1.3 Groups · Homology & Cohomology

A set of symbols, 1, α, β, … all of them different, and such that the product of any two of them (no matter in what order), or the product of any one of them into itself, belongs to the set, is said to be a group.

— Arthur Cayley, On the theory of groups (1854)

Cut an equilateral triangle out of card and lay it back in the hole it came from. In how many ways can you pick it up and put it down so that it fits again? Now glance at a clock: nine o’clock plus five hours is two o’clock. One question is about shapes, the other about counting — and yet underneath both of them lie the same four simple rules. A collection of things obeying those rules is called a group, and groups are the raw material from which homology is built.

We will play with the triangle and the clock until the shared rules jump out at you, write them down, and test them on a gallery of groups and impostors. Then come the few ideas the rest of the book leans on: abelian groups, subgroups and generators, and the structure-preserving maps called homomorphisms, with their kernels and images.

Symmetry: moves that change nothing

Let us take the triangle seriously. Imagine an equilateral triangle cut from card, resting in a triangular slot. Close your eyes while a friend lifts it out and drops it back into the slot. When you open your eyes, can you tell whether anything happened? If the card is blank, you cannot: every corner looks like every other corner. That is what it means for a move to be a symmetry — it leaves the object looking exactly as it did before.

To keep track of what your friend did, paint the corners and call them $A,$ $B$ and $C .$ Now different moves leave visibly different pictures behind, and we can count them. Your friend can:

do nothing at all — we call this move $e$ (the letter is traditionally said to come from the German Einheit, “unity”);
rotate the card by a third of a turn, $12 0^{\circ}$ anticlockwise — we call this $r;$
rotate it by two thirds of a turn, $24 0^{\circ}$ — which is the same as doing $r$ twice, so we call it $r^{2};$
flip the card over, turning it about one of the three dashed mirror lines of the slot. Call the three flips $f_{1},$ $f_{2},$ $f_{3} :$ $f_{1}$ turns the card about the vertical line through the top of the slot, $f_{2}$ about the line through the bottom-left corner of the slot, and $f_{3}$ about the line through the bottom-right corner.

Notice one detail that will matter in a minute: the mirror lines belong to the slot, not to the card. When you do $f_{1},$ you always turn the card about the vertical line, whichever painted corner happens to sit on top at that moment.

Why exactly six moves? Because a move is completely decided by two pieces of information: where corner $A$ ends up (three possible places) and whether the card ends up face up or face down (two possibilities). Once you know those, the positions of $B$ and $C$ are forced. Three times two is six. Since a move is decided by where it sends the three corners, and each of the six ways of rearranging $A, B, C$ happens exactly once, the triangle group is also the group $S_{3}$ of all permutations of three things. Try it for yourself:

The six symmetries of a triangle Press the moves · tap a table cell to replay it

face up Press a move below.

Cayley table: row

g,

column

h

holds

g \circ h

\circ

e

r

r^{2}

f_{1}

f_{2}

f_{3}

e

r

r^{2}

f_{1}

f_{2}

f_{3}

0 of 36 products discovered

Each move you make from the current position fills in one entry. Tap any cell to watch that product.

Figure 1.3.1 Every move lands the card back in its slot, so every sequence of moves is again one of the six. Each time you press a move

g

while the card shows the net move

h,

you discover one entry of the table: the result of “first

h,

then

g ”.

The arrow printed on the card reverses whenever the card is turned over. Find the two cells that show that

r

followed by

f_{1}

is not the same as

f_{1}

followed by

r .

Doing one move after another

The interesting thing about symmetries is that you can combine them. Do one move, then another: the card is still in its slot, so the combination is itself a symmetry — one of the same six. For example, if you first flip with $f_{1}$ and then rotate with $r,$ the card ends up exactly as if you had done the single flip $f_{3} .$ You can check this in Figure 1.3.1: press Reset, then $f_{1},$ then $r,$ and compare with Reset followed by $f_{3} .$

Playing with the triangle for a few minutes reveals five facts. Keep them in mind; they are the whole chapter in miniature.

Combining stays inside. Doing two symmetries in a row gives a symmetry.
There is a move that does nothing. Doing $e$ before or after any move $g$ changes nothing: $e \circ g = g \circ e = g .$
Every move can be undone. The rotation $r$ is undone by $r^{2}$ (a third of a turn plus two thirds is a full turn, which is the same as doing nothing): $r^{2} \circ r = e .$ Each flip undoes itself: turning the card over twice about the same line returns it to where it was, so $f_{1} \circ f_{1} = e .$
Brackets do not matter. If you do three moves $f,$ then $g,$ then $h,$ it makes no difference whether you first combine $f$ and $g$ into one move, or $g$ and $h :$ in symbols, $(h \circ g) \circ f = h \circ (g \circ f) .$ Both sides mean “ $f,$ then $g,$ then $h ”.$
But the order can matter. First $f_{1}$ then $r$ gives $f_{3};$ first $r$ then $f_{1}$ gives $f_{2} .$ In symbols, $r f_{1} = f_{3}$ but $f_{1} r = f_{2} .$

Here is one way to see it. Rotating and then flipping is like walking forward and then turning left; flipping and then rotating is like turning left and then walking forward. Both sequences use the same two actions, but you end up in different places. In everyday life, “first put on your socks, then your shoes” is famously different from the opposite order. Order mattering is natural, then; it is the triangle’s way of telling us that symmetries are a richer kind of arithmetic than the arithmetic of numbers.

