1.4 Abelian Groups and Formal Sums · Homology & Cohomology

Subgroups are inside a group while quotient groups are a type of collapsing of a group, generalizing the way $R$ can be wrapped around to form the circle group $R /2 π Z .$

— Keith Conrad, “Quotient Groups”

You have met clock arithmetic twice now: as a set of classes in §1.2 Equivalence and Quotients, and as a group in §1.3 Groups. Behind it stands a construction so important that homology itself is an instance of it: take a group, choose a subgroup, and declare every element of that subgroup to be zero. What survives is a new group, called a quotient. Beside it we build a second kind of group that homology needs just as much: the group of formal sums, the algebra of inventories. Put the two together and you can describe every abelian group homology will ever produce.

We go slowly, because this is the algebraic engine of the whole book. Shifted copies of a subgroup, the cosets, get collapsed to points to make quotient groups, which we watch happen to the integers, the real line and a lattice in the plane. The First Isomorphism Theorem then ties quotients to homomorphisms. After that come inventories (free abelian groups), a way to build groups from rules (generators and relations), and a theorem that gives every group homology can produce a standard name. A first look at exact sequences closes the chapter.

Shifted copies: cosets

From now on every group is abelian and written additively, as agreed in §1.3 Groups. Start with the integers $Z$ and the subgroup of multiples of three, $3 Z = {\dots, - 6, - 3, 0, 3, 6, 9, \dots} .$ Now shift it: add $1$ to every element. You get $1 + 3 Z = {\dots, - 5, - 2, 1, 4, 7, 10, \dots},$ the numbers that leave remainder $1$ when divided by three. Shift by $2$ instead and you get $2 + 3 Z = {\dots, - 4, - 1, 2, 5, 8, \dots} .$ Shift by $3$ and nothing new happens: $3 + 3 Z$ is $3 Z$ again. Shift by $4$ and you are back to $1 + 3 Z;$ shift by $- 1$ and you get $2 + 3 Z .$

So there are exactly three different shifted copies of $3 Z,$ and together they contain every integer exactly once. Each integer belongs to one copy: the one named by its remainder.

A coset usually has many names. $1 + 3 Z,$ $4 + 3 Z$ and $- 5 + 3 Z$ are three names for one and the same set. When do two names describe the same coset? Exactly when the two representatives differ by an element of $H :$

This proposition says that the cosets of $H$ are exactly the classes of the relation “ $a \sim b$ when $a - b \in H ”,$ and this is an equivalence relation in the sense of §1.2 Equivalence and Quotients. The proof is the one given there for congruence, with $n Z$ replaced by $H :$ each of the three requirements uses one property of a subgroup. It is reflexive because $a - a = 0 \in H;$ symmetric because if $a - b \in H$ then its negative $b - a$ is in $H;$ and transitive because if $a - b$ and $b - c$ are in $H,$ so is their sum $a - c .$ Therefore the cosets form a partition of $G :$ every element lies in exactly one coset, and two cosets are either identical or have nothing in common.

A picture worth keeping in mind: in the plane $R^{2},$ take $H$ to be a straight line through the origin. Its cosets are all the lines parallel to $H,$ one through each point of the plane. Every point lies on exactly one of them. You also met cosets, without the name, in §1.3 Groups: the elements that a homomorphism $φ$ sends to the same place as $a$ are exactly $a + ker φ .$

Counting cosets: Lagrange’s theorem

All cosets of $H$ have the same size as $H .$ Indeed, the rule $h \mapsto a + h$ matches the elements of $H$ one-for-one with the elements of $a + H :$ every element of $a + H$ is hit, and no two elements of $H$ are sent to the same place (if $a + h = a + h^{'}$ then $h = h^{'}).$ For a finite group, that observation is already a theorem.

For example, the subgroups of $Z /6$ found in §1.3 Groups have $1, 2, 3$ and $6$ elements — all divisors of $6 .$ There is no subgroup of $Z /6$ with four elements, and Lagrange tells us there could not be. Since the order of an element $a$ is the size of the subgroup $⟨ a ⟩,$ it follows that the order of every element of a finite group divides the order of the group.

Collapsing each coset to a point: quotient groups

Here is the decisive step. Instead of looking at the individual integers, look at the three cosets $3 Z,$ $1 + 3 Z$ and $2 + 3 Z,$ and treat each of them as a single object — one dot, standing for the whole infinite pile of integers in it. Can we add these dots?

Research on how people learn group theory singles out this step, treating a whole set as one thing, as the hard one (Dubinsky et al. 1994). It is worth doing it with your own hands.

Collapse the cosets Choose a stage · tap two integers (or two dots) to add

Tap an integer to choose

a,

then another to choose

b .

n 3

Figure 1.4.1 Stage 1: the integers, coloured by their coset of

n Z .

