- is an ideal of
*R*; - is an ideal of
*R*; - 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 *I* ∩ *J* 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 : . Indeed, if , then since I is an ideal; likewise, since J is an ideal. Hence .

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

*m***Z**+*n***Z**is the smallest ideal containing*m***Z**and*n***Z**, i.e.*d***Z**where*d*is the largest value dividing*m*and*n*; thus*m***Z**+*n***Z**= gcd(*m*,*n*)**Z**;*m***Z**∩*n***Z**= lcm(*m*,*n*)**Z**;*m***Z**·*n***Z**=*mn***Z**.

*Generalisation.* One can generalise the intersection and sum to arbitrary collections of ideals. Thus if {*I _{λ}*} is a collection of ideals, we can define ∩

Although we can’t take an infinite product of *I _{λ}*, we can take any finite product of

Another way of looking at the *n*-term product *I* = *I*_{1}*I*_{2}…*I _{n}* is that

, for .

*Another example*. Consider the ring **R**[*x*, *y*] of polynomials in *x*, *y* with real coefficients. Define the ideals and , 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;- = set of multiples of
*xy*; - = 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:

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.

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:

,

i.e.

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

if , , then .

Hence we define an

idealto be a subsetIofRsuch that:

Icontains 0;- if
x,ylies inI, thenx+ylies inI;- if
xlies inIandylies inR, thenxyandyxlie inI.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*

- In
**Z**, the only ideals are*n***Z**where*n*is a non-negative integer. - 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*. ] - In a commutative ring
*R*, for any , we can take the set of all multiples of*x*. Clearly, this is non-empty, closed under addition, and closed under multiplication by any element of*R*. - 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. - 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).

]]>

Let R be a ring. A

subringis 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*, then*T* is also a subring of *R*.

*Examples*

- We have a sequence of subrings and .
- In the case of polynomials we have: .
- For matrices, we have .
- Consider the ring
**R**[*x*] of polynomials in*x*with real coefficients. We have the subring**R**[*x*^{2}] of polynomials in*x*^{2}with coefficients as well as*S*, the set of all polynomials spanned by all monomials except*x*^{1}. - The set 2
**Z**of even integers is*not*a subring of**Z**since it doesn’t contain 1. - 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

Ris a collection of rings, let_{i}R= ∏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._{I}R_{i}

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*; - (
*r*,*s*) × (*r’*,*s’*) := (*r*×*r’*,*s*×*s’*).

One can also generalise this to the product of an arbitrary collection of rings, i.e.

, where is a collection of rings indexed by *λ*.

An element of *R* is a gigantic tuple , where .* *

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*.

