of (R, +);
• S contains 1;
• for any a, b in S, ab is in S.

We can actually simplify the above conditions to just the following:

• S contains 1;
• for any a, b in S, both a-b and ab lie in S.

Indeed, since S is non-empty and closed under subtraction, (S, +) must be a subgroup of (R, +) . The result then follows. By the way, this probably doesn’t need saying, but if T is a subring of S and S is a subring of R, thenT is also a subring of R.

Examples

1. We have a sequence of subrings $\mathbf{Z} \subset \mathbf{Z}[i] \subset \mathbf{C}$ and $\mathbf{Z} \subset \mathbf{Q} \subset \mathbf{R} \subset \mathbf{C}$.
2. In the case of polynomials we have: $\mathbf{R} \subset \mathbf{R}[x] \subset \mathbf{R}[x, y] \subset \dots$.
3. For matrices, we have $D(n, \mathbf{R}) \subset B(n, \mathbf{R}) \subset M(n, \mathbf{R})$.
4. Consider the ring R[x] of polynomials in x with real coefficients. We have the subring R[x2] of polynomials in x2 with coefficients as well as S, the set of all polynomials spanned by all monomials except x1.
5. The set 2Z of even integers is not a subring of Z since it doesn’t contain 1.
6. If R is any ring with identity=1, let R’ be the set of multiples of 1, i.e. … -(1+1+1+1), -(1+1+1), -(1+1), -1, 0, 1, 1+1, 1+1+1, 1+1+1+1, … . Clearly, multiplying m and n in R’ corresponds to integer multiplication of m and n. Hence R’ is a subring of R. Let n be the smallest positive integer for which n = 0 in R. If no such n exists, then R’ is isomorphic to Z. Otherwise, R’ is isomorphic to Z/n.

Exercise : Prove that in example 5, if R is an integral domain, then n is prime.

Note that S is a subgroup of (R, +) and since all additive groups are abelian here, we see that S is a normal subgroup of (R, +). Hence, the set S divides the ring R into cosets of the form x+S. Although true, this is actually not very useful in practice. One really prefers to write x+S when S is an “ideal” of R rather than a subring. This will be covered later.

Also, one can define a “subrng” of a rng, in which case one simply drops the condition that 1 must be contained in S.

The next topic is on ring products.

If Ri is a collection of rings, let R = ∏I Ri be the set-theoretic product. Then R is a ring under component-wise addition and component-wise multiplication. The identity element is the one where all components are 1.

The easiest non-trivial case is that of R × S, where:

• (1, 1) is the identity;
• (r, s) + (r’, s’) := (r+r’, s+s’) for any r, r’ in R and s, s’ in S;
• (rs) × (r’s’) := (r×r’s×s’).

One can also generalise this to the product of an arbitrary collection of rings, i.e. $R = \prod_\lambda R_\lambda$, where $\{R_\lambda\}$ is a collection of rings indexed by λ.

An element of R is a gigantic tuple $(r_\lambda)$, where $r_\lambda \in R_\lambda$. Beware! The ring R × S contains a copy of R × {0} and {0} × S, which appear to be isomorphic to R and S respectively. However, these two subsets are not subrings because they fail to contain the identity element!

On the other hand, ignoring the identity element, it is clear that R × {0} and {0} × S are both subrngs of R × S. Furthermore, these subrngs are themselves rings since they contain the identity element. So we have the awkward situation where a subrng which is a ring is not a subring in general. This can trip you badly if you weren’t careful.

However, if R = S, then R × R does contain a copy of R, namely the set of “diagonal entries” (r, r) for r in R.