Stage 2: sorted into rows; each row is one coset, a shifted copy of

n Z .

Stage 3: each row collapses to a single dot, and these

n

dots are the group

Z / n Z .

In stage 2, tap a representative

a

of one row and a representative

b

of another: the row containing

a + b

never depends on which representatives you tapped. That is why adding the dots makes sense.

The natural rule for adding cosets is: to add two cosets, pick a representative of each, add the representatives, and take the coset of the result: $(a + H) + (b + H) = (a + b) + H .$ But there is a worry, and it is not a small one. A coset has many representatives. If somebody else picked different representatives of the same two cosets, would they get the same answer? If not, the “rule” would not be a rule at all. (This is the question of being well defined, from §1.2 Equivalence and Quotients.) In $Z$ with $H = 3 Z,$ for example: $1 + 3 Z$ is also $4 + 3 Z,$ and $2 + 3 Z$ is also $- 1 + 3 Z .$ Adding the first names gives $1 + 2 = 3;$ adding the second names gives $4 + (- 1) = 3$ too. Lucky? No:

Everything here has already been checked: the addition is well defined; it is associative and commutative because the addition of $G$ is; $H + (a + H) = (0 + a) + H = a + H,$ so $H$ is the zero; and $(a + H) + (- a + H) = 0 + H .$ The quotient homomorphism $q$ is a homomorphism because $q (a + b) = (a + b) + H = q (a) + q (b);$ it is surjective (every coset is the coset of something); and its kernel is exactly $H,$ since $q (a) = H$ means $a + H = 0 + H,$ which means $a \in H .$

Three ways to look at $Z / n Z$

With $G = Z$ and $H = n Z,$ the quotient $Z / n Z$ is our clock arithmetic $Z / n$ (we will prove the two are isomorphic in a moment, but you can already see it in Figure 1.4.1). Daniel Siebert and Steven Williams interviewed students about $Z / n$ and found three different pictures of its elements in use; the students’ trouble came less from any one picture than from being unable to move between them (Siebert & Williams 2003). All three are correct, and you should be able to switch between them at will.

An element is an infinite set. The element we call $2$ in $Z /5 Z$ is ${\dots, - 8, - 3, 2, 7, 12, \dots} .$ This picture explains why the addition rule is what it is: add any member of one set to any member of another, and every sum lands in the same set.
An element is a shifted copy of the subgroup. The same element is $2 + 5 Z,$ “the multiples of five, moved two steps to the right”. This is the picture used in proofs: shift by $a,$ then by $b,$ and you have shifted by $a + b .$
An element is named by a representative. We write “ $2 ”$ and compute with names: $3 + 4 = 7 = 2 .$ This is the picture for fast arithmetic, and it is safe precisely because addition of cosets is well defined: any names give the same answer.

In every picture the element is one thing. In Figure 1.4.1, after the collapse, the whole infinite set is a single dot, and $Z /5 Z$ has exactly five of them.

The real line wrapped into a circle

Now take $G = R,$ the real numbers, and $H = Z,$ the integers inside them. Two real numbers lie in the same coset when they differ by a whole number: $0.3,$ $1.3,$ $2.3$ and $- 0.7$ all belong to the coset $0.3 + Z .$ Each coset contains exactly one number in the interval $0 \leq t < 1$ (the “fractional part”), so you might think $R / Z$ is just that interval. But look at what happens near the ends: $0.99$ and $0.01$ are very far apart in the interval, yet adding $0.02$ to the first gives $1.01,$ which is in the coset of $0.01 .$ In $R / Z$ the two ends of the interval are glued together, and the result is a circle.

ℝ/ℤ is a circle Press play or scrub · move the coset · drag to rotate

0.30 + Z = {\dots, - 1.70, - 0.70, 0.30, 1.30, 2.30, \dots}

The real line, with the integers in teal and one coset in gold. Press play, or scrub.

line circle

the coset t₀ + ℤ 0.30

Figure 1.4.2 The real line, with the integers marked in teal and one coset

t_{0} + Z

in gold, coils into a spring whose every turn has length

1,

and then flattens. Every gold bead ends on the same point, and every teal bead on another: each coset of

Z

has become a single point of a circle. Adding cosets is adding angles — a quarter turn plus a half turn is three quarters of a turn, whatever representatives you use.

Adding in $R / Z$ is adding fractions of a turn: $0.7 + 0.5 = 1.2,$ which lies in the coset $0.2 + Z,$ so in $R / Z$ “seven tenths of a turn plus half a turn is a fifth of a turn”. This is why $R / Z$ is called the circle group. Conrad’s sentence at the top of this chapter uses the same circle with a different circumference, $R /2 π Z :$ the real numbers with multiples of $2 π$ declared zero, which is how angles in radians behave. Gluing the ends of an interval to make a circle is also your first example of building a shape as a quotient — the subject of §2.2 Gluing Spaces Together.