]]>

- The set
**Z**of integers forms a ring under addition and multiplication, but the subset 2**Z**of even integers forms a rng. - The set
**Z**/*n***Z**of integers modulo*n*forms a ring under addition and multiplication mod*n*. - We have rings
**Q**,**R**and**C**, which are the sets of rational numbers, real numbers and complex numbers respectively. - The ring of Gaussian integers
**Z**[*i*] is the set of complex numbers of the form*a*+*bi*, where*i*= √-1 and*a*,*b*are*integers*. The ring of Gaussian numbers**Q**(*i*) is the set of numbers of the form*a*+*bi*, where*i*= √-1 and*a*,*b*are*rational*. - The set of polynomials in
*x*with real coefficients forms a ring**R**[*x*]. In fact, the set of polynomials informs a ring*x*_{1},*x*_{2}, …,*x*_{n}**R**[*x*_{1},*x*_{2}, …,*x*]. The same holds if you replace_{n}**R**with**Z**,**Q**or**C**. - The set of
*n*-tuples in**R**, given by**R**, forms a ring under component-wise addition and multiplication, i.e. and .^{n} - The set of quaternions forms a ring
**H**. This is the set of 4-tuple of reals, denoted by*a*1 +*b***i**+*c***j**+*d***k**, where*a*,*b*,*c*,*d*are real, and multiplication is defined by the following laws:**i**^{2}=**j**^{2}=**k**^{2}= -1 and**ij**= –**ji**=**k**,**jk**= –**kj**=**i**,**ki**= –**ik**=**j**. [*We will say more about this ring later*. ] - The set of
*n*×*n*square matrices with real entries forms a ring*M*(*n*,**R**). Of course we can replace real by integer, complex or rational and still obtain rings*M*(*n*,**Z**),*M*(*n*,**C**) and*M*(*n*,**Q**) respectively. - The set of upper-triangular
*n*×*n*square matrices with real entries forms a ring*B*(*n*,**R**). 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. - The set of diagonal
*n*×*n*square matrices with real entries forms a ring*D*(*n*,**R**). - 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*). - (Non-example) The set of vectors in
**R**^{3}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*(*n*,**Z**) 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*b*,*c*for which*ab*=*ca*= 0. - Worse, there are matrices
*a*≠ 0, but*a*= 0 for some^{m}*m*> 1. [*Find one for n=2!*] Thus we say an element*a*is**nilpotent**if*a*= 0 for some^{n}*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.**Z**,**Z**[*i*],**Q**,**Q**(*i*),**R**are integral domains. - An element
*a*of ring*R*is a**unit**if there exists*b*such that*ab*=*ba*= 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.**Q**,**R**,**C**are fields and**Z**,**Z**[*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* = 0*b* = 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/nis a domain iff it is a field iffnis prime.

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.

- (
*R*, +) is a group, whose identity is denoted by 0 and inverse is denoted by –*r*; - (
*distributive*) for any*r*,*s*,*t*in*R*, we have: (*r*+*s*)*t*=*rt*+*st*; - (
*distributive*) for any*r*,*s*,*t*in*R*, we have:*t*(*r*+*s*) =*tr*+*ts*; - (
*associative*) for any*r*,*s*,*t*in*R*, we have: (*r**s*)*t*=*r*(*st*).

A **ring** is a **rng** if it has an element 1 (called the **identity** or **unity**) such that 1 × *r* = *r* = *r*× 1. Thus, one sees that a rng is a ring without the **i**dentity. [ No, I did *not* just make that up. ] If *rs* = *sr* 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*:

- If 1 exists, then it is unique.
- 0 ×
*r*= 0 =*r*× 0 for any*r*. - 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 *r ^{n}*. Furthermore,

*r ^{m}* ×

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} = *a ^{3}* +

and if our ring were commutative, the order of multiplication doesn’t matter and the above simplifies to *a*^{3} + 3*a*^{2}*b* + *3ab*^{2} + *b*^{3}.

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.

*a*^{3} + *b*^{3} + *c*^{3} – 3*abc* = (*a *+ *b *+ *c*)(*a*^{2} + *b*^{2} + *c*^{2} – *ab* – *bc* – *ca*)

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

*Example of Errors*

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

… to which the teacher points out 3 mistakes.

- In equating 2
*X*^{2}– 14*XY*+ 24*Y*^{2}and 2(*X*– 4*Y*)(*X*– 3*Y*), 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.

]]>

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.

]]>Let C be a category. The

opposite categoryC^{op}is the category such that:

- Ob(C
^{op}) = Ob(C);- for any objects X, Y, Mor(X, Y) in C
^{op}is Mor(Y, X) in C.

Hence in *Grp*^{op}, a morphism *G* → *H* is actually a group homomorphism *H* → *G*. With this concept, we see that a contravariant functor *F* : *C* → *D* is precisely the same as a covariant functor *F* : *C*^{op} → *D* or *F* : *C* →* D*^{op}.

Now, what would a product in *Grp*^{op} 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 i

_{1}: G → T, i_{2}: H → T such that:

- for any group S with homomorphisms j
_{1}: G → S, j_{2}: H → S there is a unique f : T → S such that fi_{1}= j_{1}, fi_{2}= j_{2}.

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:

, where .