## Basic Algebra I.2 (Examples + Basic Properties of Rings)

Examples

1. The set Z of integers forms a ring under addition and multiplication, but the subset 2Z of even integers forms a rng.
2. The set Z/nZ of integers modulo n forms a ring under addition and multiplication mod n.
3. We have rings QR and C, which are the sets of rational numbers, real numbers and complex numbers respectively.
4. The ring of Gaussian integers Z[i] is the set of complex numbers of the form a +bi, where i = √-1 and ab are integers. The ring of Gaussian numbers Q(i) is the set of numbers of the form a +bi, where i = √-1 and ab are rational.
5. The set of polynomials in x with real coefficients forms a ring R[x]. In fact, the set of polynomials in x1x2, …, xn forms a ring R[x1x2, …, xn]. The same holds if you replace R with ZQ or C.
6. The set of n-tuples in R, given by Rn, forms a ring under component-wise addition and multiplication, i.e. $(x_1, ..., x_n) + (y_1, ..., y_n) := (x_1+y_1, ..., x_n+ y_n)$ and $(x_1, ..., x_n) \times (y_1, ..., y_n) := (x_1 y_1, ..., x_n y_n)$.
7. The set of quaternions forms a ring H. This is the set of 4-tuple of reals, denoted by a1 + bicjdk, where abcd are real, and multiplication is defined by the following laws: i2j2k2 = -1 and ij = –jikjk = –kjiki = –ikj. [ We will say more about this ring later. ]
8. The set of n × n square matrices with real entries forms a ring M(nR). Of course we can replace real by integer, complex or rational and still obtain rings M(nZ),M(nC) and M(nQ) respectively.
9. The set of upper-triangular n × n square matrices with real entries forms a ringB(nR). We can replace R by Z …, ok, you get the drill. The set of strictly upper-triangular n × n square matrices (i.e. whose diagonal entries are zero) with real entries only forms a rng.
10. The set of diagonal n × n square matrices with real entries forms a ring D(nR).
11. If (G, +) is an abelian group, the set of endomorphisms of G (i.e. group homomorphisms G → G) gives a ring End(G), where multiplication corresponds to composition and addition is pointwise-addition, i.e. (f+f’)(g) := f(g) + f’(g).
12. (Non-example) The set of vectors in R3 does not form a rng under vector addition and cross product since the latter is not associative. There are some interesting examples of non-associative algebras, like this case, but we won’t go into that now.

Exercise : take a minute to identify, for each ring, the identity element and to check whether it is commutative.

Armed with our arsenal of examples, it is clear that rings (even commutative ones) come in many shapes and sizes.

• In the case of matrices M(nZ) clearly we can have non-zero matrices which multiply to zero. Thus we say an element a of ring R is a zero-divisor if there are non-zero bc for which abca = 0.
• Worse, there are matrices a ≠ 0, but am = 0 for some m > 1. [ Find one for n=2! ] Thus we say an element a is nilpotent if an = 0 for some n.

Note that by definition, 0 is a zero-divisor and also nilpotent, unless R = {0}, the trivial ring.

• A commutative ring which has no non-zero divisors is called an integral domain (or just domain for lazy people). E.g. ZZ[i], QQ(i), R are integral domains.
• An element a of ring R is a unit if there exists b such that abba = 1. Exercise : prove that the set of units in R forms a group with identity 1.
• A ring in which every non-zero element is a unit is called a division ring. E.g. H is a division ring.
• A commutative division ring is called a field. E.g. QRC are fields and ZZ[i], R[x], Z[x] are integral domains which are not fields. Is Q(i) a field?

Exercise. Find a ring R with elements a and b such that ab = 1 but ba≠1. Hence the condition ba=1 for “unit” is not a superfluous requirement. Hint: look at the ring End(G) for a product of infinitely many copies of Z.

However, prove that if ab = 1 and ca = 1, then b=c.

In short, the left and right inverses must match if they exist. But one can exist without the other.

Note that every field is an integral domain.  Indeed, if a ≠ 0, then ab =  ba = 1 for some b. Then noting that multiplication is commutative, if ca = 0, we get c = cab = 0b = 0, and so a is not a zero-divisor.

Exercise. Prove that a finite integral domain is a field. Hence, prove that the ring Z/n is a domain iff it is a field iff n is prime.

## Basic Algebra I.1 (Enter the Rings)

In this series, we will cover some common algebraic structures – other than groups, which had been amply covered in the Group Theory series. However, due to the considerable depth of many of the topics, we can only provide a brief overview of each of them. Thus, if the Group Theory series were a 2-hour movie, each of the following topics is more like a 5-minute preview. In some cases, it’s closer to a 30-second teaser; in fact, the topic of commutative rings is so deep that we can barely even begin to do it justice here.

As prerequisites, the reader is assumed to be familiar with Chapters I – VI of group theory, and preferably some exposure to Chapter XII as well. Abstract diagrams and universal properties are not so critical here, but they’re indispensable by the time we reach commutative algebra.