One more observation, which will come back much later in the book: the rotations keep the card face up and the arrow on it turning anticlockwise, while the flips turn the card face down and reverse the arrow. A consistent “direction of turning” is called an orientation. A torus has one and a Klein bottle does not, and homology can tell the difference (§2.4 Manifolds and Surfaces, §3.4 Computing Homology).

Counting in circles: clock arithmetic

Our second example comes from counting rather than from shapes. If it is 9 o’clock now, what time will it be in 5 hours? Not 14 o’clock — on an ordinary clock the hand goes past 12 and starts again, so the answer is 2 o’clock. We have just computed “ $9 + 5 = 2 ”,$ which is perfectly correct on a clock.

Mathematicians make two small changes to the clock. First, they write $0$ at the top instead of $12$ — on a clock, 12 o’clock and 0 o’clock are the same position, and $0$ is the more natural name, as we will see. Second, they allow any number of hours, not just twelve. A clock with $n$ hours has the positions $0, 1, 2, \dots, n - 1,$ and we add two positions by walking around the clock: start at $0,$ walk $a$ steps, then walk $b$ more steps. In practice: add the two numbers as usual, and if the answer is $n$ or more, subtract $n .$ (When adding more than two numbers, keep subtracting $n$ until you land among $0, \dots, n - 1;$ in other words, take the remainder after dividing by $n .)$

The symbol $Z$ (a blackboard-bold Z, from the German Zahlen, “numbers”) stands for the integers: the whole numbers $0, 1, 2, \dots$ together with their negatives $- 1, - 2, \dots .$ The notation $Z / n$ is a reminder of where clock arithmetic comes from: the integers, with every multiple of $n$ declared to be equal to $0 .$ You met this idea in §1.2 Equivalence and Quotients, where $Z / n$ appeared as a set of equivalence classes — the hours “1 o’clock”, “13 o’clock” and “25 o’clock” are all the same class. Here we are adding those classes.

Clock arithmetic Change n, a and b · tap a cell of the table

7 + 8 = 15 = 12 + 3

, so in

Z /12

7 + 8 = 3

Inverse of

7

- 7 = 5

, because

7 + 5 = 12 = 0

on this clock.

Addition table of

Z /12

— tap a cell

+

clock size

n

a

b

show inverses

Figure 1.3.2 The gold arc walks

a

steps, the teal arc

b

more; when the walk passes

0

it simply keeps going round. In the addition table every value has its own colour. Notice the diagonal stripes, and notice that every row and every column contains every colour exactly once. Switch on show inverses: each number is joined to the number that cancels it.

Look at the five facts from the triangle again. Four of them hold on the clock too:

Adding stays on the clock. The sum of two positions is a position.
There is a number that does nothing: $a + 0 = 0 + a = a .$
Every step can be undone. Walking $a$ steps is undone by walking $n - a$ more, because $a + (n - a) = n,$ which is $0$ on the clock. On a 12-hour clock, $7 + 5 = 0,$ so the number that undoes $7$ is $5 .$ We write it as $- 7 :$ on this clock, $- 7 = 5 .$
Brackets do not matter: $(a + b) + c = a + (b + c),$ because both are “walk $a,$ then $b,$ then $c ”.$

The fifth fact is different: on a clock, the order never matters, since walking $a$ and then $b$ steps gets you to the same place as walking $b$ and then $a .$ In the table this shows up as a mirror symmetry across the diagonal.

The smallest interesting clock, $Z /2,$ deserves a special mention. It has just two positions, $0$ and $1,$ and the one surprising rule $1 + 1 = 0 .$ It is the arithmetic of even and odd: read $0$ as “even” and $1$ as “odd”, and the rules $0 + 0 = 0,$ $0 + 1 = 1,$ $1 + 1 = 0$ say “even plus even is even, even plus odd is odd, odd plus odd is even”. You will also see $Z /2$ written $F_{2}$ (“the field with two elements”). It is the first number system homology will use (§3.2 Chains and the Boundary Operator), precisely because in it $1 = - 1$ and every question of sign disappears.

Finally, notice that the three rotations of the triangle behave exactly like $Z /3 .$ Rotating by $a$ thirds of a turn and then by $b$ more thirds is rotating by $a + b$ thirds, wrapping past a full turn: $r^{a} \circ r^{b} = r^{a + b},$ with the exponent computed on a 3-hour clock. For instance $r^{2} \circ r^{2} = r^{4} = r^{1} .$ Two worlds, one structure: that is what the next section is about.