Collapsing a lattice

For a two-dimensional example, take $G = Z^{2},$ the points of the plane whose two coordinates are integers, added coordinate by coordinate: $(1, 2) + (3, - 1) = (4, 1) .$ (This is the integer lattice; we will meet it again as the direct sum $Z \oplus Z .)$ For $H,$ take all integer combinations of the two vectors $v = (2, 1)$ and $w = (1, 2) :$ $H = ⟨ v, w ⟩ = {m (2, 1) + n (1, 2) : m, n \in Z} .$ How many cosets does $H$ have, and what group is $Z^{2} / H ?$ Explore before reading on.

Collapsing a lattice Drag the tips of v and w · or tap points · try the presets

Drag the arrow tips, or tap a point to move

H = ⟨(2, 1), (1, 2)⟩

det (2112) = 3

Number of cosets (colours)

= ∣ det ∣ = 3

. The shaded tile contains exactly one point of each colour (ringed).

Smith normal form:

(1003)

Z^{2} / H ≅ Z /3

Collapse each colour to one dot: this is the quotient group.

Figure 1.4.3 Every lattice point is coloured by its coset of

H;

the points of

H

itself are the large glowing ones. The shaded parallelogram spanned by

v

and

w

contains exactly one point of each colour (ringed), and its translates tile the plane, so the number of cosets equals its area, the absolute value of the determinant. Drag the vectors until they line up (determinant zero): suddenly there are infinitely many cosets, and the quotient contains a copy of

Z .

With $v = (2, 1)$ and $w = (1, 2)$ there are three colours. The shaded tile, the parallelogram with corners $0, v, v + w, w,$ has area $det (2112) = 2 \cdot 2 - 1 \cdot 1 = 3.$ (For a $2 \times 2$ matrix, the determinant $det (p r q s) = p s - q r$ is the signed area of the parallelogram spanned by its columns; §1.5 Linear Algebra explains why.) Why should the number of cosets equal the area? The translates of the tile cover the plane without overlapping, and each tile contains exactly one point of each coset. A large region of area $A$ contains about $A$ lattice points and about $A /3$ tiles, so each tile must hold $3$ points: three cosets. And a group with three elements must be $Z /3$ (by Lagrange, each nonzero element has order dividing $3,$ so it generates everything). So $Z^{2} / ⟨(2, 1), (1, 2)⟩ ≅ Z /3.$ Other choices give other answers: $v = (2, 0),$ $w = (0, 2)$ gives four cosets, forming the group we will call $Z /2 \oplus Z /2$ in the section on direct sums; $v = (2, 0),$ $w = (0, 3)$ gives six, forming $Z /6 .$ When the determinant is zero the quotient is infinite. How to name the quotient in general is the subject of the sections on direct sums and on generators and relations; the figure already uses the method.

The First Isomorphism Theorem

Quotients and homomorphisms are two sides of one coin. Every quotient comes with a homomorphism, $q : G \to G / H,$ whose kernel is $H .$ Conversely, every homomorphism comes with a quotient: in §1.3 Groups we saw that the elements a homomorphism $φ$ sends to the same point form a coset of its kernel, $a + ker φ .$ Collapse those cosets, and the homomorphism stops losing information.

(The bar in $φ ˉ,$ read “phi bar”, marks the new map that $φ$ induces on the quotient.) Before the proof, watch it happen.

The First Isomorphism Theorem Step through · choose k

φ : Z /12 \to Z /12, φ (x) = 3 x

The homomorphism

φ (x) = k x

, k =

Figure 1.4.4 The homomorphism

φ (x) = k x

Z /12 .

Colour each element by where it lands; elements of the same colour form a coset of the kernel (gold). Pile each coset above the point it lands on, collapse each pile to one dot, and what remains matches the image perfectly, one dot to one point. Collapse what

φ

cannot see, and what remains is exactly what it reaches.

(By Lagrange, $∣ G / ker φ ∣ = ∣ G ∣/∣ ker φ ∣,$ and by the theorem this is $∣ im φ ∣ .)$ This is the pattern you noticed in Figure 1.3.6. Its counterpart for vector spaces, “dimension of the kernel plus dimension of the image equals dimension of the domain”, is the rank–nullity theorem of §1.5 Linear Algebra, and it will be the reason the Euler characteristic can be computed from homology.

Side by side: direct sums

Quotients make groups smaller. The opposite construction puts two groups side by side.