Ok, let’s begin. Start with the definition of rings, which are algebraic structures with addition and multiplication.

Definition. A rng (not a misspelling!) is a set R together with operations + and × (for brevity, r × s is often denoted by rs) such that the following hold.

1. (R, +) is a group, whose identity is denoted by 0 and inverse is denoted by –r;
2. (distributive) for any rst in R, we have: (rs)trtst;
3. (distributive) for any rst in R, we have: t(rs) = trts;
4. (associative) for any rst in R, we have: (rs)tr(st).

ring is a rng if it has an element 1 (called the identity or unity) such that 1 × rrr× 1. Thus, one sees that a rng is a ring without the identity. [ No, I did not just make that up. ]  If rssr for all r and s, then we say the rng/ring is commutative.

Our focus will be on rings rather than rngs; hence if we forget to mention, assume R is a ring and not merely a rng.

Take a minute out to prove the following basic properties for a rng R:

1. If 1 exists, then it is unique.
2. 0 × r = 0 = r × 0 for any r.
3. For any r and s, (-r) × s = -(r × s) = r × (-s), and (-r) × (-s) = r × s.

Notation : multiplying an element r by itself n>0 times, we get r × … × r, denoted rn. Furthermore, r0 is defined to be 1 when r≠0. From this, we have:

rm × rn = rm+n and  (rm)n = rmn

for any non-negative integers m and n.

Note : for rings, one can multiply and expand expressions as per normal, as long as one is careful with the order of multiplication. E.g. for any elements a, b of the ring,

(a+b)3 = a3 + a2b + aba + ab2 + ba2 + bab + b2a + b3,

and if our ring were commutative, the order of multiplication doesn’t matter and the above simplifies to a3 + 3a2b + 3ab2 + b3.

Note that the symbols 3, 5 and -4 etc refer to 1+1+1, 1+1+1+1+1 and -(1+1+1+1) respectively, where 1 is the unity in R, i.e. these are not integers but specific elements of the ring R. Seemingly weird things can then happen, e.g. in the ring Z/10 of integers modulo 10 (see next section), we have 20 = 0, 25 = 15 = 5 etc. Certainly 20 ≠ 0 as integers, but when we use these symbols to denote elements of R, 20 = 0 can happen.

Anyway, expansion works in commutative rings! Similarly, factoring generally works, e.g.

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2abbcca)

holds in commutative rings since we can expand the RHS by merely using the above laws.

Example of Errors A student wants to solve 2X2 – 14XY + 24Y2 = 0 in the ring R. He reasons as follows: “The expression is equivalent to 2(X – 4Y)(X – 3Y) = 0. Hence, one of the terms 2, X-4, Y-3 is equal to zero. Since 2 ≠ 0, we must have X = 4Y or X = 3Y.” …

… to which the teacher points out 3 mistakes.

• In equating 2X2 – 14XY + 24Y2 and 2(X – 4Y)(X – 3Y), he assumed XY = YX.
• Also 2=0 can occur in a ring.
• Finally, it is not true that ab = 0 implies a = 0 or b = 0.

We will go through these properties in greater detail in the next section.

## Group Theory ∞ (Epilogue)

That concludes the end of the series of notes on Group Theory.

Has it been successful? I don’t know, but I’m reasonably pleased with the way the notes turn out, except the fact that there’s a huge disparity between the level of the first chapter and the last. In Chapter I, a reasonably motivated pre-university student should be able to understand everything. On the other hand, Chapter XII is at the level of an advanced undergraduate (probably 3rd or 4th year).

In other words, the beginning reader should not expect to be able to comprehend everything from alpha to omega. That being said, I’d urge the reader to at least persist until Chapter VIII – the Sylow theorems, which are exceptionally alluring even to someone who’s not particularly inclined towards group theory.

The main point of the notes is not to write another textbook on group theory. [ With about a thousand such texts available, does the world really need another one? ] Rather, it’s to document things which I wish someone had told me, the little moments of epiphany where the light bulb went off and I wanted to yell to the textbook, “why the heck didn’t you say so?”. Another purpose is to collect the best bits & proofs from various books. For example, while Herstein’s Topics in Algebra is a great book all-in-all, its approach to Sylow theorems is horribly dull. Hence I’ve used the approach in Hungerford’s Algebra instead.