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

(*g*_{1}*h*_{1}*g*_{2}*h*_{2}) * (*g _{3}h_{3}*) = (

The homomorphisms *i*_{1} : *G* → *T* and *i*_{2} : *H* → *T* are defined by taking *g* to (*ge _{H}*) and

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 objectis an object X such that for any object Y, there is a unique morphism X → Y. Aterminal objectis 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:

Mor* _{Set}*(

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

- an object of
*C*(*X*) is a arbitrary function*φ*:*X*→*G*, where*G*is some group; - a morphism from (
*φ*:*X*→*G*) to (*ψ*:*X*→*H*) is a*group homomorphism f*:*G*→*H*such that*f*=*φ**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* : *X* → *F*, such that for any object *φ* : *X* → *G* there is a unique homomorphism *f* :* F* → *G* such that *fi* = *φ*. 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*,*σ*_{1},*σ*_{2}), where*P*is a group,*σ*_{1}:*P*→*G*and*σ*_{2}:*P*→*H*are group homomorphisms; - a morphism (
*P*,*σ*_{1},*σ*_{2}) → (*Q*,*ρ*_{1},*ρ*_{2}) is a group homomorphism*f*:*P*→*Q*such that*ρ*_{1}*f*=*σ*_{1 }and*ρ*_{2}*f*=*σ*_{2}.

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 *π*_{1} : *P* → *G* and *π*_{2} : *P* → *H* such that for any other triplet (*Q*, *ρ*_{1}, *ρ*_{2}), there is a unique *f* : *Q* → *P* such that *π*_{1}*f* = *ρ*_{1} and *π*_{2}*f* = *ρ*_{2}. 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.

]]>

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(1
_{X}) = 1_{F(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* : *C* → *D* and *G* : *D* → *E* are both functors of categories, then so is *GF* : *C* → *E*. 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

subcategoryD is given by subset Ob(D) of Ob(C), and for any objects X, Y in D, a subset Mor_{D}(X, Y) of Mor_{C}(X, Y), such that

- for any object X in D, 1
_{X}lies in Mor_{D}(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 subcategoryif Mor_{D}(X, Y) = Mor_{C}(X, Y) for any objects X, Y of D (i.e. D inherits the full morphism set from C).

*Examples*

- In
*Set*, we have the full subcategory*FinSet*, whose objects are finite sets and morphisms are functions*f*:*X*→*Y*. - In
*FinSet*, we have the (non-full) subcategorywhose objects are finite sets and morphisms are*FinSet*_{0}*bijections**f*:*X*→*Y*. Thus, Mor(*X*,*Y*) = Ø unless |*X*|=|*Y*|. *Ab*is a full subcategory of*Grp*.- We have the
**forgetful functor***Grp*→*Set*which takes a group to its underlying set and a group homomorphism to the underlying set function: thus, the functor forgets the group structure. - We have a functor
*Grp*→*Grp*which takes*G*to*G*×*G*since any homomorphism*G*→*H*induces*G*×*G*→*H*×*H*. - We have a functor
*Set*→*Ab*which takes a set*X*to (|*X*| copies of**Z**), since a function*X*→*Y*gives a group homomorphism*P*→_{X}*P*._{Y} - The free-group functor
*F*:*Set*→*Grp*takes*X*to*F*(*X*). - Abelianisation also gives a functor
*F*:*Grp*→*Ab*since any homomorphism of groups*G*→*H*induces a homomorphism*G*^{ab}→*H*^{ab}. - We do
*not*have a functor*Grp*→*Grp*which takes*G*to its automorphism group Aut(*G*). The reason is that a group homomorphism*G*→*H*does not induce a homomorphism Aut(*G*) → Aut(*H*). - 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*Ab*→*Ab*which takes*H*to Hom(*G*,*H*). Any group homomorphism*φ*:*H*→*H’*will induce a map Hom(*G*,*H*) → Hom(*G*,*H’*) by composing with*φ*(i.e.*f*:*G→H*maps to*φf*:*G*→*H’*). 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* : *G* → *G’* gives us a group homomorphism in the *reverse* direction:

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

This inspires us to consider another class of functors.