The rules of the game: what a group is

We have seen two very different worlds — moves of a card, positions on a clock — obeying the same four rules. The great trick of modern mathematics is to stop asking what the things are and to keep only the rules they obey. Whatever we then prove from the rules alone will be true in every world that obeys them, at once.

Let us read this slowly, symbol by symbol.

The star $*$ (read “star”) is a placeholder. In a particular group it might be $+,$ or $\times,$ or $\circ .$ A rule that combines two elements into one like this is called a binary operation.
“ $a$ in $G ”$ is often written $a \in G,$ read “ $a$ is an element of $G ”,$ as in §1.1 Sets and Functions.
Closure says the operation never throws you out of the set. (Many books build this into the word “operation” and do not list it separately. We list it because it is the first thing to check, and it is often the first thing to fail.)
Associativity says that when you combine three elements, it does not matter which two you combine first. This is what allows us to write $a * b * c,$ or $ab c,$ with no brackets at all. It is not automatic: subtraction of numbers is not associative, since $(5 - 3) - 1 = 1$ but $5 - (3 - 1) = 3 .$
The identity is the element that does nothing — the move $e,$ the number $0$ on the clock.
$a^{- 1}$ is read “ $a$ inverse”. It is just a name for the element that undoes $a;$ it does not mean $1/ a$ unless the operation happens to be multiplication of numbers.

Most of the time we write the operation as plain juxtaposition, $ab$ instead of $a * b,$ and speak of “the group $G ”$ even though, strictly, the group is the set together with its operation. When the operation could be in doubt, we write the pair: $(Z, +)$ is “the integers under addition”.

Your first proofs about groups

Here is a first taste of how much follows from the four rules alone. The definition says there is an identity element and an inverse of each element. Could there be two different identities? Could an element have two different inverses? The rules say no.

Here is a second fact, which explains the experiment with the triangle. How do you undo “first $h,$ then $g ”?$ You must undo the last move first: undo $g,$ then undo $h .$ In symbols, $(g h)^{- 1} = h^{- 1} g^{- 1} .$ To check it, combine: $(g h) (h^{- 1} g^{- 1}) = g (h h^{- 1}) g^{- 1} = g e g^{- 1} = g g^{- 1} = e,$ so $h^{- 1} g^{- 1}$ undoes $g h,$ and by uniqueness it is the inverse (Judson 2025, Prop 3.2.12). This is often called the socks-and-shoes rule: you put on socks, then shoes, but you take off the shoes first.

Cayley tables and the Sudoku property

For a finite group we can write down the whole operation as a table, like the one you filled in for the triangle: the entry in row $g$ and column $h$ is $g h .$ Such a table is called a Cayley table. In both of our tables you may have noticed that every element appears exactly once in every row and exactly once in every column, like the digits of a Sudoku. It is no coincidence. Cayley, in the 1854 paper that introduced such tables, already observed that “each line as each column of the square will contain all the symbols” (Cayley 1854). Here is why.

Now look at row $a$ of a Cayley table: its entries are $ab$ for the different elements $b .$ If two of them were equal, say $ab = a c$ with $b \neq = c,$ the cancellation law would be violated. So no element repeats in a row; and since a row of a table with $n$ columns has $n$ entries, each of the $n$ elements appears exactly once. The same argument works for columns. A table with this property is called a Latin square.

A little history Galois and Cayley

Évariste Galois 1811 – 1832

Arthur Cayley 1821 – 1895

The word “group” (groupe) was first used in this sense by Évariste Galois, who used groups of permutations to decide which polynomial equations can be solved by a formula. A paper he sent to the Academy was lost when Fourier, its secretary, died. Galois died after a duel in May 1832, aged twenty; the night before, he wrote to his friend Auguste Chevalier summarising his discoveries (Neumann 2011), and his papers waited until 1846 for print. Twenty-two years after the duel Arthur Cayley, then a practising lawyer, gave the first abstract definition of a group (this chapter’s epigraph) and displayed his groups as tables, which now carry his name. In his fourteen years at the bar he published about 250 papers. He paid his debt in a footnote: “The idea of a group as applied to permutations or substitutions is due to Galois” (Cayley 1854).

A gallery of groups — and some impostors

A definition earns its keep through its examples. Here are the groups we will meet again and again; after them come some impostors, which teach just as much.