Please leave comments. Error corrections are always welcome and appreciated. Suggestions on what works best for a set of notes like this will also help – though there’s no guarantee they will be implemented.

## Group Theory XII.5 (More Category Theory)

Since a category is really a bunch of abstract objects and arrows between them, we can reverse them by duality.

Let C be a category. The opposite category Cop is the category such that:

• Ob(Cop) = Ob(C);
• for any objects X, Y, Mor(X, Y) in Cop is Mor(Y, X) in C.

Hence in Grpop, a morphism GH is actually a group homomorphism HG. With this concept, we see that a contravariant functor F : CD is precisely the same as a covariant functor FCop → D or F : C Dop.

Now, what would a product in Grpop look like? Let’s call this the coproduct of groups G and H. Unwinding the definition, we get:

Let G, H be groups. The coproduct G*H is a group T, together with group homomorphisms i1 : G → T, i2 : H → T such that:

• for any group S with homomorphisms j1 : G → S, j2 : H → S there is a unique f : T → S such that fi1 = j1, fi­2 = j2. In the case of groups, we also call it the free product of G and H. Its definition is as follows: take the set T of all strings of the following form: $g_1 h_1 g_2 h_2 \dots g_m h_m$, where $g_i \in G, h_i \in H$.

Define the product operation on T by concatenation and simplification, e.g.:

(g1h1g2h2) * (g3h3) = (g1h1g2h2g3h3),   (g1h1g2h2) * (eh2-1g4h4) = (g1h1(g2g4)h4).

The homomorphisms i1G → T and i2H → T are defined by taking g to (geH) and h to (eGh) respectively. It follows that for any group homomorphisms  j1 : GS and j2 : HS the resulting f is uniquely defined by: $f: T \to S, \,\, f((g_1 h_1 g_2 h_2 \dots g_m h_m)) = j_1(g_1) j_2(h_1) j_1(g_2) j_2(h_2) \dots j_1(g_m) j_2(h_m).$

Exercise : compute the coproducts in the categories Ab and Set. Don’t worry: these are much simpler.

Finally, we look at particularly simple objects in the category C.

In category C, an initial object is an object X such that for any object Y, there is a unique morphism X → Y. A terminal object is an object Y such that for any object X, there is a unique morphism X → Y.

For example, in Set, the empty set is an initial object, while the singleton set is a terminal object. In both Grp and Ab, the trivial group 1 is both initial and terminal. It is easy to show that any two initial or two terminal objects are isomorphic.

In what follows, we shall interpret universal properties as initial / terminal objects in the category of appropriate diagrams.

To fix ideas, consider a set X and the free group F(X) on X. We know that:

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

Let’s fix X and consider the category C(X) as follows:

• an object of C(X) is a arbitrary function φ :XG, where G is some group;
• a morphism from (φX → G) to (ψX → H) is a group homomorphism f : GH such that fφ = ψ. Beware!! Till now, we’re used to categories whose underlying objects are sets of some form; this is indeed the case for Set, Grp and Ab. The reader may take some time to get accustomed to the fact that an object of C(X) is not a set, but some function. Furthermore, a morphism between two such functions is a group homomorphism which makes the diagram commute.

Now consider an initial object in this category. This comprises of a group F and a function i : XF, such that for any object φXG there is a unique homomorphism f : FG such that fiφ. Thus, the initial object in this category is the free group on X.

Let’s briefly look at one more example: the group product G × H. We consider the category in which:

• an object is a triplet (P, σ1σ2), where P is a group, σ1P → G and σ2 : PH are group homomorphisms;
• a morphism (Pσ1σ2) → (Qρ1ρ2) is a group homomorphism f : PQ such that ρ1fσ1 and ρ2fσ2. Note that an object in this category comprises of a group and pair of morphisms. Now a terminal object in this category is a P with π1P → G and π2P → H such that for any other triplet (Qρ1ρ2), there is a unique f : QP such that π1f = ρ1 and π2fρ2. But this is precisely the universal property of the product.

Thus, the terminal object in this category is the product of G and H.

Exercise : for each of the other universal properties, try to construct a suitable category C such that the object with the universal property is precisely an initial / terminal object in C.

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