This deck presents the four group axioms as the upgrade from a monoid, then proves that the identity and inverses are unique and covers the cancellation law and the socks-and-shoes rule. It distinguishes the order of a group from the order of an element, and abelian from non-abelian groups, and works the standard first examples: the additive integers, the integers mod n, the units mod n, the symmetric group, and the dihedral symmetries of a square. It targets the real traps: forgetting closure or inverses when calling a subset a group, assuming every group is abelian, confusing the order of a group with the order of an element, and thinking that the integers mod n form a group under multiplication.
Subject: Foundations of Higher Mathematics · 109 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you can:
1. State the four group axioms precisely and check them for a candidate group.
2. Prove that the identity and inverses are unique, and that the cancellation law holds.
3. Compute the order of a group and the order of an element, and tell the two apart.
4. Recognize the standard first examples: the additive integers, the integers modulo n, the units modulo n, and the symmetric and dihedral groups.
5. Distinguish abelian from non-abelian groups and read structure off a Cayley table.
Warm-up
Discussion prompt
Before we open Groups: Axioms & First Examples: without looking back, what was the main idea of Binary Operations, Semigroups & Monoids, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
Builds the algebraic ladder magma to semigroup to monoid from the ground up: closure, associativity, and identity as separate axioms you must actually check. Targets the misconceptions that closure is automatic, that every operation is associative or commutative, and that the identity is the same thing as an absorbing element.
Concept
A monoid gives you an associative operation with an identity, but it makes no promise that you can undo anything. A group adds exactly that missing power: every element can be reversed.
So a group is a monoid in which every element has an inverse. That single extra axiom separates arithmetic you can only build up from arithmetic you can also take apart.
group — A set G with a binary operation that is closed, associative, has a two-sided identity element, and in which every element has a two-sided inverse.
Counterexample
Discussion prompt
A monoid gives you an associative operation with an identity, but it makes no promise that you can undo anything. A group adds exactly that missing power: every element can be reversed.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Intuition
Think of the elements as moves you can make: rotate a square, shuffle a deck, add a number. The operation means doing one move and then another.
The identity is the do-nothing move. The inverse axiom says every move can be undone by another move that is itself in the set. That reversibility is the entire idea.
This is why groups are the mathematics of symmetry: a symmetry is a transformation you can always reverse and still land on the same object.
Analogy
Discussion prompt
Explain A group is a set of reversible moves by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Think of the elements as moves you can make: rotate a square, shuffle a deck, add a number. The operation means doing one move and then another.
Concept
The first axiom is closure: combining two elements never takes you outside the set.
\[ \forall a,b \in G, \quad a \ast b \in G \]
Closure is often hidden inside the phrase a binary operation on G, since a genuine operation on G must return a value in G. Check it anyway: it is the axiom people quietly assume.
Explain it
Discussion prompt
Explain Axiom G1: closure to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
The first axiom is closure: combining two elements never takes you outside the set.
Concept
The second axiom is associativity: when you combine three elements, the grouping does not matter.
\[ \forall a,b,c \in G, \quad (a \ast b) \ast c = a \ast (b \ast c) \]
Associativity is what lets us drop parentheses and write a long product unambiguously. Notice it says nothing about the order of the elements, only about how they are grouped.
Concept
The third axiom guarantees a do-nothing element.
identity element — An element e of G that, combined with any element on either side, changes nothing.
\[ \exists\, e \in G \ \text{such that}\ \forall a \in G,\ e \ast a = a \ast e = a \]
The identity must work on both sides at once. We will prove shortly that a group can have only one such element.
Concept
The fourth axiom is the one that upgrades a monoid to a group: every element can be undone.
inverse of a — An element that combines with a on both sides to give the identity. It must itself be an element of G.
\[ \forall a \in G\ \exists\, b \in G \ \text{such that}\ a \ast b = b \ast a = e \]
The inverse must return the identity from both sides, and it must live inside G. This is precisely the axiom that fails for the natural numbers under addition.
Concept
Putting the four axioms together gives the definition you should be able to recite on demand.
\[ \textbf{(G1)}\ a\ast b \in G \qquad \textbf{(G2)}\ (a\ast b)\ast c = a\ast(b\ast c) \]
\[ \textbf{(G3)}\ e\ast a = a\ast e = a \qquad \textbf{(G4)}\ a\ast a^{-1} = a^{-1}\ast a = e \]
If in addition the operation is commutative, the group is called abelian. Commutativity is a bonus property, not part of the core definition.
Concept
Two notations are standard, and switching between them is a common source of confusion.
In multiplicative notation we write the operation as juxtaposition, the identity as one, and the inverse with a small negative-one exponent. In additive notation we write plus, the identity as zero, and the inverse as negative a.
| role | multiplicative | additive |
|---|---|---|
| operation | a b | a + b |
| identity | 1 | 0 |
| inverse of a | inverse of a | negative a |
By convention, additive notation is reserved for abelian groups. We will use multiplicative notation for the general theory and additive notation for the integers and the modular groups.
Pattern
Step through it
Step through Two notations for one idea one row at a time. What is driving the change, and what would the row after the last one be?
Concept
The size of a group has a name and a notation.
order of a group — The number of elements in G, written with vertical bars. A group may be finite or infinite.
\[ |G| = \text{the number of elements of } G \]
The integers under addition form an infinite group. The integers modulo n form a finite group of order n. The order is the first fact you report about a finite group.
Definition probe
Sort into buckets
Every line below is part of the definition of inverse of a or of order of a group — one or the other, never both. Put each where it belongs.