$(Z, +)$ , the integers under addition. The identity is $0;$ the inverse of $a$ is $- a .$ This is the most important group in this book: homology will count the hole in a circle with one copy of $Z,$ and the two independent holes of a torus with two.
$(Z / n, +)$ , clock arithmetic with $n$ hours. The identity is $0;$ the inverse of $a$ is $n - a$ (and $0$ is its own inverse).
$(R, +)$ , the real numbers — all the points of the number line, including fractions and numbers like $2$ and $π$ — under addition.
$(R ∖ {0}, \times)$ , the nonzero real numbers under multiplication. (The symbol $∖$ means “remove”: $R ∖ {0}$ is “the real numbers with $0$ removed”.) Here the identity is $1,$ and the inverse of $a$ is $1/ a .$
$({+ 1, - 1}, \times)$ , the two signs under multiplication: $(- 1) \times (- 1) = + 1 .$
$D_{3}$ , the six symmetries of the triangle, under composition. The name means “dihedral group”: di-hedral, “two faces”, because a flip shows the other face of the card.
The rotations of a square, ${e, ρ, ρ^{2}, ρ^{3}},$ where $ρ$ (the Greek letter rho) is a quarter turn.
The symmetries of a rectangle that is not a square: do nothing ( $e),$ flip top-to-bottom ( $h,$ turning about the horizontal axis), flip left-to-right ( $v,$ about the vertical axis), and a half turn ( $t).$
Braids on three strands. Hang three strings from a bar and let them cross over and under each other on the way down to a second bar. Two braids combine by hanging one below the other, the braid with straight strings does nothing, and every braid is undone by its reflection in a horizontal mirror. Unlike the symmetries above, this group is infinite: twisting two strands around each other again and again never brings you back to the start. Braids, and the knots they close up into, return in §3.7 Knots and Links.
The trivial group ${0},$ with a single element and the only possible rule $0 + 0 = 0 .$ It is the smallest group there is, and it will be the answer to many homology questions (“no holes here”), so it is usually written plain $0 .$

The number of elements of a group $G$ is called its order and written $∣ G ∣ .$ So $∣ D_{3} ∣ = 6,$ $∣ Z / n ∣ = n,$ and $Z$ and $R$ have infinite order.

And here are some impostors — pairs of a set and an operation that fail at least one rule. Each failure teaches something.

$(N, +)$ , the natural numbers $0, 1, 2, \dots$ under addition, has closure, associativity and an identity, but no inverses: there is no natural number $x$ with $3 + x = 0 .$ You can count up, but never undo.
$(Z, -)$ , the integers under subtraction, is not associative, and has no element that does nothing on both sides ( $a - 0 = a,$ but $0 - a = - a).$
$(R, \times)$ fails only because of one element: $0$ has no inverse, since $0 \times x = 0$ is never $1 .$ Remove $0$ and you get a group.
The odd integers under addition are not even closed: $3 + 5 = 8$ is even.
$(Z /6 ∖ {0}, \times)$ , the nonzero numbers on a 6-hour clock under clock multiplication, is not closed either: $2 \times 3 = 6 = 0,$ which has been removed. Strangely, $(Z /5 ∖ {0}, \times)$ is a group. The difference is that $5$ is prime and $6$ is not.

Group or impostor? Pick a candidate · predict · tap to check

(N, +)

the natural numbers 0, 1, 2, … under addition

Predict each axiom, then tap it to check.

Test bench — choose elements:

a

2

b

5

c

1

2 + 5 = 7

in the set

(2 + 5) + 1 = 8

2 + (5 + 1) = 8

same

5 + 2 = 7

same as

a + b

Figure 1.3.3 For each candidate, guess which of the four rules hold before you tap them. Then use the test bench to try the operation on elements of your choice: hunt for a pair whose combination leaves the set, a triple where the brackets matter, and a pair where the order matters. Only a candidate that passes all four checks is a group.

When order doesn’t matter: abelian groups

The clock and the triangle differ in one important respect: on the clock the order of addition never matters, while for the triangle it sometimes does. Groups of the first kind have a name.

A little history Abel’s pamphlet

Niels Henrik Abel 1802 – 1829

The word honours the Norwegian mathematician Niels Henrik Abel, who, like Galois, died young and changed the theory of equations. In 1824, poor as he always was, he proved that the general equation of the fifth degree cannot be solved by a formula, and printed the proof at his own expense, squeezed onto six pages to save money. He sent a copy to Gauss, in whose papers it was found after his death, unopened. At Christmas 1828 Abel fell seriously ill on a sledge journey to visit his fiancée, and he died in April 1829, aged twenty-six; two days later his friend August Crelle wrote to tell him that a post had been found for him in Berlin. The word is usually written with a small “a”, as if it were an ordinary word — which, in algebra, it now is.

Most of our examples are abelian: $Z,$ $Z / n,$ $R,$ $R ∖ {0},$ the two signs, the rotations of a square, the symmetries of a rectangle. The triangle group $D_{3}$ is not: $r f_{1} = f_{3}$ but $f_{1} r = f_{2} .$ One way to see why is to follow the arrow on the card. A flip reverses the arrow, so “flip, rotate anticlockwise, flip back” is the same as rotating clockwise: $f_{1} r f_{1} = r^{2} = r^{- 1} .$ If $D_{3}$ were abelian, the left side would simplify to $f_{1} f_{1} r = r,$ and we would get $r = r^{- 1},$ which is false. In fact $D_{3}$ is the smallest non-abelian group: every group with five or fewer elements is abelian.