Think of two dials that turn independently: the first coordinate lives on one dial, the second on the other, and an addition turns each dial by its own amount. Its zero is $(0, 0)$ and $- (g, h) = (- g, - h) .$ Some examples:

$Z \oplus Z = Z^{2}$ is the integer lattice of the previous section.
$Z /2 \oplus Z /2$ has four elements, $(0, 0), (1, 0), (0, 1), (1, 1),$ each of which added to itself gives $(0, 0) .$ It is the group of symmetries of a rectangle from §1.3 Groups, with $h \mapsto (1, 0),$ $v \mapsto (0, 1),$ $t \mapsto (1, 1) .$
$Z \oplus Z /2$ is like a ladder with two rails: a copy of the integers on each rail, where adding $(0, 1)$ hops to the other rail and hopping twice brings you back.

A direct sum inherits orders in a simple way: the multiple $k \cdot (g, h)$ is zero exactly when $k g = 0$ and $k h = 0 .$ So the order of $(g, h)$ is the smallest number that is a multiple of both orders: their least common multiple, $lcm .$ For instance in $Z /4 \oplus Z /6$ the element $(1, 1)$ has order $lcm (4, 6) = 12 .$

When are two clocks one clock?

Here is a puzzle that shows why we must be careful with names. Both $Z /2 \oplus Z /3$ and $Z /6$ have six elements. Are they the same group? And $Z /2 \oplus Z /2$ and $Z /4,$ with four elements each?

Two clocks ticking together Step or play · change m and n

after 0 steps:

0 \cdot (1, 1) = (0, 0)

Both hands are back at 0 after

lcm (2, 3) = 6

steps, having visited 6 of the 6 pairs.

Every pair is reached, so

(1, 1)

generates everything:

Z /2 \oplus Z /3 ≅ Z /6

m 2

n 3

Figure 1.4.5 Each step adds

(1, 1) :

both hands advance one hour. The grid records the pairs visited; leaving it on one side re-enters on the other, as on a torus. With

m = 2

and

n = 3

the walk visits all six cells before coming home, so

(1, 1)

generates everything. With

m = n = 2

it comes home after two steps, having seen only half the cells.

So $Z /2 \oplus Z /3 ≅ Z /6$ and $Z /3 \oplus Z /4 ≅ Z /12,$ while $Z /2 \oplus Z /2 \neq ≅ Z /4$ and $Z /2 \oplus Z /4 \neq ≅ Z /8 .$ The same group can wear very different names — $Z /6$ “is” $Z /2 \oplus Z /3$ — which is why we will need a standard way of naming abelian groups.

Formal sums: free abelian groups

Imagine a market stall that sells apples, bananas and cherries. At the end of the day the stallholder writes down what changed hands: “two apples sold, four bananas sold, one cherry returned”. A compact way to write this is $2 a + 4 b - c,$ where the letters are names of the three kinds of fruit, not numbers. This is a formal sum: “formal” because it is not evaluated — $a + b$ does not simplify to anything, it simply records one apple and one banana. (The fruit-stand picture is borrowed from Michael Postol’s notes in the further reading.)

Formal sums obey the arithmetic of inventories:

they are added and subtracted kind by kind: $(2 a + 4 b - c) + (a - 4 b + 3 c) = 3 a + 2 c$ — the bananas cancel, and a term with coefficient $0$ is simply not written;
the empty inventory, with every coefficient zero, is written $0;$
negative coefficients are debts, and every inventory can be cancelled by its negative;
the order in which the items are listed never matters.

In other words, inventories form an abelian group.

If $S = {a, b, c}$ then $Z [S]$ is just $Z^{3}$ in disguise: the formal sum $2 a - b + 3 c$ corresponds to the triple of coefficients $(2, - 1, 3),$ and adding formal sums is adding triples. The only difference is that the coordinates have names. That difference matters for us, because the names will be the pieces of a shape. In the next figure the three symbols are drawn as the three edges of a triangle, each with a chosen direction.

Inventories of edges Step the coefficients · choose + or − · try the presets

first inventory

2 a - b + 3 c

2a −1b 3c

second inventory

a + b - 3 c

1a 1b −3c

result

3 a

(2 + 1) a + (- 1 + 1) b + (3 + (- 3)) c

Figure 1.4.6 The symbols

a, b, c

are the edges of a triangle, each with an arrow. A coefficient is drawn as the thickness of an edge, and a negative coefficient reverses its arrow. Sums are computed coefficient by coefficient; terms with coefficient zero vanish (dashed). The preset “around the triangle” produces

a + b - c :

walking once around the triangle uses

a

and

b

forwards and

c

backwards. Soon such formal sums will be called chains, and those that go around a loop, cycles.

Look at $a + b - c$ in the figure: it records a walk once around the triangle, with the minus sign saying that edge $c$ is traversed against its arrow. This is precisely how §3.2 Chains and the Boundary Operator will describe a loop in a shape: as a formal sum of edges, with signs for direction. And $2 (a + b - c)$ goes around twice.