Ranking
Put in order
Put the moves of Worked example: the integers under addition are a group into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The sum of two integers is an integer, so addition never leaves the set.
Worked example
Let us certify the most familiar group of all: the integers under addition.
\[ (\mathbb{Z}, +) \]
Check closure
Why: The sum of two integers is an integer, so addition never leaves the set.
\[ a, b \in \mathbb{Z} \implies a + b \in \mathbb{Z} \]
Check associativity
Why: Integer addition is associative; regrouping the terms does not change the total.
\[ (a + b) + c = a + (b + c) \]
Exhibit the identity
Why: Zero added to any integer returns it unchanged, on both sides, so 0 is the identity.
\[ 0 + a = a + 0 = a \]
Exhibit inverses
Why: The negative of an integer is again an integer, and the two sum to zero on both sides.
\[ a + (-a) = (-a) + a = 0 \]
Verify all four axioms hold
Why: Closure, associativity, the identity 0, and the inverse negative a are each satisfied, and addition is commutative, so the integers under addition form an abelian group of infinite order.
Picture it
Animation
Shows: Each line of the worked example "the integers under addition are a group", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The sum of two integers is an integer, so addition never leaves the set.
Concept
One property splits the group world in two.
abelian group — A group whose operation is commutative: the order in which you combine two elements never changes the result.
\[ \forall a,b \in G, \quad a \ast b = b \ast a \]
A group where some pair fails to commute is called non-abelian. The smallest non-abelian group has six elements, and we will meet it as the symmetries of a triangle.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of group, identity element, inverse of a, order of a group, abelian group as Groups: Axioms & First Examples uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Intuition
Abelian means the moves do not interfere: doing move a then move b lands you exactly where doing b then a would.
Adding numbers is abelian, since the order of the two numbers is irrelevant. But composing physical motions usually is not: flip a card and then turn it, versus turn it and then flip it, generally disagree.
So commutativity is a genuine restriction. When a source calls a group abelian, it is telling you that reordering is always safe.
Step zero
Discussion prompt
Worked example: the integers mod 6 under addition — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Confirm closure through the remainder
Answer:
Worked example
Now a finite group: the integers modulo 6, where we add and then take the remainder after dividing by 6.
\[ \mathbb{Z}/6\mathbb{Z} = \{0,1,2,3,4,5\} \]
Confirm closure through the remainder
Why: Adding two residues and reducing mod 6 always lands back among the remainders 0 through 5.
\[ (a + b) \bmod 6 \in \{0,1,2,3,4,5\} \]
Identify the identity
Why: Adding 0 changes nothing, so 0 is the identity element.
Find each inverse
Why: For a nonzero residue a, the element 6 minus a sums with it to 6, which reduces to 0; and 0 is its own inverse.
\[ a + (6 - a) \equiv 0 \pmod 6 \]
Lay out the Cayley table
Why: Each row and column is a permutation of all six residues, which confirms closure and shows every element appears exactly once.
| + | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| 0 | 0 | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 2 | 3 | 4 | 5 | 0 |
| 2 | 2 | 3 | 4 | 5 | 0 | 1 |
| 3 | 3 | 4 | 5 | 0 | 1 | 2 |
| 4 | 4 | 5 | 0 | 1 | 2 | 3 |
| 5 | 5 | 0 | 1 | 2 | 3 | 4 |
Verify it is an abelian group of order 6
Why: The table is symmetric across its main diagonal, 0 is the identity, every row contains a 0 so every element has an inverse, and associativity is inherited from integer addition. So it is an abelian group of order 6.
Picture it
Animation
Shows: Each line of the worked example "the integers mod 6 under addition", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The table is symmetric across its main diagonal, 0 is the identity, every row contains a 0 so every element has an inverse, and associativity is inherited from integer addition. So it is an abelian group of order 6.
Concept
The axioms demand at least one identity. In fact there is exactly one, and the argument is a one-line proof worth memorizing.
Suppose two elements both acted as identities. Combine them: read from one side the product is the first, read from the other side it is the second, so the two are equal.
\[ e = e \ast e' = e' \]
So we may speak of the identity of a group, not merely an identity. Uniqueness is what makes the notation well defined.
Estimation
Predict first
We prove the claim in full: a group has exactly one identity element.
Commit before you compute: what does Worked example: the identity is unique come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the two identities are equal
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Both computations describe the single product of e with e-prime, so e-prime equals e.
Worked example
We prove the claim in full: a group has exactly one identity element.
Assume two identities exist
Why: Suppose e and e-prime both satisfy the identity axiom on both sides. We show they must coincide.
\[ e \ast a = a \ast e = a, \qquad e' \ast a = a \ast e' = a \]
Read the product using e as an identity
Why: Since e is an identity, combining it with e-prime leaves e-prime unchanged.
\[ e \ast e' = e' \]
Read the same product using e-prime as an identity
Why: Since e-prime is an identity, combining e with it leaves e unchanged.
\[ e \ast e' = e \]
Verify the two identities are equal
Why: Both computations describe the single product of e with e-prime, so e-prime equals e. A group has exactly one identity element.
\[ e' = e \ast e' = e \]
Picture it
Animation
Shows: Each line of the worked example "the identity is unique", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Both computations describe the single product of e with e-prime, so e-prime equals e. A group has exactly one identity element.
Concept
The same style of argument pins down inverses. Once the identity is fixed, each element has exactly one inverse.