Additive notation

For abelian groups there is a special, very convenient way of writing. Since the operation behaves like addition, we write it as addition.

Notation Additive notation for abelian groups

In a general group	In an abelian group, written additively	Read
$ab$	$a + b$	“a plus b”
$e$	$0$	“zero”
$a^{- 1}$	$- a$	“minus a”
$a b^{- 1}$	$a - b$	“a minus b”
$a^{k} = aa \dots a$	$k a = a + a + \dots + a$	“k times a”

We use $+$ only for abelian groups. Seeing a plus sign is a promise that the order of the terms does not matter, so $a + b - c = - c + b + a .$

Be careful with the expression $k a :$ here $k$ is an ordinary integer (how many copies of $a$ to add), while $a$ is an element of the group. For negative $k$ it means adding copies of $- a :$ $(- 3) a = - a - a - a .$ In $Z /12,$ for example, $5 \cdot 7 = 7 + 7 + 7 + 7 + 7 = 35 = 11 .$

Groups inside groups

Inside the six symmetries of the triangle sit the three rotations ${e, r, r^{2}} .$ They form a little group of their own: combining two rotations gives a rotation, the do-nothing move is a rotation (by $0^{\circ}),$ and the inverse of a rotation is a rotation. A group living inside a bigger group like this is called a subgroup.

(Associativity never needs checking for a subgroup: it holds for all elements of $G,$ so in particular for those of $H .)$ Some examples, in additive notation where the group is abelian:

The even integers ${\dots, - 4, - 2, 0, 2, 4, \dots}$ form a subgroup of $Z,$ written $2 Z :$ $0$ is even, the sum of two even numbers is even, and the negative of an even number is even.
More generally, the multiples of any fixed number $n,$ $n Z = {\dots, - 2 n, - n, 0, n, 2 n, \dots},$ form a subgroup of $Z .$
${0, 2, 4}$ is a subgroup of $Z /6,$ and so is ${0, 3}$ (on a 6-hour clock, $3 + 3 = 0).$
In $D_{3} :$ the rotations ${e, r, r^{2}},$ and also each ${e, f_{1}},$ ${e, f_{2}},$ ${e, f_{3}} .$
Every group has two boring subgroups: the trivial subgroup ${e}$ and the whole group $G .$

And some subsets that are not subgroups:

The odd integers: they do not contain $0$ (and are not closed).
The natural numbers inside $Z :$ they contain $0$ and are closed, but $3$ has no inverse among them.
${0, 1, 2}$ inside $Z /6 :$ $1 + 2 = 3$ is missing.

Generators and cyclic groups

Pick an element $a$ of an abelian group and ask: what is the smallest subgroup that contains it? It must contain $a + a = 2 a,$ and $3 a,$ and so on; it must contain $0,$ and $- a,$ $- 2 a,$ and so on. And the collection of all these multiples is already a subgroup (the sum of two multiples of $a$ is a multiple of $a).$ So the answer is $⟨ a ⟩ = {k a : k \in Z} = {\dots, - 2 a, - a, 0, a, 2 a, 3 a, \dots},$ read “the subgroup generated by $a ”.$ (In multiplicative notation it consists of the powers $a^{k} .)$ The colon inside the braces is read “such that” or “where”: “all $k a,$ where $k$ is an integer”.

$Z$ is cyclic: every integer is a multiple of $1,$ so $Z = ⟨ 1 ⟩ .$ (It is also $⟨ - 1 ⟩,$ but not $⟨ 2 ⟩,$ which is only the even numbers.) Every clock $Z / n$ is cyclic, generated by $1 :$ every position is reached by repeatedly stepping one hour. The rotations of the triangle are cyclic, generated by $r .$

$D_{3}$ is not cyclic. A rotation generates only the rotations, and a flip $f$ generates only ${e, f} .$ But two elements suffice: using the products you found in Figure 1.3.1, $r f_{1} = f_{3}$ and $r^{2} f_{1} = f_{2},$ so $D_{3} = {e, r, r^{2}, f_{1}, r f_{1}, r^{2} f_{1}}$ is generated by $r$ and $f_{1} .$

The order of an element

On a clock, keep adding the same number and you must eventually come back to $0 .$ On a 12-hour clock, starting from $0$ and adding $3$ each time gives $3, 6, 9, 12 = 0 :$ four steps. That count will matter in homology. A Klein bottle carries a loop that is not the edge of any region, yet the same loop traced twice is one: in homology it behaves like a clock hand that returns after two steps (§3.4 Computing Homology).

