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The four conditions that define a group, the six examples this course keeps using, and why $ab = ba$ is deliberately not among the axioms.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state the group axioms, check a set with an operation against them one at a time, name the identity and the inverse of an element in each of the standard examples, count the elements of a dihedral group, decide whether a group is abelian as a question separate from whether it is a group at all, and prove from the axioms alone that a group has only one identity.
Addition on the integers, multiplication on the non-zero rationals, composition of functions: you have used all three for years, and you have used the same three facts about each of them without naming them. Brackets do not matter. There is an element that changes nothing. Everything can be undone. This lesson names those facts, and from here on they are the only thing assumed.
A binary operation on a set $G$ assigns to each ordered pair $a, b$ of elements an element written $ab$. A group is a set with a binary operation that is associative, has an identity, and gives every element an inverse. A group is abelian when the operation also commutes. The order of a group, written $|G|$, is how many elements it has; a group with finitely many is a finite group. The trivial group has one element, the identity alone.
A group is a set $G$ together with an operation that takes two elements of $G$ and returns one, subject to:
That is the whole definition. What is not there matters as much as what is: nothing says $ab = ba$. A group in which it does hold is called abelian, after Abel, and plenty of the important groups are not abelian — the symmetries of a square are not, the permutations of three letters are not, and invertible matrices are not.
The operation is written multiplicatively, $ab$, whatever it actually is; in an abelian group it is often written $a + b$ instead, with identity $0$ and inverse $-a$. The notation is a convention and carries no mathematics.
An algebraic fact is about the operation, not about what the elements happen to be. The same group turns up as rotations of a square, as residues under addition and as matrices, and a theorem proved once from the axioms holds in all of them at once.
Another way: picture
Hold a square by its centre. Turn it a quarter turn, a half turn, three quarters, or not at all: four rotations. Flip it about either diagonal or either of the two lines through opposite edge midpoints: four reflections. Eight motions in all, and doing one and then another always gives one of the eight. That is a group with eight elements — and a quarter turn followed by a flip is not the same motion as the flip followed by the quarter turn, which is a non-abelian group you can hold in your hand.
Another way: steps
To check a set with an operation is a group: 1. Closure: take two elements, combine them, and confirm the result is still in the set. 2. Associativity: usually inherited, because the operation is really addition, multiplication or composition of functions. 3. Identity: find the one element that changes nothing. 4. Inverses: for a general element, produce the element that undoes it — and check it is in the set. Then ask, separately, whether the operation commutes.
Six groups do almost all the work in a first course, and it is worth being able to write down each one's elements, operation, identity and order without thinking.
| Group | Elements | Operation | Identity | Order |
|---|---|---|---|---|
| $\mathbb{Z}$ | all integers | addition | $0$ | infinite |
| $\mathbb{Z}_n$ | $0, 1, \ldots, n-1$ | addition modulo $n$ | $0$ | $n$ |
| $U(n)$ | residues coprime to $n$ | multiplication modulo $n$ | $1$ | $\phi(n)$ |
| $D_n$ | symmetries of a regular $n$-gon | composition | do nothing | $2n$ |
| $S_n$ | permutations of $n$ letters | composition | the identity map | $n!$ |
| $GL_2(\mathbb{R})$ | invertible $2 \times 2$ matrices | matrix product | the identity matrix | infinite |
The first three are abelian; the last three are not, once $n \ge 3$.
$U(n)$ deserves a word, because closure is not obvious. If $a$ and $b$ are coprime to $n$ then so is $ab$, so the product stays inside; and an inverse exists exactly because $\gcd(a, n) = 1$ lets Bézout write $ax + ny = 1$, which says $ax \equiv 1 \pmod n$. That is the whole reason the condition is coprime rather than non-zero.
Forgetting to check closure. The odd integers are not closed under addition; the reflections of a square are not closed under composition. Neither is a group, and in both cases every other axiom looks fine.
Treating commutativity as an axiom. It is not one. Writing $ab = ba$ in the middle of a proof about a general group is the single commonest error in this subject, and it is silent: the line looks like algebra.
Checking the identity on one element. The identity has to work for every element, from both sides. A matrix can leave one vector alone without being the identity matrix.
Producing an inverse that is outside the set. $2$ has an inverse in the rationals and not in the integers, so the integers under multiplication are not a group. The inverse must live in $G$.
Thinking a bigger set is more likely to be a group. Enlarging a set can break closure and can add elements with no inverse. The non-zero rationals form a group under multiplication; adding $0$ back destroys it.
Elements $\{0, 1, 2, 3, 4, 5\}$, operation addition then take the remainder on division by $6$. Closure is immediate: a remainder modulo $6$ is one of those six numbers.
Closure.
Associativity is inherited from addition of integers, and $0$ changes nothing.
Associativity and identity.
The inverse of $a$ is $6 - a$ for $a \ne 0$, and $0$ is its own: $2 + 4 = 6 \equiv 0$.
Inverses, and the group is abelian because addition is.
Closure holds, multiplication is associative, and $1$ is an identity, so three of the four conditions pass.
Three out of four is not enough.
But $2b = 1$ has no integer solution, so $2$ has no inverse. The axiom fails for every integer except $1$ and $-1$.
One failed axiom is fatal.
Restricting to $\{1, -1\}$ does give a group of order $2$, and widening to the non-zero rationals also does.
The same operation, two different sets, two different answers.
Closure: a product of two positive rationals is a positive rational. Associativity is inherited. The identity is $1$.
Three conditions, quickly.
Inverses: the reciprocal of $p/q$ is $q/p$, which is positive and rational.
The fourth condition.
So it is a group, and an abelian one. Note that the positive integers under multiplication are not, for exactly the reason the whole integers were not.
Match each condition in the definition of a group to the sentence that states it.
| $(ab)c = a(bc)$ for all $a$, $b$, $c$ | $ab$ is again an element of the set | there is an $e$ with $ea = ae = a$ for every $a$ | for every $a$ there is a $b$ with $ab = ba = e$ | $ab = ba$ for all $a$ and $b$ | |
|---|---|---|---|---|---|
| Closure | |||||
| Associativity | |||||
| Identity | |||||
| Inverses |
Count the symmetries of a regular polygon with $9$ sides.
| How many | |
|---|---|
| Rotations | |
| Reflections | |
| Symmetries altogether |
What is the natural numbers under addition?
Select every statement that is true in every group.
This task has no paper form; do it on a device.
How many elements does the group of symmetries of a regular polygon with $12$ sides have?
Answer:
Build the proof that a group has only one identity element.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
What is the natural numbers under addition?
You can check the group axioms on a set with an operation and say which one fails when one does. Say in your own words why commutativity is not on the list, and name a group that does not have it. Next: the Cayley table, which is the whole of a finite group written down at once.
9. Your turn: is the set of positive rationals a group under multiplication?, step 3