The trick is to sandwich a hypothetical second inverse between an element and its first inverse, and let associativity collapse the expression.
\[ b = b \ast e = b \ast (a \ast c) = (b \ast a) \ast c = e \ast c = c \]
So the phrase the inverse of a is justified, and we may safely give it a single notation.
Hypothesis
Predict first
Worked example: the inverse of an element is unique is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Start from b and insert the identity
Why: Combining b with the identity changes nothing, so we may write e in place of it.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Here is that computation slowed down, one associativity move at a time.
\[ \text{Suppose } a \ast b = b \ast a = e \text{ and } a \ast c = c \ast a = e. \]
Start from b and insert the identity
Why: Combining b with the identity changes nothing, so we may write e in place of it.
\[ b = b \ast e \]
Replace the identity by a combined with c
Why: Because c is an inverse of a, the product of a with c equals e; substitute it in.
\[ b = b \ast (a \ast c) \]
Reassociate the triple product
Why: Associativity lets us regroup so that b meets a first.
\[ b \ast (a \ast c) = (b \ast a) \ast c \]
Collapse using that b is an inverse of a
Why: The product of b with a equals e, and e combined with c is just c.
\[ (b \ast a) \ast c = e \ast c = c \]
Verify both inverses coincide
Why: Chaining the equalities gives b equals c, so any two inverses of a are equal: the inverse is unique.
\[ b = c \]
Picture it
Animation
Shows: Each line of the worked example "the inverse of an element is unique", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Chaining the equalities gives b equals c, so any two inverses of a are equal: the inverse is unique.
Intuition
Uniqueness is what lets us drop the article a and say the inverse. Without it, inverse notation would be ambiguous.
It also means undoing a move has only one answer. There is never a choice of how to reverse an element; the reversal is forced.
This is quietly powerful: many later proofs apply the inverse to both sides of an equation, and they only make sense because that inverse is unambiguous.
Concept
In a group you can cancel a common factor, exactly as in ordinary algebra, and the reason is inverses.
cancellation law — If a combined with b equals a combined with c, then b equals c; and likewise on the right. Equal products sharing a factor force the other factors equal.
\[ a \ast b = a \ast c \implies b = c \]
Cancellation is why every row and every column of a finite group table contains each element exactly once: a repeat would violate it.
Step zero
Discussion prompt
Worked example: proving left cancellation — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Combine both sides with the inverse of a on the left
Answer:
Worked example
Prove the left cancellation law directly from the axioms.
\[ \text{Given } a \ast b = a \ast c, \text{ show } b = c. \]
Combine both sides with the inverse of a on the left
Why: a has an inverse in the group; applying it to both sides on the same side keeps the equation balanced.
\[ a^{-1} \ast (a \ast b) = a^{-1} \ast (a \ast c) \]
Reassociate each side
Why: Associativity lets the inverse of a meet a first on both sides.
\[ (a^{-1} \ast a) \ast b = (a^{-1} \ast a) \ast c \]
Simplify the inverse pair
Why: The inverse of a combined with a is the identity.
\[ e \ast b = e \ast c \]
Verify the cancellation
Why: The identity leaves each side unchanged, giving b equals c. The identical argument on the right proves right cancellation.
\[ b = c \]
Picture it
Animation
Shows: Each line of the worked example "proving left cancellation", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The identity leaves each side unchanged, giving b equals c. The identical argument on the right proves right cancellation.
Concept
The inverse of a product reverses the order of the factors. The picture is dressing: socks first, then shoes; to undo, take off the shoes first, then the socks.
\[ (a \ast b)^{-1} = b^{-1} \ast a^{-1} \]
In an abelian group the order would not matter, but in general it does. Getting this backwards is one of the most common slips in a first proof course.
Missing information
Discussion prompt
We verify the rule by checking that the proposed inverse actually works on both sides.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Combine a times b with the candidate and let the inner inverse pair cancel first.
Worked example
We verify the rule by checking that the proposed inverse actually works on both sides.
\[ \text{Claim: } (a \ast b)^{-1} = b^{-1} \ast a^{-1} \]
Multiply on the right
Why: Combine a times b with the candidate and let the inner inverse pair cancel first.
\[ (a \ast b)(b^{-1} \ast a^{-1}) = a(b \ast b^{-1})a^{-1} = a \ast a^{-1} = e \]
Multiply on the left
Why: Combine the candidate with a times b in the other order; again the inner pair cancels first.
\[ (b^{-1} \ast a^{-1})(a \ast b) = b^{-1}(a^{-1} \ast a)b = b^{-1} \ast b = e \]
Verify by uniqueness
Why: The candidate returns the identity from both sides, so it is an inverse of a times b; and inverses are unique, so it is the inverse. The reversed order is essential.
\[ (a \ast b)^{-1} = b^{-1} \ast a^{-1} \]
Picture it
Animation
Shows: Each line of the worked example "the socks-and-shoes rule", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The candidate returns the identity from both sides, so it is an inverse of a times b; and inverses are unique, so it is the inverse. The reversed order is essential.
Trap
Claiming the natural numbers under addition form a group because addition is closed, associative, and has the identity 0.
\[ (\mathbb{N}, +): \quad 0 + a = a, \quad (a+b)+c = a+(b+c) \]
Three axioms hold, so it looks finished. But ask for the inverse of 3: you need a natural number that adds to 3 to give 0, and there is none.
Check the inverse axiom explicitly. The only candidate for the inverse of 3 is negative 3, which is not a natural number.
\[ 3 + x = 0 \implies x = -3 \notin \mathbb{N} \]
So the naturals under addition are a monoid but not a group. To get a group you must enlarge the set to the integers, which supply the missing inverses.
Concept
A warning about the axioms people skip. An operation must actually stay inside the set, and it must actually be associative; do not take either for granted.
Subtraction on the natural numbers escapes the set: 3 minus 5 is negative 2, which is not a natural number. So subtraction is not even a valid operation there.
\[ 3 - 5 = -2 \notin \mathbb{N} \]
And subtraction on the integers is closed but not associative, so it fails a different axiom. Two operations, two different broken axioms.
Ranking
Put in order
Put the moves of Worked example: the integers under subtraction are not a group into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Pick 8, 5, 2 and compute both groupings; if they differ, associativity fails.
Worked example
Test whether the integers under subtraction form a group. Closure holds, so the danger is elsewhere.
\[ (\mathbb{Z}, -) \]
Test associativity on a concrete triple
Why: Pick 8, 5, 2 and compute both groupings; if they differ, associativity fails.
\[ (8 - 5) - 2 = 3 - 2 = 1 \]
Compute the other grouping
Why: Move the parentheses to the right and evaluate.
\[ 8 - (5 - 2) = 8 - 3 = 5 \]
Compare the two results
Why: One grouping gives 1 and the other gives 5, so the two are not equal.
\[ 1 \ne 5 \]
Verify the axiom fails
Why: A single counterexample defeats a universal statement, so subtraction on the integers is not associative and the integers under subtraction are not a group.
Picture it
Animation
Shows: Each line of the worked example "the integers under subtraction are not a group", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: A single counterexample defeats a universal statement, so subtraction on the integers is not associative and the integers under subtraction are not a group.
Concept
Beside the size of the whole group, each individual element carries a number of its own.
order of an element — The smallest positive number of times you must combine an element with itself to reach the identity. If no such number exists, the element has infinite order.
\[ \operatorname{ord}(a) = \min\{\, n > 0 : a^{n} = e \,\} \]
In additive notation the power of a means adding a to itself repeatedly. The order asks: how many steps until you first return home to the identity?
Intuition
Picture walking around a circular track by a fixed step. The order of that step is how many strides bring you exactly back to the start for the very first time.
In the integers modulo 6, stepping by 2 goes 2, 4, 0: three steps to return, so the element 2 has order 3. Stepping by 1 needs all six steps.
The identity itself has order 1, since it is already home. Every other element has order at least 2.
Estimation
Predict first
Compute the order of every element of the integers modulo 6 under addition.
Commit before you compute: what does Worked example: orders of the elements of the integers mod 6 come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify against the divisors of 6
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Every order listed divides the group order 6, matching Lagrange's theorem in advance.
Worked example
Compute the order of every element of the integers modulo 6 under addition.
\[ \mathbb{Z}/6\mathbb{Z} = \{0,1,2,3,4,5\} \]
Orders of 0 and 3
Why: 0 is already the identity, so its order is 1. Doubling 3 gives 6, which is 0, so 3 has order 2.
\[ 3 + 3 = 6 \equiv 0 \pmod 6 \]
Orders of 2 and 4
Why: Adding 2 three times gives 6 which is 0; adding 4 three times gives 12 which is 0. Both have order 3.
\[ 2+2+2 = 6 \equiv 0, \quad 4+4+4 = 12 \equiv 0 \pmod 6 \]
Orders of 1 and 5
Why: Adding 1 six times first reaches 0; adding 5 six times reaches 30, which is 0. Both have order 6.
\[ 6 \cdot 1 = 6 \equiv 0, \quad 6 \cdot 5 = 30 \equiv 0 \pmod 6 \]
Collect the orders
Why: Read off the pattern: the order of a residue equals 6 divided by its greatest common divisor with 6.
| element | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| order | 1 | 6 | 3 | 2 | 3 | 6 |
Verify against the divisors of 6
Why: Every order listed divides the group order 6, matching Lagrange's theorem in advance. The orders are 1, 6, 3, 2, 3, 6.
Picture it
Animation
Shows: Each line of the worked example "orders of the elements of the integers mod 6", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Every order listed divides the group order 6, matching Lagrange's theorem in advance. The orders are 1, 6, 3, 2, 3, 6.
Trap
Concluding that because the group has order 6, every element must also have order 6.
\[ |\mathbb{Z}/6\mathbb{Z}| = 6 \ \Rightarrow\ \operatorname{ord}(a) = 6\, ? \]
This confuses two different numbers. The group order counts elements; the element order counts steps back to the identity.
Keep the two notions separate. The group order is a single number for the whole group; each element has its own order, and they vary.
\[ \operatorname{ord}(2) = 3, \quad \operatorname{ord}(3) = 2, \quad \operatorname{ord}(1) = 6 \]
Element orders always divide the group order, but they need not equal it. Only the generators reach the full group order of 6.
Break the constraint
Discussion prompt
The rule this trap just fixed:
Keep the two notions separate. The group order is a single number for the whole group; each element has its own order, and they vary.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Concept
Sometimes a single element, combined with itself over and over, produces the entire group. Such an element is called a generator.
cyclic group — A group in which every element is a power of one fixed element g. That element generates the group, and we write the group as the set of powers of g.
\[ \langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\} = G \]
The integers modulo n are cyclic, generated by 1. A generator of a finite cyclic group is exactly an element whose order equals the group order.
Step zero
Discussion prompt
Worked example: the generators of the integers mod 6 — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: A generator must have order 6
Answer:
Worked example
Find every generator of the integers modulo 6 under addition.
\[ \mathbb{Z}/6\mathbb{Z} \]
A generator must have order 6
Why: To sweep out all six elements by repeated addition, the element must first return to 0 only after six steps.
Read orders from the earlier table
Why: Only the residues 1 and 5 have order 6; the others return home sooner and miss some elements.
\[ \operatorname{ord}(1) = 6, \quad \operatorname{ord}(5) = 6 \]
Confirm 1 generates the group
Why: Repeatedly adding 1 lists 1, 2, 3, 4, 5, 0, which is the whole group.
\[ \langle 1 \rangle = \{1,2,3,4,5,0\} \]
Verify 5 generates and count them
Why: Repeatedly adding 5 lists 5, 4, 3, 2, 1, 0, again the whole group. So there are exactly two generators, 1 and 5, matching the count of residues coprime to 6.
\[ \langle 5 \rangle = \{5,4,3,2,1,0\} \]
Picture it
Animation
Shows: Each line of the worked example "the generators of the integers mod 6", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Repeatedly adding 5 lists 5, 4, 3, 2, 1, 0, again the whole group. So there are exactly two generators, 1 and 5, matching the count of residues coprime to 6.
Concept
The integers modulo n do not form a group under multiplication, because some elements have no multiplicative inverse. The fix is to keep only the ones that do.
the units mod n — The residues that have a multiplicative inverse modulo n, which are exactly those coprime to n. They form a group under multiplication.
\[ (\mathbb{Z}/n\mathbb{Z})^{\ast} = \{\, a : \gcd(a,n) = 1 \,\} \]
A residue is invertible modulo n exactly when it shares no factor with n. This is why the non-units must be removed before multiplication becomes a group operation.
Intuition
Think of the units as the residues you are allowed to divide by. Multiplying by a unit can always be undone, because its inverse is waiting in the set.
Modulo 5, every nonzero residue is a unit, because 5 is prime and shares no factor with 1, 2, 3, or 4. Modulo 6, only 1 and 5 are units; 2, 3, and 4 all share a factor with 6.
So the size of the unit group depends delicately on the factorization of n, and it is counted by Euler's totient function.
Picture it
Figure (svg): The Cayley table of the units modulo 5 under multiplication, a five-by-five grid with the operation symbol in the corner, the elements 1 to 4 as row and column headers, and each cell holding the product reduced modulo 5.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Certify the units modulo 5 as a group and build their multiplication table.
Worked example
Certify the units modulo 5 as a group and build their multiplication table.
\[ (\mathbb{Z}/5\mathbb{Z})^{\ast} = \{1,2,3,4\} \]
Confirm every element is a unit
Why: 5 is prime, so each of 1, 2, 3, 4 is coprime to 5 and therefore invertible; multiplication mod 5 stays within the set.
Build the multiplication table mod 5
Why: Multiply each pair and reduce modulo 5; for instance 2 times 3 is 6 which is 1, and 3 times 4 is 12 which is 2.
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| 1 | 1 | 2 | 3 | 4 |
| 2 | 2 | 4 | 1 | 3 |
| 3 | 3 | 1 | 4 | 2 |
| 4 | 4 | 3 | 2 | 1 |
Figure (svg): The Cayley table of the units modulo 5 under multiplication, a five-by-five grid with the operation symbol in the corner, the elements 1 to 4 as row and column headers, and each cell holding the product reduced modulo 5.
Locate each inverse in the table
Why: Find the identity 1 in each row: 1 pairs with 1, 2 with 3, 3 with 2, and 4 with 4.
\[ 2 \cdot 3 \equiv 1, \quad 4 \cdot 4 \equiv 1 \pmod 5 \]
Verify the group axioms from the table
Why: Every row and column is a permutation of the four units, 1 is the identity, each row contains a 1 so inverses exist, and associativity comes from integer multiplication. So the units mod 5 form an abelian group of order 4.
Pattern
Step through it
Step through Worked example: the units mod 5 and their Cayley table one row at a time. What is driving the change, and what would the row after the last one be?
Trap
Claiming the integers modulo n form a group under multiplication, by analogy with the additive case.
\[ (\mathbb{Z}/n\mathbb{Z}, \times): \quad \{0, 1, \dots, n-1\} \]
But 0 can never have a multiplicative inverse: zero times anything is zero, never 1. And when n is composite, other non-units fail as well.
Remove every element without an inverse. At minimum 0 must go, and modulo a composite n the elements sharing a factor with n go too.
\[ 2 \cdot x \equiv 1 \pmod 6 \ \text{ has no solution} \]
What remains is the unit group, which is a group under multiplication. The lesson: closure and inverses must be checked, never imported from the additive case.
Concept
A rich family of finite groups comes from rearranging objects. A permutation of a set is a bijection from the set to itself, and permutations can be composed.
symmetric group — The set of all permutations of n objects, with composition as the operation. It has n factorial elements and is written S sub n.
\[ |S_n| = n! \]
The identity is the permutation that fixes everything; the inverse of a permutation is the rearrangement that undoes it. For n at least 3, this group is non-abelian.
Intuition
Write a permutation in cycle notation: the cycle sending 1 to 2, 2 to 3, and 3 back to 1 is written as those three numbers listed in order.
To compose two permutations we apply them one after another. Our convention is to apply the right-hand one first, exactly like composing functions.
Because doing rearrangement B then A can scramble things differently from A then B, composition of permutations generally does not commute. That is the source of non-abelian behavior.
Missing information
Discussion prompt
Show that the symmetric group on three symbols is non-abelian by finding two permutations that disagree when composed in the two orders.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Apply the swap of 2 and 3 first, then the swap of 1 and 2, tracking where each symbol lands.
Worked example
Show that the symmetric group on three symbols is non-abelian by finding two permutations that disagree when composed in the two orders.
\[ a = (1\ 2), \qquad b = (2\ 3) \]
Compose a after b, applying b first
Why: Apply the swap of 2 and 3 first, then the swap of 1 and 2, tracking where each symbol lands.
\[ a \circ b: \quad 1 \to 2,\ 2 \to 3,\ 3 \to 1 \]
Name that result
Why: The composite sends 1 to 2, 2 to 3, and 3 to 1, which is the three-cycle.
\[ a \circ b = (1\ 2\ 3) \]
Compose b after a, applying a first
Why: Now apply the swap of 1 and 2 first, then the swap of 2 and 3.
\[ b \circ a: \quad 1 \to 3,\ 3 \to 2,\ 2 \to 1 \]
Name that result
Why: This composite sends 1 to 3, 3 to 2, and 2 to 1, the opposite three-cycle.
\[ b \circ a = (1\ 3\ 2) \]
Verify the two composites disagree
Why: The two results are different three-cycles, so a and b do not commute. The symmetric group on three symbols is non-abelian, and with its six elements it is the smallest such group.
\[ (1\ 2\ 3) \ne (1\ 3\ 2) \]
Picture it
Animation
Shows: Each line of the worked example "the symmetric group on three symbols is non-abelian", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The two results are different three-cycles, so a and b do not commute. The symmetric group on three symbols is non-abelian, and with its six elements it is the smallest such group.
Trap
Assuming that a combined with b always equals b combined with a, and using that freely inside a proof.
\[ a \ast b = b \ast a \quad \text{(assumed for every group)} \]
This silently imports commutativity. It is true in the integers and in every modular group, so it feels universal, but it is an extra hypothesis.
Only assume commutativity when the group is known to be abelian. In general, keep the order of factors exactly as written.
\[ (1\ 2)(2\ 3) = (1\ 2\ 3) \ne (1\ 3\ 2) = (2\ 3)(1\ 2) \]
The symmetries of a triangle already break commutativity. A correct general proof never reorders a product unless an axiom or a stated hypothesis allows it.
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Concept
Another concrete family: the symmetries of a regular polygon. A symmetry is a rigid motion that maps the polygon onto itself.
dihedral group — The group of symmetries of a regular n-gon: n rotations and n reflections, with composition as the operation. It has 2n elements and is written D sub n.
\[ |D_n| = 2n \]
These groups are non-abelian for n at least 3. They are the standard first example of symmetry turned into algebra.
Picture it
Figure (svg): A square with dashed lines marking its four reflection axes, one vertical, one horizontal, and the two diagonals, together with a curved red arrow indicating a quarter-turn rotation.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Take the square. Its symmetry group is the dihedral group on four vertices, with eight elements: four rotations and four reflections.
Concept
Take the square. Its symmetry group is the dihedral group on four vertices, with eight elements: four rotations and four reflections.
Figure (svg): A square with dashed lines marking its four reflection axes, one vertical, one horizontal, and the two diagonals, together with a curved red arrow indicating a quarter-turn rotation.
The four rotations turn the square by nothing, a quarter turn, a half turn, and three quarters of a turn. The four reflections flip it across the vertical axis, the horizontal axis, and the two diagonals.
\[ D_4 = \{\, e,\ r,\ r^2,\ r^3,\ s,\ rs,\ r^2 s,\ r^3 s \,\} \]
Here r is the quarter-turn rotation and s is a single reflection. Every symmetry is either a rotation, or a reflection followed by a rotation.
Estimation
Predict first
Compute the order of the quarter-turn rotation in the symmetries of the square.
Commit before you compute: what does Worked example: the order of the quarter-turn rotation come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify no smaller power works
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. We checked that r, its square, and its cube are each different from the identity, and the fourth power is the identity, so the order is exactly 4, matching a quarter of a full turn.
Worked example
Compute the order of the quarter-turn rotation in the symmetries of the square.
\[ r = \text{rotation by a quarter turn in } D_4 \]
Apply r once and twice
Why: One quarter turn is not the identity; two quarter turns give a half turn, still not the identity.
\[ r^2 = \text{half turn} \ne e \]
Apply r three and four times
Why: Three quarter turns give a three-quarter turn; the fourth quarter turn completes a full revolution, returning every vertex home.
\[ r^4 = \text{full turn} = e \]
Read off the order
Why: The smallest positive power of r equal to the identity is the fourth.
\[ \operatorname{ord}(r) = 4 \]
Verify no smaller power works
Why: We checked that r, its square, and its cube are each different from the identity, and the fourth power is the identity, so the order is exactly 4, matching a quarter of a full turn.
Picture it
Animation
Shows: Each line of the worked example "the order of the quarter-turn rotation", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: We checked that r, its square, and its cube are each different from the identity, and the fourth power is the identity, so the order is exactly 4, matching a quarter of a full turn.
Step zero
Discussion prompt
Worked example: a rotation and a reflection do not commute — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: State the defining relation
Answer:
Worked example
Confirm the symmetries of the square are non-abelian using the rotation r and a reflection s.
\[ r = \text{quarter turn}, \quad s = \text{a reflection}, \quad s^2 = e \]
State the defining relation
Why: In every dihedral group, sliding a reflection past a rotation inverts the rotation.
\[ s r = r^{-1} s = r^{3} s \]
Compare the two orders
Why: Rotation then reflection gives r followed by s; the relation shows reflection then rotation gives r cubed followed by s instead.
\[ r s \quad \text{versus} \quad s r = r^{3} s \]
Check that they differ
Why: Since r is not equal to r cubed in the group, the products r s and r cubed s are different elements.
\[ r \ne r^{3} \implies r s \ne s r \]
Verify non-commutativity
Why: The two products disagree, so r and s do not commute and the dihedral group on the square is non-abelian.
Picture it
Animation
Shows: Each line of the worked example "a rotation and a reflection do not commute", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Since r is not equal to r cubed in the group, the products r s and r cubed s are different elements.
Concept
One more landmark, this time infinite and non-abelian: the invertible matrices.
general linear group — The set of invertible n by n matrices with real entries, under matrix multiplication. It is written GL sub n of the reals.
\[ GL_n(\mathbb{R}) = \{\, A : \det A \ne 0 \,\} \]
Matrix multiplication is associative, the identity matrix is the identity element, and an invertible matrix has an inverse matrix. Matrix multiplication is generally not commutative, so this group is non-abelian for n at least 2.
Ranking
Put in order
Put the moves of Worked example: the Klein four-group into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The product of any two distinct non-identity elements is the third one; this keeps the operation closed and commutative.
Worked example
Not every group of order 4 is cyclic. Consider the four-element group in which every non-identity element is its own inverse.
\[ V = \{\, e,\ a,\ b,\ c \,\}, \quad a^2 = b^2 = c^2 = e \]
State the products of distinct elements
Why: The product of any two distinct non-identity elements is the third one; this keeps the operation closed and commutative.
\[ ab = c, \quad bc = a, \quad ca = b \]
Build the Cayley table
Why: Fill each cell from the rules above; the diagonal is all identities because every element squares to e.
| * | e | a | b | c |
|---|---|---|---|---|
| e | e | a | b | c |
| a | a | e | c | b |
| b | b | c | e | a |
| c | c | b | a | e |
Read off the structure
Why: Every row and column is a permutation of the four elements, so closure and inverses hold; the identity is e and every element is its own inverse.
Verify it is an abelian group of order 4
Why: The table is symmetric, associativity can be checked case by case, and no element has order 4, so this Klein four-group is abelian, of order 4, and not cyclic.
Picture it
Animation
Shows: Each line of the worked example "the Klein four-group", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The table is symmetric, associativity can be checked case by case, and no element has order 4, so this Klein four-group is abelian, of order 4, and not cyclic.
Concept
The Klein four-group and the integers modulo 4 both have four elements, yet they are genuinely different groups.
In the integers modulo 4 the element 1 has order 4: it takes four steps to return to 0, so that group is cyclic. In the Klein four-group every non-identity element has order 2, so no single element generates it.
\[ \operatorname{ord}(1) = 4 \ \text{in}\ \mathbb{Z}/4\mathbb{Z}; \quad \operatorname{ord}(a) = 2 \ \text{in}\ V \]
So the order of a group does not determine the group. The pattern of element orders is a finer fingerprint, and it already separates these two.
Concept
One pattern from the modular example is no accident and deserves naming, even before its proof.
In every finite group, the order of each element divides the order of the group. This is a corollary of Lagrange's theorem, the subject of the next deck.
\[ \operatorname{ord}(a) \ \text{divides}\ |G| \]
It is a powerful sanity check: an element of order 5 cannot live in a group of order 6, and every element order in the integers modulo 6 was among 1, 2, 3, and 6.
Concept
The tables we have been building all share a striking feature, and it is forced by the axioms.
Latin square — A square grid in which every symbol appears exactly once in each row and exactly once in each column.
In a finite group, every row and every column of the Cayley table is a permutation of the elements. This follows directly from the cancellation law: a repeat in a row would mean two equal products sharing a left factor.
\[ a x_1 = a x_2 \implies x_1 = x_2 \]
Explain it
Discussion prompt
Explain A Cayley table is a Latin square to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
The tables we have been building all share a striking feature, and it is forced by the axioms.
Intuition
Step back and the examples rhyme. The additive integers, modular residues, permutations, polygon symmetries, and invertible matrices are all collections of reversible transformations closed under composition.
That is the deep content of a group: it is the algebra of symmetry, the structure-preserving bijections of some object gathered into one set with composition.
Whenever a system has transformations you can compose and undo, a group is hiding inside it. Finding that group is often how a hard problem becomes tractable.
Analogy
Discussion prompt
Explain A group is the algebra of symmetry by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Whenever a system has transformations you can compose and undo, a group is hiding inside it. Finding that group is often how a hard problem becomes tractable.
Concept
The connection to computer science is direct. A permutation is exactly a reordering of data, and the symmetric group is the space of all such reorderings.
Shuffling a deck, permuting array indices, and the legal moves of a Rubik's cube are all group elements: each is reversible and they compose. The cube's roughly 43 quintillion states are the elements of one finite group.
Symmetry groups also drive fast algorithms and error-correcting codes, where knowing the group of a problem prunes the search or repairs corrupted data.
Counterexample
Discussion prompt
The connection to computer science is direct. A permutation is exactly a reordering of data, and the symmetric group is the space of all such reorderings.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Symmetry groups also drive fast algorithms and error-correcting codes, where knowing the group of a problem prunes the search or repairs corrupted data.
Pattern
To decide whether a set with an operation is a group, run the same four checks every time, in this order.
1. Closure
Why: Confirm the operation applied to any two elements lands back in the set. This is the check most often skipped.
2. Associativity
Why: Confirm that regrouping does not change the result, or cite that it is inherited from a known associative operation.
3. Identity
Why: Exhibit a specific two-sided identity element that lives in the set.
4. Inverses
Why: Show every element has a two-sided inverse that is itself in the set. Fail any one check and it is not a group.
Pattern
For a finite group given by a table, four features can be read off directly.
Find the identity
Why: The identity is the element whose row and column copy the header exactly, unchanged.
Check the Latin-square property
Why: Every element must appear once per row and once per column; a repeat means it is not a group.
Read inverses from the identity
Why: For each element, the position of the identity within its row points to its inverse.
Test commutativity by symmetry
Why: The group is abelian exactly when the table is symmetric across its main diagonal.
Edge cases
Discussion prompt
Reading a Cayley table works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
For a finite group given by a table, four features can be read off directly.
Elimination
Eliminate the wrong options
Which group axiom fails for the natural numbers under addition?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: D
Why: Addition of naturals is closed and associative, and 0 is an identity, but a positive number like 3 has no natural number that adds to it to give 0. The inverse axiom fails, so this is a monoid, not a group.
Check
Think through the four axioms for the natural numbers under addition before choosing.
Check your understanding
Which group axiom fails for the natural numbers under addition?
Answer: D
Why: Addition of naturals is closed and associative, and 0 is an identity, but a positive number like 3 has no natural number that adds to it to give 0. The inverse axiom fails, so this is a monoid, not a group.
Check
Add the element to itself repeatedly and count the steps to reach 0.
\[ \text{In } \mathbb{Z}/12\mathbb{Z}, \ \text{find the order of } 8. \]
Check your understanding
In the integers modulo 12 under addition, what is the order of the element 8?
Answer: A
Why: The order is the smallest positive k with k copies of 8 summing to 0 modulo 12. Since 8, then 16 which is 4, then 24 which is 0 appear at k equal to 3, the order is 3, equal to 12 divided by the gcd of 8 and 12.
Check
Recall which operations are commutative and which are not.
Check your understanding
Which of these groups is non-abelian?
Answer: B
Why: Every modular additive group and the additive integers are commutative, and the units modulo 5 are commutative too. The symmetric group on three symbols contains two transpositions whose composites differ by order, so it is the non-abelian one.
Prediction
Predict first
In a general group, the inverse of the product a times b equals which expression?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: The inverse of b, times the inverse of a
Why: Undoing a and then b requires reversing the order: apply the inverse of b first, then the inverse of a. Checking that product against a times b returns the identity from both sides, so the reversed order is correct.
Check
Remember the socks-and-shoes rule and whether the group is assumed abelian.
Check your understanding
In a general group, the inverse of the product a times b equals which expression?
Answer: A
Why: Undoing a and then b requires reversing the order: apply the inverse of b first, then the inverse of a. Checking that product against a times b returns the identity from both sides, so the reversed order is correct.
Commit first
Predict first
Which residues make up the group of units modulo 8 under multiplication?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: 1, 3, 5, 7
Why: The units modulo 8 are the residues coprime to 8, namely the odd numbers 1, 3, 5, and 7. Each has a multiplicative inverse modulo 8, and together they form a group of order 4.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
The units are the residues coprime to the modulus.
Check your understanding
Which residues make up the group of units modulo 8 under multiplication?
Answer: A
Why: The units modulo 8 are the residues coprime to 8, namely the odd numbers 1, 3, 5, and 7. Each has a multiplicative inverse modulo 8, and together they form a group of order 4.
Prediction
Predict first
Under the stated operation, which of these is a group?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: The units modulo 5 under multiplication
Why: The units modulo 5 are 1, 2, 3, 4 with every element invertible, so they form a group. Each of the other three fails an axiom: either missing inverses or missing closure.
Check
Run the four-step test on each candidate.
Check your understanding
Under the stated operation, which of these is a group?
Answer: A
Why: The units modulo 5 are 1, 2, 3, 4 with every element invertible, so they form a group. Each of the other three fails an axiom: either missing inverses or missing closure.
Elimination
Eliminate the wrong options
In the Cayley table of a finite group, why does every element appear exactly once in each row?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: If an element repeated in a row, two products with the same left factor would be equal, and cancellation would force the two other factors to be equal, a contradiction. So cancellation makes each row a permutation of the elements.
Check
Think about which axiom forces the row pattern.
Check your understanding
In the Cayley table of a finite group, why does every element appear exactly once in each row?
Answer: A
Why: If an element repeated in a row, two products with the same left factor would be equal, and cancellation would force the two other factors to be equal, a contradiction. So cancellation makes each row a permutation of the elements.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — The four-step test: is this a group? · Reading a Cayley table · From monoid to group: the one missing power · A group is a set of reversible moves · Axiom G1: closure. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You can state the four group axioms, closure, associativity, identity, and inverses, and check them for any candidate.
You proved that the identity and inverses are unique, and derived the cancellation law and the reversed-order rule for the inverse of a product.
You can compute the order of a group and the order of an element, keep the two apart, and recognize cyclic groups and their generators.
And you met the landmark examples: the additive integers, the integers modulo n, the units modulo n, the symmetric groups, the dihedral symmetries of a polygon, and the invertible matrices, along with abelian versus non-abelian and the Cayley table as a Latin square.
| group | order | abelian? |
|---|---|---|
| integers under addition | infinite | yes |
| integers mod n, addition | n | yes |
| units mod 5 | 4 | yes |
| symmetric group on 3 | 6 | no |
| dihedral group of the square | 8 | no |
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