The order of $a$ is exactly the number of elements of $⟨ a ⟩ :$ the walk $0, a, 2 a, \dots$ visits new elements until it first returns to $0,$ and then repeats. In $Z,$ every element except $0$ has infinite order — adding $3$ to itself never gives $0 .$ In $D_{3},$ the flips have order $2$ and $r,$ $r^{2}$ have order $3 .$

In $Z / n$ there is a neat formula. Write $g cd (n, k),$ the greatest common divisor, for the largest whole number that divides both $n$ and $k;$ for example $g cd (12, 9) = 3$ and $g cd (12, 5) = 1 .$ Then the order of $k$ in $Z / n$ is $order (k) = \frac{n}{g cd ( n , k )} .$ (The walk $0, k, 2 k, \dots$ returns to $0$ the first time it reaches a common multiple of $k$ and $n,$ and the smallest such is $nk / g cd (n, k),$ reached after $n / g cd (n, k)$ steps; Judson (2025, Thm 4.1.13) gives a careful proof.) In particular, $k$ generates all of $Z / n$ exactly when $g cd (n, k) = 1 .$ On a 12-hour clock the generators are $1, 5, 7, 11 .$

Walking around a clock Tap a dot to choose k · change n

⟨ 5 ⟩ = {0, 5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7}

The order of

5

is the number of dots visited:

order (5) = \frac{12}{g cd ( 12 , 5 )} = \frac{12}{1} = 12

5

generates all of

Z /12

All subgroups of

Z /12

show all generators

n

k

Figure 1.3.4 Starting at

0

and repeatedly adding

k

traces a star that closes up when it returns to

0 .

The dots it visits form the subgroup

⟨ k ⟩ .

Which

k

visit every dot? What happens when

n

is prime? The chips list every subgroup of

Z / n :

there is exactly one for each divisor of

n .

In a finite group every element has finite order, since the walk $e, a, a^{2}, \dots$ must eventually repeat. In 1902 William Burnside asked about the converse. Suppose a group is generated by finitely many elements and every element has finite order; must the group be finite? In general the answer is no, as Evgeny Golod and Igor Shafarevich showed in 1964. But Burnside also asked a sharper version, with one fixed order for everything, and one small case of it has resisted for more than a century.

Open problem The Burnside problem for exponent 5

Open since 1902 Posed by William Burnside, 1902

Take two letters $a$ and $b,$ form all words in them, and impose only the rule that $w^{5} = 1$ for every word $w .$ Is the resulting group, called $B (2, 5),$ finite?

Every element of $B (2, 5)$ has order $1$ or $5,$ and two elements generate the whole group. The question is whether these two facts force the group to be finite.

What is known. With $w^{2} = 1,$ $w^{3} = 1,$ $w^{4} = 1$ or $w^{6} = 1$ in place of $w^{5} = 1$ the group is finite; for large odd exponents it is infinite (Novikov and Adian, 1968). The largest finite group of this kind with exponent $5$ is known, with $5^{34}$ elements (Havas, Wall and Wamsley, 1974), and Zelmanov proved that such a largest finite group always exists, work recognised with a Fields Medal in 1994. $B (2, 5)$ is finite exactly when it is that group, and nobody knows whether it is.

A way in. Start small. If every word squares to $1,$ show that $ab = ba$ (expand $(ab)^{2} = 1)$ and that the group has exactly $4$ elements. With cubes instead of squares the group has $27$ elements; try to list them.

(Burnside 1902; Havas, Wall & Wamsley 1974) Status as of October 2026 All open problems

All the subgroups of the integers

Which subsets of $Z$ are subgroups? We have found $n Z$ for every $n .$ Could there be stranger ones — say, the smallest subgroup containing both $4$ and $6 ?$ Try it: that subgroup must contain $6 - 4 = 2,$ hence every even number; and anything built from $4$ s and $6$ s by adding and subtracting is even, so it contains nothing odd. It is $2 Z$ all over again. That is no accident.

(For $4$ and $6,$ the proof’s $n$ is $2 = g cd (4, 6) .)$ The theorem will pay off in the next chapter, which collapses a subgroup of $Z$ to zero: since every subgroup is some $n Z,$ the clocks $Z / n$ (and $Z$ itself, for $n = 0)$ are the only possible results.

Maps that respect the rules: homomorphisms

Every integer is either even or odd, and there are well-known rules: even plus odd is odd, odd plus odd is even. Look at what those rules let you do. To decide whether $a + b$ is even you never need $a$ and $b$ themselves — only whether they are even. Write $p (m)$ for the parity of $m :$ $p (m) = 0$ if $m$ is even and $p (m) = 1$ if $m$ is odd. Then the parity rules say $p (a + b) = p (a) + p (b) for all integers a, b,$ where the addition on the right is the addition of $Z /2 .$ The function $p : Z \to Z /2$ translates the arithmetic of $Z$ faithfully into the arithmetic of $Z /2 .$ It loses a great deal of information — it forgets everything about a number except its parity — but what it keeps, it keeps correctly.

The same is true of a clock. If $c (m)$ is the clock reading after $m$ hours, then $c (a + b) = c (a) + c (b) :$ waiting $a$ hours and then $b$ more moves the hand as the two waits combined. And it is true of the triangle: let $s (g) = + 1$ if the symmetry $g$ leaves the card face up (a rotation), and $s (g) = - 1$ if it turns the card over (a flip). Turning the card over twice leaves it face up, so $s (g h) = s (g) s (h) .$ This function $s : D_{3} \to {+ 1, - 1}$ is called the sign. In each case, combining and then translating gives the same answer as translating and then combining.

The Greek letter $φ$ is read “phi” (rhymes with “fie”). The word comes from the Greek for “same shape”: a homomorphism carries the shape of the operation across. On the left of the equation we combine in $G$ and then apply $φ;$ on the right we apply $φ$ and then combine in $H .$ A homomorphism is a map for which the two routes agree.

Two consequences follow immediately, and both are used constantly. In additive notation:

$φ (0) = 0 .$ Indeed $φ (0) = φ (0 + 0) = φ (0) + φ (0);$ subtracting $φ (0)$ from both sides leaves $0 = φ (0) .$
$φ (- a) = - φ (a) .$ Indeed $φ (a) + φ (- a) = φ (a + (- a)) = φ (0) = 0,$ so $φ (- a)$ is the inverse of $φ (a) .$

Wrapping the integers around a clock Drag to rotate · change n

φ : Z \to Z /4, m \mapsto m mod 4

ker φ = 4 Z = {\dots, - 8, - 4, 0, 4, 8, \dots}

φ^{- 1} (1) = {\dots, - 7, - 3, 1, 5, 9, \dots}

n

(integers per turn) 4

highlight what lands on

a

Figure 1.3.5 The number line is coiled into a spring with

n

integers per turn; looking straight down, each integer lands on one of the

n

positions of the dial below. This is the homomorphism

φ : Z \to Z / n .

The gold beads are everything that lands on

0

— the multiples of

n

— and they stack up in a single column. Choose a nonzero

a

to light up, in teal, the integers that land on

a :

another column, the gold one shifted by

a .

Kernel and image

Every homomorphism comes with two subgroups that describe it. Figure 1.3.5 shows both: the gold column is one, and the fact that every position of the dial is hit is the other.

The kernel lives in the starting group $G;$ the image lives in the target $H .$ (The image is the same as the image of a function from §1.1 Sets and Functions; the kernel is the preimage of $0 .)$ Let us compute them for our examples.

Parity $p : Z \to Z /2 :$ the kernel is the even integers $2 Z;$ the image is all of $Z /2 .$
Reduction $Z \to Z / n :$ the kernel is $n Z,$ the gold column in Figure 1.3.5; the image is all of $Z / n .$
The sign $s : D_{3} \to {+ 1, - 1} :$ the kernel is everything sent to the identity $+ 1,$ namely the rotations ${e, r, r^{2}};$ the image is ${+ 1, - 1} .$
Doubling on a 6-hour clock, $φ : Z /6 \to Z /6,$ $φ (x) = 2 x :$ the values are $0, 2, 4, 0, 2, 4$ for $x = 0, 1, 2, 3, 4, 5 .$ So $ker φ = {0, 3}$ and $im φ = {0, 2, 4} .$

A homomorphism under the lens Change m, n and φ(1) · hover the top dots

φ : Z /6 \to Z /6, φ (x) = 2 x

ker φ = {0, 3}

im φ = {0, 2, 4}

not injective (kernel is bigger than 0) not surjective

∣ ker φ ∣ \cdot ∣ im φ ∣ = 2 \cdot 3 = 6

Hover or tap a top dot to see which elements share its image.

m (domain ℤ/m) 6

n (target ℤ/n) 6

choose

φ (1)

Figure 1.3.6 A homomorphism from

Z / m

is completely decided by where it sends

1,

since every element is

1 + 1 + \dots + 1 .

When

m

and

n

differ, only some choices are allowed (try

m = 4,

n = 6 :

the greyed-out choices would not respect

m \cdot 1 = 0).

The kernel glows gold and the image teal. Hover a dot to see everything that lands in the same place: always a copy of the kernel, shifted. Notice that the number of elements in the kernel times the number in the image is always

m .

The kernel detects injectivity

A function is injective (one-to-one) if different inputs always give different outputs. For a general function, checking this means comparing every pair of inputs. For a homomorphism there is a shortcut: you only have to look at what lands on $0 .$

The observation in the proof is worth more than the theorem. It says that the elements landing on the same point as $a$ are exactly the elements $a + k$ with $k$ in the kernel — a copy of the kernel, shifted to $a .$ You saw these shifted copies as the teal column in Figure 1.3.5 and as the highlighted dots in Figure 1.3.6. The kernel measures precisely how far $φ$ is from being injective. In the next chapter, these shifted copies get a name — cosets — and become the elements of a new group.

Same group, different names: isomorphism

Write out the table of the rotations of a square, ${e, ρ, ρ^{2}, ρ^{3}},$ and next to it the addition table of $Z /4 .$ Rename $e \mapsto 0,$ $ρ \mapsto 1,$ $ρ^{2} \mapsto 2,$ $ρ^{3} \mapsto 3,$ and the first table turns into the second, entry for entry. The two groups are made of different things — turns of a square, hours on a clock — but as groups they are indistinguishable.

An isomorphism is exactly a renaming: a way of matching up the elements of $G$ and $H$ one for one so that the table of $G$ becomes the table of $H .$ (The inverse of an isomorphism is automatically an isomorphism too, so the relationship goes both ways.) Some examples:

The rotations of a square $≅ Z /4,$ by $ρ^{k} \mapsto k .$
The two signs ${+ 1, - 1}$ under multiplication $≅ Z /2,$ by $+ 1 \mapsto 0,$ $- 1 \mapsto 1 .$
The rotations of the triangle ${e, r, r^{2}} ≅ Z /3 .$
$(Z /5 ∖ {0}, \times) ≅ Z /4 :$ the powers of $2$ on a 5-hour clock are $2^{0} = 1,$ $2^{1} = 2,$ $2^{2} = 4,$ $2^{3} = 8 = 3,$ so the renaming $k \mapsto 2^{k}$ turns clock addition mod 4 into clock multiplication mod 5.
$Z ≅ 2 Z,$ by $x \mapsto 2 x .$ An infinite group can be isomorphic to a part of itself — the integers and the even integers are “the same group”, merely with every name doubled.

An isomorphism is a renaming Switch the group · try the renamings

Its own table (names)

\circ

e

ρ

ρ^{2}

ρ^{3}

e

e

ρ

ρ^{2}

ρ^{3}

ρ

ρ

ρ^{2}

ρ^{3}

e

ρ^{2}

ρ^{2}

ρ^{3}

e

ρ

ρ^{3}

ρ^{3}

e

ρ

ρ^{2}

e \mapsto 0

ρ \mapsto 1

ρ^{2} \mapsto 2

ρ^{3} \mapsto 3

After renaming, compared with

Z /4

+

All 16 entries agree with

Z /4

: this renaming is an isomorphism.

Figure 1.3.7 The first table is a group’s own table, coloured by the number each element is renamed to. In the second, the renamed table is compared cell by cell with the addition table of

Z /4 :

green where they agree, rose where they do not. For the square, two of the six renamings work. For the rectangle, none does.

How can we be sure that no renaming turns the rectangle’s table into that of $Z /4 ?$ Trying all six is one way, but there is a better one. An isomorphism preserves every property that can be stated in terms of the operation — being abelian, the number of subgroups, the orders of elements (Judson 2025, Thm 9.1.6). In $Z /4$ the element $1$ has order $4 .$ In the rectangle group every element has order $1$ or $2 :$ every symmetry of a rectangle, done twice, is $e .$ An isomorphism would have to send $1$ to an element of order $4,$ and there is none. So the two groups are not isomorphic, even though both have four elements.

Similarly, $Z /6$ and $D_{3}$ both have six elements, but $Z /6$ is abelian and $D_{3}$ is not, so they cannot be isomorphic. This is the strategy of invariants, the same one topology uses to prove that two shapes are different (§0.1 The Shape of a Question): find a property that isomorphic groups must share, and show that it differs.

That completes the vocabulary, and leaves one question hanging. The warning “ $Z / n$ is not a subgroup of $Z ”$ said that clock arithmetic is the integers with the multiples of $n$ declared equal to zero. How do you do that to any group and any subgroup, and get a group back? The answer, the quotient group, opens the next chapter. It is the piece of algebra on which the definition of homology rests.

Exercises

Summary

Recap

A group is a set with an operation that is closed and associative, has an identity element, and has inverses. The symmetries of a triangle and the arithmetic of a clock are both groups.
Identities and inverses are unique; $(g h)^{- 1} = h^{- 1} g^{- 1};$ the cancellation law makes every Cayley table a Latin square.
A group is abelian if $ab = ba$ always. $D_{3}$ is not abelian. Abelian groups are written additively, and all the groups of homology are abelian.
Subgroups are groups inside groups; $⟨ a ⟩$ is the subgroup generated by $a;$ cyclic groups have a single generator; the order of $k$ in $Z / n$ is $n / g cd (n, k);$ every subgroup of $Z$ is some $n Z .$
A homomorphism respects the operation. Its kernel (what is crushed to $0)$ and image (what is reached) are subgroups, and it is injective exactly when its kernel is ${0} .$ Elements with the same image differ by an element of the kernel.
An isomorphism is a bijective homomorphism — a renaming. Isomorphic groups share every structural property, which is how we prove two groups are different.

Groups