Let C, D be categories. A

contravariant functorF 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(1
_{X}) = 1_{F(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*, -) :*C*→*Set*, takes*Y*to the set Mor(*X*,*Y*); - Mor(-,
*Y*) :*C*→*Set*(contravariant), takes*X*to the set Mor(*X*,*Y*).

]]>

*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* : *X* → *F*(*X*) gives rise to the following correspondence:

Mor* _{Set}*(

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

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

Mor* _{Set}*(

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*):

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 2*a* + 3*b* – 4*c* in *F’*(*X*) (where *a*, *b*, *c* in *X*) gets mapped to the element * f(a)^{2}f(b)^{3}f(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):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 *π* : *G* → *F”*(*G*) such that:

Mor* _{Grp}*(

where *g* : *F”*(*G*) → *H* corresponds to *g π* :

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

Mor

_{Set}(G, X) = Mor_{Grp}(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 *ρ*_{1} : *G* → *K* and *ρ*_{2} : *H* → *K* be two fixed group homomorphisms. We wish to find the group *P*, together with homomorphisms *π*_{1} : *P* → *G* and *π*_{2} : *P* → *H* such that *ρ*_{1}*π*_{1} = *ρ*_{2}*π*_{2}, and

- for any group
*Q*and*σ*_{1}:*Q*→*G*and*σ*_{2}:*Q*→*H*such that*ρ*_{1}*σ*_{1}=*ρ*_{2}*σ*_{2}, there is a unique*f*:*Q*→*P*for which*π*_{1}*f*=*σ*_{1}and*π*_{2}*f*=*σ*_{2}.

We denote this group by *G* ×* _{K} H*.

Exercise : prove that the group

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

]]>

So we extract the essence of the remaining concepts.

A

categoryCcomprises of the following data:

- a collection of objects Ob(
C),- for any
X,Yin Ob(C), a set Mor(X,Y) (whose elements are calledmorphisms), and- for any
X,Y,Zin Ob(C), a function Mor(Y,Z) × Mor(X,Y) → Mor(X,Z) (calledcompositionof morphisms).Notation: elements of Mor(

X,Y) are denoted byf:X→Y. Since the definition is so abstract,fmay not even be a function, although it’s certainly helpful to keep that example in mind. For brevity, composition off:X→Yandg:Y→Zis denoted bygf:X→Z. We also require the conditions:

- composing
f:W→X,g:X→Y,h:Y→Zis associative:h(gf) = (hg)f.- for any object Y there is an
identitymorphism 1in Mor(_{Y}Y, Y) such that 1=_{Y}ffandg1= g for any_{Y}f:X→Yandg:Y→Z.

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

Following our observation above regarding inverses, we say *f* : *X* → *Y* is an **isomorphism** if there exists *g* : *Y* → *X* such that *gf* = 1* _{X}* and

–

Following this definition, we can define the **product** of objects *G* and *H* to be a triplet (*P*, *π*_{1}, *π*_{2}), where *π*_{1} : *P* → *G* and *π*_{2} : *P* → *H* are morphisms such that:

- for any such triplet (
*Q*,*σ*_{1},*σ*_{2}) where*Q*is an object and*σ*_{1}:*Q*→*X*and*σ*_{2}:*Q*→*Y*are morphisms, there is a*unique*morphism*f*:*Q*→*P*such that*π*_{1}*f*=*σ*_{1}and*π*_{2}*f*=*σ*_{2}.

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* → (*π*_{1}*f*, *π*_{2}*f*). 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.

The nice thing about this notation is that it gives us:

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

since set product is associative. The correspondence is likewise given by *f* → (*π*_{1}*f*, *π*_{2}*f*, *π*_{3}*f*) for some *π*_{1}, *π*_{2}, *π*_{3 }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:

*Exercise: write out the categorical argument in full detail*.