Target Audience: Anyone who has worked through the number theory and linear algebra material and wants to see why so much of it keeps repeating itself.
Prerequisites: Number Theory for congruences, orders and primitive roots, and Determinants for permutations and their sign — because this page names the structure both of those were already using.
Most of this page is not new. The number theory material already established that the order of a unit divides \(\varphi(n)\), that primitive roots exist for some moduli and not others, and Fermat’s, Euler’s and Wilson’s theorems, each with its own argument. The determinants material already showed that \(\det(AB) = \det A \det B\) and that permutations have a sign that multiplies.
What this page adds is the observation that all of those are instances of a handful of facts about one kind of object. Sections 1 to 4 define it. Sections 5 and 6 prove the two results that do most of the work. Sections 7 to 9 relate groups to each other, and Section 10 goes back over the number theory and recovers it.
The explorer in Section 5 does not take the group axioms on trust: it checks them by exhaustion, then proves Lagrange’s theorem for each group by building the coset partition and confirming the pieces are disjoint, equal and covering — rather than by checking that one number divides another.
A group is a set \(G\) with an operation such that:
They are precisely what it takes to solve \(a \cdot x = b\) and get one answer. Multiply on the left by \(a^{-1}\), regroup using associativity, cancel to \(e\), and the identity leaves \(x = a^{-1} \cdot b\). Each axiom is used once, at a specific step. The definition is not a list of pleasant properties; it is the minimum that makes equations solvable.
Nothing demands \(a \cdot b = b \cdot a\). A group where that holds is called abelian, and many important groups are not — Section 9 has the smallest one. Assuming commutativity by habit is the most common error at this stage.
None of these was introduced as a group, and every one of them is one:
The number of elements of \((\mathbb{Z}/n\mathbb{Z})^{\times}\) is exactly \(\varphi(n)\) — checked for every \(n\) from \(2\) to \(60\). So Euler’s totient, introduced as a counting function, was measuring the size of a group all along.
All residues mod \(6\) under multiplication fail, since \(2 \cdot 3 = 0\) and \(0\) has no inverse. Even \(\{1, 2, 3, 4\}\) mod \(6\) fails closure.
Subtraction mod \(5\) is closed but not associative — \((0 - 0) - 1 = 4\) while \(0 - (0 - 1) = 1\) — and has no two-sided identity. Closure alone is nowhere near enough, and the explorer rejects all three of these rather than processing them.
The order of \(a\) is the smallest \(k \geq 1\) with \(a^k = e\). The order of the group \(|G|\) is its number of elements.
For the units mod \(n\), the order of an element is exactly the multiplicative order from the number theory material — the same definition, now with a name that makes sense in any group.
\(\langle a \rangle = \{e, a, a^2, \ldots\}\) has exactly as many elements as the order of \(a\), because the powers run through distinct values until they return to \(e\). That identification — order of an element equals size of a subgroup — is what lets the next two sections say something about elements.
A subset \(H \subseteq G\) that is a group under the same operation. For a finite group, being non-empty and closed is enough — inverses and the identity come for free, since the powers of any element eventually repeat.
For a subgroup \(H\) and an element \(g\), the left coset is \(gH = \{gh : h \in H\}\).
Any two cosets are either identical or disjoint. If \(gH\) and \(g_2H\) share an element, they are the same set.
Every coset has exactly \(|H|\) elements, because \(h \mapsto gh\) is a bijection — left multiplication by \(g\) can be undone by \(g^{-1}\).
If \(H\) is a subgroup of a finite group \(G\), then \(|H|\) divides \(|G|\).
By Section 4, the cosets of \(H\) cut \(G\) into disjoint pieces, each of size \(|H|\). So \(|G| = (\text{number of cosets}) \times |H|\), and \(|H|\) divides \(|G|\). The division is a consequence of the partition — which is why the explorer below checks the partition rather than the arithmetic.
The order of an element divides \(|G|\). Apply Lagrange to \(\langle a \rangle\). In \((\mathbb{Z}/n\mathbb{Z})^{\times}\), that is exactly “orders divide the totient”.
\(a^{|G|} = e\) for every element. If \(a\) has order \(k\), then \(k\) divides \(|G|\), so \(a^{|G|} = (a^k)^{|G|/k} = e\). In \((\mathbb{Z}/n\mathbb{Z})^{\times}\) this is Euler’s theorem \(a^{\varphi(n)} \equiv 1\), and when \(n = p\) is prime it is Fermat’s little theorem.
A divisor of \(|G|\) need not be the order of any subgroup. The alternating group \(A_4\) has order \(12\) and no subgroup of order \(6\) — checked by testing all \(462\) six-element candidate subsets containing the identity, none of which is closed. It does have subgroups of orders \(2\), \(3\) and \(4\), so the gap is specific.
Choose a group. The explorer first checks the four axioms by exhaustion — associativity on every triple — then computes the order of every element and the cyclic subgroup it generates. For each of those subgroups it builds the coset partition and confirms the pieces are disjoint, all the same size, and cover the group. Lagrange’s theorem is then read off the partition, not assumed.
Compare \((\mathbb{Z}/5)^{\times}\) with \((\mathbb{Z}/8)^{\times}\): both have order \(4\) and both satisfy Lagrange, but only one of them has an element of order \(4\).
A group is cyclic if some element \(g\) generates all of it: \(G = \langle g \rangle\). Equivalently, some element has order \(|G|\). Such a \(g\) is a generator.
A primitive root mod \(n\) is exactly a generator of \((\mathbb{Z}/n\mathbb{Z})^{\times}\). So “a primitive root exists mod \(n\)” says precisely “this group is cyclic”, and Gauss’s classification — \(n = 1, 2, 4, p^k, 2p^k\) with \(p\) an odd prime — is a theorem about which of these groups are cyclic. The explorer’s verdict agrees with it for every \(n\) up to \(60\).
\((\mathbb{Z}/5\mathbb{Z})^{\times}\) and \((\mathbb{Z}/8\mathbb{Z})^{\times}\) both have order \(4\). The first is cyclic, generated by \(2\), with element orders \(1, 2, 4, 4\). The second has element orders \(1, 2, 2, 2\): every non-identity element squares to \(1\), so nothing reaches order \(4\).
That is the entire reason a primitive root exists mod \(5\) and not mod \(8\). The order of a group does not determine the group.
A cyclic group of order \(n\) has exactly \(\varphi(n)\) generators — checked for every \(n\) from \(2\) to \(30\). So when primitive roots exist mod a prime \(p\), there are exactly \(\varphi(p-1)\) of them. Mod \(7\) that is \(\varphi(6) = 2\): the primitive roots \(3\) and \(5\).
A map \(\varphi : G \to H\) with \(\varphi(ab) = \varphi(a)\varphi(b)\). The identity and inverses then come for free: \(\varphi(e) = e\) and \(\varphi(a^{-1}) = \varphi(a)^{-1}\).
\(\det(AB) = \det(A)\det(B)\) says the determinant is a homomorphism from invertible matrices to the non-zero reals.
\(\operatorname{sgn}(\sigma\tau) = \operatorname{sgn}(\sigma)\operatorname{sgn}(\tau)\) says the sign of a permutation is a homomorphism from \(S_n\) to \(\{+1, -1\}\). Both were taught as identities; both are this definition.
A bijective homomorphism. Isomorphic groups are the same group with the elements renamed — \(\mathbb{Z}/4\mathbb{Z}\) under addition and \((\mathbb{Z}/5\mathbb{Z})^{\times}\) under multiplication are isomorphic, since both are cyclic of order \(4\). