The algebra in part 1 manipulates numbers using rules like a + b = b + a. Abstract algebra asks which rules matter. Take a few of them as the starting point, and everything that obeys them behaves the same way, whether it is numbers, rotations of a triangle, or remainders on a clock. Prove something once and it holds for all of them.
This post builds up to three structures: groups, rings and fields. It uses the proof methods from part 2. The last section is a map of the advanced subjects that come after.
How to use this post
The layout is the same: the idea, how to do it, worked examples, the usual mistake, practice. The structures here are new, so write the definitions out by hand once and then test every example against them.
1. Clock arithmetic
Modular arithmetic works with remainders. We write a ≡ b (mod n), read "a is congruent to b modulo n", when n divides a − b. On a 12-hour clock, 9 + 5 = 14, which shows as 2, so 14 ≡ 2 (mod 12).
How to do it
- To reduce a number, divide by n and keep the remainder: 17 mod 12 = 5.
- Addition and multiplication can be done before or after reducing. You may reduce at any point.
- To solve an equation like 3x ≡ 4 (mod 7), try the values 0 to 6, since nothing else is possible.
- A number a has a multiplicative inverse mod n (a number b with a × b ≡ 1) exactly when a and n share no common factor above 1.
Example 1: 7 × 8 mod 10. 7 × 8 = 56, and 56 mod 10 = 6.
Example 2: solve 3x ≡ 4 (mod 7). Try x = 0, 1, ..., 6. Only x = 6 works: 3 × 6 = 18 = 14 + 4.
Example 3: inverses mod 12.
- 5 × 5 = 25 = 24 + 1, so 5 is its own inverse mod 12. (5 and 12 share no factor.)
- 4 has no inverse mod 12: 4x ≡ 1 has no solution, since 4 and 12 share the factor 4.
Example 4: big powers. Find 2¹⁰⁰ mod 7.
- 2³ = 8, which is 1 mod 7.
- 100 = 3 × 33 + 1, so 2¹⁰⁰ = (2³)³³ × 2.
- That is 1³³ × 2 = 2 mod 7.
The usual mistake: cancelling a factor that is not invertible. Mod 6, 2 × 3 ≡ 2 × 0, since both sides are 0. But you cannot cancel the 2 to get 3 ≡ 0. Cancellation is only safe when the factor has an inverse.
Check yourself
What is 7 × 8 mod 10?
Practice
Find 100 mod 7. Then find 3 to the power 100 mod 5. Then solve 5x ≡ 3 (mod 8). Then decide whether 6x ≡ 1 (mod 9) has a solution.
Show the steps and the answer
- 98 = 14 × 7, so 100 mod 7 = 2.
- 3⁴ = 81 = 80 + 1, which is 1 mod 5. Then 3¹⁰⁰ = (3⁴)²⁵, which is 1²⁵ = 1 mod 5.
- Try x = 0 to 7. The inverse of 5 mod 8 is 5 (25 = 24 + 1), so multiply both sides by 5: x ≡ 15 ≡ 7 (mod 8). Check: 5 × 7 = 35 = 32 + 3.
- No. 6 and 9 share the factor 3, so 6 has no inverse mod 9. Every multiple of 6, mod 9, is 0, 3 or 6, never 1.
Answers: 2; 1; x = 7; no solution.
2. Groups
A group is a set G with an operation (written here as ∘) that obeys four rules:
- Closure: combining two elements gives an element of G.
- Associativity: (a ∘ b) ∘ c = a ∘ (b ∘ c).
- Identity: some element e has e ∘ a = a ∘ e = a for every a.
- Inverses: every a has an element a⁻¹ with a ∘ a⁻¹ = e.
Nothing says a ∘ b equals b ∘ a. A group where it always does is called abelian.
How to do it: to test a candidate, check the four rules in order. One failure is enough to say it is not a group.
Example 1: the integers under addition. Closure yes, identity 0, the inverse of a is −a. A group.
Example 2: the clock {0, 1, 2, 3} under addition mod 4. The table shows every sum. Each row and each column contains every element exactly once, a feature of every group table.
| + | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| 0 | 0 | 1 | 2 | 3 |
| 1 | 1 | 2 | 3 | 0 |
| 2 | 2 | 3 | 0 | 1 |
| 3 | 3 | 0 | 1 | 2 |
Example 3: the non-zero residues mod 5 under multiplication.
- Closure: products of 1, 2, 3, 4 mod 5 never give 0.
- Identity: 1.
- Inverses: 1 and 4 are their own inverses (4 × 4 = 16 ≡ 1), and 2 and 3 are inverses of each other (2 × 3 = 6 ≡ 1).
- It is a group.
Example 4: what fails.
- The integers under multiplication: 2 has no inverse among the integers.
- The positive integers under addition: there is no identity 0, so no inverses either.
- The non-zero residues mod 4 under multiplication: 2 × 2 = 4 ≡ 0, which is not in the set. Closure fails.
Example 5: swapping things, where order matters. Label the corners of a triangle 1, 2, 3. A symmetry moves the corners around, and combining two symmetries gives another one. There are 6 symmetries in total (3 rotations and 3 reflections), and together they form a group.
- Let a swap corners 1 and 2, and let b swap corners 2 and 3.
- Do b first, then a: corner 1 goes to 1 then 2. Corner 2 goes to 3 then 3. Corner 3 goes to 2 then 1. So 1 → 2, 2 → 3, 3 → 1, a rotation.
- Do a first, then b: corner 1 goes to 2 then 3. Corner 2 goes to 1 then 1. Corner 3 goes to 3 then 2. So 1 → 3, 2 → 1, 3 → 2, the opposite rotation.
- The two answers differ, so this group is not abelian.
A first proof: the identity is unique. Suppose e and f are both identities. Then e ∘ f = f, because e is an identity, and e ∘ f = e, because f is an identity. So e = f. This is the direct proof from part 2 applied to a new object.
The usual mistake: assuming a ∘ b = b ∘ a. It holds for addition of numbers but not for the triangle's symmetries or for matrices (part 3).
Check yourself
Which of these is a group?
Practice
Show that {0, 1, 2} under addition mod 3 is a group, and give the inverse of each element. Then decide whether the even integers under addition form a group, and whether the odd integers do.
Show the steps and the answer
- Closure: the sum mod 3 is always 0, 1 or 2. Identity: 0. Inverses: 0 is its own inverse, 1 and 2 are inverses of each other (1 + 2 = 3 ≡ 0). Associativity is inherited from ordinary addition.
- The even integers: a sum of evens is even (part 2), the identity 0 is even, and the inverse of 2k is −2k, which is even. A group.
- The odd integers: not closed, since 1 + 1 = 2 is even. There is also no identity, since 0 is not odd. Not a group.
3. Subgroups, orders and Lagrange's theorem
A subgroup is a subset of a group that is itself a group under the same operation. The order of a group is its number of elements. The order of an element g is the smallest positive k for which applying g to itself k times gives the identity.
Lagrange's theorem. In a finite group, the size of every subgroup divides the size of the whole group. The proof splits the group into equal-sized pieces, so the sizes divide evenly. Two quick uses:
- Ruling out subgroups. A group of order 15 cannot have a subgroup of order 6, because 6 does not divide 15.
- Orders of elements. The order of any element divides the order of the group, because an element generates a subgroup.
Example 1: the clock mod 12 under addition.
- The element 8 gives 8, 16 ≡ 4, 24 ≡ 0, so 8 has order 3 and generates the subgroup
{0, 4, 8}. Three divides 12. - The element 5 has order 12: multiples of 5 mod 12 hit every residue before returning to 0. So 5 generates the whole group.
- In general the order of a in the clock mod n is n ÷ gcd(a, n). The subgroup sizes of the clock mod 12 are exactly the divisors of 12: 1, 2, 3, 4, 6 and 12.
Example 2: the triangle's 6 symmetries. Lagrange says a subgroup can only have 1, 2, 3 or 6 elements. A subgroup of order 4 or 5 is impossible. A check of every subset confirms that the sizes that actually occur are 1, 2, 3 and 6.
Example 3: prime order. If a group has p elements and p is prime, every non-identity element has order p, so it generates the whole group. Such a group has no subgroups other than the trivial ones.
The usual mistake: reading Lagrange's theorem backwards. It does not say that a group of order n has a subgroup of every size dividing n. The group of even permutations of four objects has 12 elements but no subgroup of order 6, as a check of every candidate subset confirms.
Check yourself
A finite group has 15 elements. Which subgroup size is impossible?
Practice
Find the order of 3 in the clock mod 12. Then count how many elements generate the whole clock mod 10. Then decide whether a group of order 7 can have a subgroup of order 3.
Show the steps and the answer
- Multiples of 3: 3, 6, 9, 12 ≡ 0. The order is 4. (Formula: 12 ÷ gcd(3, 12) = 12 ÷ 3 = 4.)
- An element a generates the clock mod 10 when gcd(a, 10) = 1. Those are 1, 3, 7 and 9, so 4 elements.
- No. 3 does not divide 7.
Answers: 4; 4; no.
4. Rings and fields
A ring has two operations, addition and multiplication. Under addition it is an abelian group. Multiplication is associative, and the distributive law a × (b + c) = a × b + a × c holds. The integers, the clock mod n and the 2 × 2 matrices from part 3 are rings. In matrices, multiplication is not commutative, so a ring does not have to be.
A field is a ring where multiplication is commutative, 1 is not 0, and every non-zero element has a multiplicative inverse. In a field you can add, subtract, multiply and divide by anything except 0. The rationals, the reals and the complex numbers are fields. The integers are not, since 2 has no inverse.
The clock mod n as a ring. It is a field exactly when n is prime.
Example 1: mod 7 is a field. Every non-zero residue has an inverse: 1 × 1, 2 × 4 = 8, 3 × 5 = 15, 6 × 6 = 36, and each product is 1 mod 7. In general, if p is prime then every number from 1 to p − 1 shares no factor with p, so it has an inverse.
Example 2: mod 6 is not a field. 2 × 3 = 6 ≡ 0, though neither factor is 0. These are called zero divisors, and they can only appear when n is not prime. Mod 8 has them too: 2 × 4 ≡ 0.
Fermat's little theorem. If p is prime and a is not a multiple of p, then a to the power p − 1 is congruent to 1 mod p. Examples: 2⁶ = 64 ≡ 1 (mod 7), 3⁴ = 81 ≡ 1 (mod 5), and 5⁶ ≡ 1 (mod 7).
The proof follows from section 3. The non-zero residues mod p form a group under multiplication of order p − 1. By Lagrange, the order of a divides p − 1, so raising a to the power p − 1 gives the identity, which is 1.
Where this is used. Modular arithmetic and Fermat's little theorem are the starting point of public-key cryptography such as RSA, and finite fields appear in error-correcting codes.
The usual mistake: assuming a × b = 0 forces a = 0 or b = 0. That holds in a field, but fails mod 6 and fails for matrices. It is the reason cancelling needs an inverse.
Check yourself
Which of these is a field?
Practice
Find the inverse of 4 mod 7. Then show that the integers mod 8 are not a field. Then use Fermat's little theorem to find 3 to the power 100 mod 7.
Show the steps and the answer
- 4 × 2 = 8 ≡ 1 (mod 7), so the inverse of 4 is 2.
- 2 × 4 = 8 ≡ 0 (mod 8) with neither factor 0. In a field that cannot happen, so mod 8 is not a field.
- By Fermat, 3⁶ ≡ 1 (mod 7). Since 100 = 6 × 16 + 4, 3¹⁰⁰ = (3⁶)¹⁶ × 3⁴ ≡ 3⁴ = 81. And 81 − 77 = 4, so the answer is 4.
Answers: 2; 2 × 4 ≡ 0; 4.
5. A map of what comes after
Most advanced subjects take a standard route. This table gives a plain description of each and the usual advice on what to know first. The advice to take real analysis before topology, and to learn abstract algebra before full number theory, comes from a forum thread cited in the sources. The other rows, including differential geometry, follow the usual course order. None of it is a formal rule, and people disagree on details.
| Subject | What it studies | What to know first |
|---|---|---|
| Topology | Shapes and spaces where only nearness matters, not distance. A coffee mug and a doughnut count as the same shape, since each has one hole | Real analysis |
| Complex analysis | Calculus with complex numbers, which has unexpectedly strong results | Calculus, some analysis |
| Number theory | Primes, congruences, and equations in whole numbers | Abstract algebra, and analysis for the deeper parts |
| Measure theory and probability | A rigorous base for integration and probability | Real analysis |
| Differential geometry | Curves and surfaces studied with calculus | Multivariable calculus, linear algebra |
| Galois theory | Uses groups and fields to explain why general fifth-degree equations have no solution formula | This post and linear algebra |
Some pairs of subjects have no fixed order. People disagree on whether linear algebra or abstract algebra comes first. One experienced poster finished Gallian's abstract algebra before taking linear algebra and found it enriched vector spaces. Another forum answer, in the prerequisites thread, advises doing linear algebra before abstract algebra and calls it as foundational as calculus. Either path works.
The mistakes in one table
| Mistake | Why it fails | Fix |
|---|---|---|
| Cancelling a factor mod n | The factor may have no inverse | Check that it shares no factor with n |
| Assuming a ∘ b equals b ∘ a | Only abelian groups work that way | Test with a small example |
| Assuming a product of non-zero things is non-zero | Mod 6, 2 × 3 is 0 | That is only guaranteed in a field |
| Reading Lagrange backwards | It rules sizes out, it does not promise them | Use it to exclude, not to construct |
| Forgetting the identity check | Positive integers under addition fail it | Check all four group rules |
| Assuming ab = 0 means a = 0 or b = 0 | True in fields, false elsewhere | Check whether the ring is a field |
Where to go next
Joseph Gallian's Contemporary Abstract Algebra is described as very readable with many exercises, and Charles Pinter's A Book of Abstract Algebra is another common first book. Dummit and Foote's Abstract Algebra goes deeper. For topology, James Munkres' text is the standard. Tristan Needham's Visual Complex Analysis builds geometric intuition for complex analysis, and Gareth and Mary Jones' Elementary Number Theory is a gentle start in number theory.
Sources and further reading
- Joseph A. Gallian, Contemporary Abstract Algebra.
- Charles C. Pinter, A Book of Abstract Algebra.
- David S. Dummit and Richard M. Foote, Abstract Algebra.
- James R. Munkres, Topology.
- Tristan Needham, Visual Complex Analysis.
- Gareth A. Jones and J. Mary Jones, Elementary Number Theory.
- A self-learner's reading map of math books and a forum thread on prerequisites for advanced mathematics, which are where the suggested orderings and book picks above come from.
- A forum thread on recommended texts for self-study, for the comments on Gallian and Dummit and Foote.