Bases and rank

The symbols $a, b, c$ have a special property: every element of $Z [a, b, c]$ can be written as an integer combination of them in exactly one way. They are not the only building blocks with that property, and when we compute homology we will often trade one set of building blocks for a more convenient one. So the property needs a name.

Take $Z [a, b] .$ The pair ${a, b}$ is a basis, of course. So is ${a, a + b} :$ any element $p a + q b$ equals $(p - q) a + q (a + b),$ and in only one way. But ${2 a, b}$ is not a basis — you cannot reach $a$ itself, because $n \cdot 2 a + mb = a$ would need $2 n = 1 .$ And ${a, b, a + b}$ is not a basis either: it reaches everything, but $a + b$ can be written in two ways (as $1 \cdot (a + b),$ or as $1 \cdot a + 1 \cdot b).$

Is the rank well defined — could one free group have bases of different sizes? No: every basis of $Z^{r}$ has exactly $r$ elements. Here is a short reason. If $e_{1}, \dots, e_{r}$ is a basis of $F,$ then the quotient $F /2 F$ (where $2 F$ is the subgroup of doubles $2 x)$ consists of the combinations $n_{1} e_{1} + \dots + n_{r} e_{r}$ with each $n_{i}$ only mattering mod $2,$ so it has exactly $2^{r}$ elements. Since $F /2 F$ was defined without mentioning any basis, $2^{r},$ and hence $r,$ cannot depend on the basis.

Defining a homomorphism on a basis

Free abelian groups have one more property that makes them indispensable: it is very easy to define homomorphisms out of them.

This is well defined because each element of $F$ has exactly one expression in the basis, and it respects addition because coefficients add. In words: say where each basis element goes, and the rest is forced. In homology the boundary map is defined in just this way — by saying what the boundary of each single edge or triangle is — and then extended to all formal sums (the exercises give a first taste).

Finally, a fact we will use without proof: every subgroup of $Z^{r}$ is itself free, of rank at most $r$ . For example the subgroup $2 Z \oplus 0$ of $Z^{2}$ is free with basis ${(2, 0)} .$ In homology this guarantees that the cycles and the boundaries, which are subgroups of a free group of chains, are free too.

Generators and relations

A free abelian group is built from generators that obey no rules except commutativity. To build other groups, we add rules. The notation $⟨ a ∣ 3 a = 0 ⟩$ (read “the group generated by $a,$ subject to $3 a = 0 ”)$ means: start with all the integer multiples of a symbol $a,$ and impose the single rule $3 a = 0$ together with everything it implies ( $6 a = 0,$ $4 a = a,$ and so on), but nothing more. The result has the three elements $0, a, 2 a :$ it is $Z /3 .$ Think of generators as Lego bricks and relations as the rules about which constructions count as “nothing”.

$⟨ a, b ∣ 2 a = 0 ⟩ ≅ Z /2 \oplus Z :$ $a$ is a twist that undoes itself when doubled, and $b$ is free.
$⟨ a, b ∣ a - b = 0 ⟩ ≅ Z :$ the relation says $a = b,$ so there is really one generator.
$⟨ a, b ∣ ⟩ ≅ Z^{2} :$ no relations at all gives the free abelian group.

Precisely: a relation is a formal sum declared to be zero, and declaring things zero is exactly what a quotient does.

For example, $⟨ a, b ∣ 2 a + b = 0, a + 2 b = 0 ⟩$ has relation matrix $(2112) :$ it is $Z^{2}$ modulo the integer combinations of $(2, 1)$ and $(1, 2),$ which is exactly the lattice quotient of Figure 1.4.3. We found that it has three elements. Here are two ways to see which group it is.

By substitution. The second relation says $a = - 2 b .$ Substitute into the first: $2 (- 2 b) + b = - 3 b = 0,$ so $3 b = 0 .$ Every element can be written using $b$ alone, and the only rule left is $3 b = 0 .$ So the group is $⟨ b ∣ 3 b = 0 ⟩ ≅ Z /3 .$

By matrix operations. Certain changes to the relation matrix never change the group it presents:

column operations — swap two columns, multiply a column by $- 1,$ or add an integer multiple of one column to another — replace the relations by an equivalent list (each new relation follows from the old ones and vice versa, so the same things are declared zero);
row operations of the same three kinds correspond to choosing new generators (for instance, using $a^{'} = a + 2 b$ in place of $a).$

Using these operations, every integer matrix can be brought to a diagonal form in which each diagonal entry divides the next, called its Smith normal form, after Henry John Stephen Smith, who studied it in 1861 (Smith 1861); the general algorithm is in §1.5 Linear Algebra. For our matrix: swap the two columns; subtract twice the first row from the second; subtract twice the first column from the second; multiply the second row by $- 1 :$ $(2112) \to (1221) \to (10 2 - 3) \to (10 0 - 3) \to (1003) .$ A diagonal matrix is easy to read: the new generators $a^{'}, b^{'}$ are subject only to $1 \cdot a^{'} = 0$ and $3 b^{'} = 0 .$ The first relation kills $a^{'}$ entirely, and the second makes $b^{'}$ a 3-hour clock: the group is $0 \oplus Z /3 = Z /3 .$ In general, a diagonal entry $d \geq 1$ contributes a factor $Z / d$ (nothing at all when $d = 1),$ while a diagonal entry $0,$ or a generator with no diagonal entry at all, contributes a free $Z .$

From relations to a group Edit the matrix · step through · try the presets

generators relations

rel. 1rel. 2

a

b

⟨ a, b ∣ 2 a + b = 0, a + 2 b = 0 ⟩

2	1
1	2

Start: one row for each generator, one column for each relation.

invariant factors:

1, 3

(each 1 contributes nothing)

≅ Z /3

rank (free part): 0 torsion:

Z /3

torsion has 3 elements

Figure 1.4.7 Type any small relation matrix: one row per generator, one column per relation. The stepper diagonalises it by integer row and column operations — each step changes the presentation but never the group — until each diagonal entry divides the next. Then the group can be read off. The presets include the presentations that homology will produce for the torus, the Klein bottle and the projective plane.

The shape of every finite answer

We now have a machine: relations in, diagonal matrix out, group read off. Two questions remain. Does the machine accept every group homology will hand it? And could two different-looking answers, such as $Z /6$ and $Z /2 \oplus Z /3,$ secretly be the same group? The first needs a word. A group is finitely generated if some finite set of its elements generates it. Every finite group is, $Z^{r}$ is, and so is every homology group of a shape built from finitely many pieces. Such groups always have a finite presentation, and the Smith normal form then turns that presentation into a standard name. The result is a complete catalogue, in which every such group appears exactly once.

Why it is true. Existence is what you just watched: present $G$ by generators and relations (a finite list suffices, by the fact about subgroups of $Z^{m}$ quoted earlier), and diagonalise the relation matrix. Uniqueness — that two different lists can never give isomorphic groups — needs a separate argument, by counting how many elements of each order the group has; we will not prove it here. A complete proof is in Judson (2025, Thm 13.1.10), which lists the clocks by prime powers instead (as $Z /4 \oplus Z /3$ rather than $Z /12);$ the theorem on $Z / m \oplus Z / n$ above converts one list into the other.

The theorem says that every finitely generated abelian group is “some free directions and some clocks”: a lattice $Z^{r}$ of independent infinite directions, together with a finite collection of clocks. The condition $d_{1} ∣ d_{2} ∣ \dots$ is what makes the name unique: without it, $Z /2 \oplus Z /3$ and $Z /6$ would be two names for one group.

Torsion

The two halves of the catalogue behave very differently. Add a nonzero element of $Z^{r}$ to itself again and again and it marches off for ever; do the same to an element of a clock $Z / d$ and it comes home to $0 .$ The part that comes home has a name.

In $Z^{r} \oplus Z / d_{1} \oplus \dots \oplus Z / d_{k},$ an element is torsion exactly when its $Z^{r}$ part is zero (a nonzero integer vector never becomes zero when multiplied). So the torsion subgroup is the finite part $Z / d_{1} \oplus \dots \oplus Z / d_{k},$ and the classification can be read as $G ≅ free part Z^{r} \oplus torsion T (G) .$ For example, in $Z \oplus Z /3$ the torsion subgroup is $0 \oplus Z /3$ and the rank is $1 .$

Now the payoff. In Part III we will compute the first homology groups of some surfaces. Here, on trust, are the answers, already in classified form:

Shape	First homology $H_{1}$	Rank (holes)	Torsion
circle	$Z$	1	$0$
sphere	$0$	0	$0$
torus	$Z^{2}$	2	$0$
Klein bottle	$Z \oplus Z /2$	1	$Z /2$
projective plane	$Z /2$	0	$Z /2$

The rank will be called the Betti number $b_{1} :$ it counts independent loops that bound nothing. The torsion is subtler. On the Klein bottle there is a loop that does not bound anything, but which, travelled twice, does: an element of order two. That is the $Z /2,$ the homological trace of the bottle’s one-sidedness, and §3.4 Computing Homology finds it by exactly the matrix steps of Figure 1.4.7.

A little history Why “torsion”?

The word comes from topology, not algebra. Poincaré discovered torsion in 1900 (Poincaré 1900) and chose the name because, as his translator John Stillwell explains, it occurs only in shapes “such as the Möbius band, that are non-orientable and hence ‘twisted onto themselves’ in some way”. (For closed surfaces this is exactly right: a surface has torsion in its first homology precisely when it cannot be oriented. In higher dimensions torsion can also appear in orientable shapes.) Stillwell continues: “When Emmy Noether built the Betti numbers and torsion numbers into the homology groups in 1926, the word ‘torsion’ took up residence in algebra, much to the mystification of group theory students who were not informed of its origin in topology” (Stillwell 2010). (Noether’s talk was in January 1925; Stillwell’s date is that of the printed report.) Before Noether, Betti numbers and torsion coefficients were read off from tables of incidence numbers; after her, they were invariants of groups. As Hatcher puts it, “It is a curious historical fact that homology was not thought of originally as a sequence of groups, but rather as Betti numbers and torsion coefficients” (Hatcher 2002, p. 130).

A first look at exact sequences

We end with a piece of bookkeeping that will become one of the main tools of Part III. Often several groups are linked by a chain of homomorphisms, $A f B g C,$ and the question is how two subgroups of the middle group compare: the image of $f,$ everything $f$ can reach, and the kernel of $g,$ everything $g$ sends to zero.

Read it as: what $f$ produces is exactly what $g$ kills. Two things are packed into the equality. First, $im f \subseteq ker g :$ everything $f$ produces is killed by $g,$ or in other words $g (f (a)) = 0$ for all $a$ — doing $f$ and then $g$ gives zero. Second, $ker g \subseteq im f :$ nothing else is killed by $g .$ Two special cases are worth memorising (here $0$ is the trivial group, and the maps to and from it are the only possible ones):

$0 \to A f B$ is exact at $A$ exactly when $f$ is injective: the image of the map from $0$ is ${0},$ so exactness says $ker f = {0} .$
$B g C \to 0$ is exact at $C$ exactly when $g$ is surjective: the map $C \to 0$ kills everything, so exactness says $im g = C .$

Short exact sequences

By the First Isomorphism Theorem, $C ≅ B / ker g = B / f (A) .$ So a short exact sequence says: $B$ contains a copy of $A,$ and collapsing that copy leaves $C$ . The fundamental example is $0 \to Z \times n Z mod n Z / n \to 0.$ Multiplying by $n$ is injective (for $n \neq = 0);$ its image is $n Z;$ the kernel of reduction mod $n$ is also $n Z;$ and reduction is surjective.

A short exact sequence Change n

Figure 1.4.8

0 \to Z \times n Z mod n Z / n \to 0 .

The teal arrows show multiplication by

n :

no two integers land together (injective), and what they reach, the multiples of

n,

is exactly what the gold arrows send to

0 .

Every point of

Z / n

is reached (surjective).

A short exact sequence does not always determine its middle group. Both $0 \to Z /2 \times 2 Z /4 mod 2 Z /2 \to 0 and 0 \to Z /2 \to Z /2 \oplus Z /2 \to Z /2 \to 0$ are exact (in the second, the maps are $x \mapsto (x, 0)$ and $(x, y) \mapsto y),$ yet $Z /4 \neq ≅ Z /2 \oplus Z /2 .$ Knowing the two ends tells you the size of the middle, but not always its structure — a subtlety that will matter when homology is assembled from pieces.

The second sequence is the simplest kind: the middle is just the two ends side by side, $B ≅ A \oplus C,$ with the obvious maps. Such a sequence is said to split. If $C$ is free abelian, every short exact sequence ending in $C$ splits: pick, for each basis element of $C,$ one element of $B$ that $g$ sends to it, and those choices define a copy of $C$ inside $B$ next to $f (A) .$ The first sequence shows that $Z /2,$ which is not free, can fail. Is freeness the whole story?

Problem Whitehead’s problem

Neither provable nor disprovable Posed by J. H. C. Whitehead, about 1950; shown undecidable by Saharon Shelah, 1974

Call an abelian group $A$ a Whitehead group if every short exact sequence $0 \to Z \to B \to A \to 0$ splits. Every free abelian group is a Whitehead group. Is every Whitehead group free?

The question sounds like a routine exercise about the sequences on this page. Its answer is that the usual axioms of mathematics cannot decide it.

How it was settled. Shelah showed that the answer depends on the axioms of set theory. If every set is constructible (Gödel’s axiom $V = L),$ every Whitehead group is free; under Martin’s axiom with the continuum hypothesis false, there is a Whitehead group that is not free. It was the first problem from ordinary algebra shown to be undecidable.

Before and after. For countable groups the answer is yes: every countable Whitehead group is free (Karl Stein, 1951). Shelah later showed that even the continuum hypothesis does not decide the uncountable case.

A way in. Show that $Z / n$ is not a Whitehead group, using the sequence $0 \to Z \times n Z \to Z / n \to 0 .$ Deduce that a Whitehead group has no elements of finite order, and then that every finitely generated Whitehead group is free.

(Shelah 1974; Eklof 1976) Status as of October 2026 All open problems

Finally, what if a sequence is not exact? Suppose only that doing $f$ and then $g$ gives zero, so that $im f \subseteq ker g .$ Then the quotient $ker g / im f$ measures exactly how far the sequence is from being exact: it is zero precisely when the sequence is exact. Try it.

Exact, or not? Choose n and the two multipliers

im f = {0, 4}

ker g = {0, 4}

Always

im f \subseteq ker g

here, since

g (f (x)) = 2 \cdot 4 x = 8 x = 0

Z /8

im f = ker g

: the sequence is exact at

B

f = × g = ×

Figure 1.4.9 Three copies of

Z / n,

joined by multiplication maps whose composite is zero. In the middle, the image of

f

is filled teal and the kernel of

g

is ringed. When every ringed dot is teal the sequence is exact. When some are not (rose), the leftover

ker g / im f

is a nonzero group. Try

n = 8

with

f = \times 4

and

g = \times 2,

then change

g

\times 4 .

Exercises

Exercise A boundary teaser

A triangle has vertices $0, 1, 2$ and edges $a$ from $0$ to $1,$ $b$ from $1$ to $2,$ and $c$ from $0$ to $2$ (as in Figure 1.4.6). Define a homomorphism $\partial$ from $Z [a, b, c]$ to the free abelian group on the vertices $v_{0}, v_{1}, v_{2}$ by saying where each edge goes: an edge from $p$ to $q$ goes to $v_{q} - v_{p}$ (“end minus start”). Compute $\partial (a + b - c)$ and $\partial (a + b) .$ What do the answers mean?

Solution

By the rule for homomorphisms on a basis, $\partial (a + b - c) = (v_{1} - v_{0}) + (v_{2} - v_{1}) - (v_{2} - v_{0}) = 0,$ and $\partial (a + b) = (v_{1} - v_{0}) + (v_{2} - v_{1}) = v_{2} - v_{0} .$ The walk $a + b - c$ goes once around the triangle and has no end and no start: its “boundary” is zero, so it lies in the kernel of $\partial .$ The walk $a + b$ starts at $v_{0}$ and ends at $v_{2},$ and its boundary records exactly that. This $\partial$ is the boundary map of §3.2 Chains and the Boundary Operator, and its kernel will be called the cycles.

Summary

Recap

A coset $a + H$ is a shifted copy of a subgroup; $a + H = b + H$ exactly when $a - b \in H .$ The cosets partition $G$ into pieces of equal size (Lagrange).
The quotient group $G / H$ has the cosets as its elements, each treated as one object, added through representatives — which is well defined. It is $G$ with $H$ declared zero: $Z / n Z$ is clock arithmetic, $R / Z$ is a circle, and $Z^{2} / H$ has $∣ det ∣$ elements.
First Isomorphism Theorem: $G / ker φ ≅ im φ .$ Collapse what $φ$ cannot see, and what remains is what it reaches.
Direct sums put groups side by side; $Z / m \oplus Z / n ≅ Z / mn$ exactly when $g cd (m, n) = 1 .$
A free abelian group $Z [S]$ consists of formal sums — inventories — of the symbols in $S .$ Homomorphisms out of it are decided by where the basis goes. Chains will be formal sums of edges and triangles.
Groups can be presented by generators and relations; integer row and column operations diagonalise the relation matrix (the Smith normal form) and reveal the group.
Classification: every finitely generated abelian group is $Z^{r} \oplus Z / d_{1} \oplus \dots \oplus Z / d_{k}$ with $d_{1} ∣ \dots ∣ d_{k} .$ The rank $r$ will count holes; the torsion detects twisting.
A sequence is exact where image equals kernel; short exact sequences $0 \to A \to B \to C \to 0$ say $C ≅ B / A .$ Homology will measure the failure of exactness.

Abelian Groups and Formal Sums

Shifted copies: cosets

Counting cosets: Lagrange’s theorem

Collapsing each coset to a point: quotient groups

Three ways to look at $Z / n Z$

The real line wrapped into a circle

Collapsing a lattice

The First Isomorphism Theorem

Side by side: direct sums

When are two clocks one clock?

Formal sums: free abelian groups

Bases and rank

Defining a homomorphism on a basis

Generators and relations

The shape of every finite answer

Torsion

A first look at exact sequences

Short exact sequences

Exercises

Summary

Further reading

References

Shifted copies: cosets

Counting cosets: Lagrange’s theorem

Collapsing each coset to a point: quotient groups

Three ways to look at Z/nZ

The real line wrapped into a circle

Collapsing a lattice

The First Isomorphism Theorem

Side by side: direct sums

When are two clocks one clock?

Formal sums: free abelian groups

Bases and rank

Defining a homomorphism on a basis

Generators and relations

The shape of every finite answer

Torsion

A first look at exact sequences

Short exact sequences

Exercises

Summary

Further reading

References

Three ways to look at $Z / n Z$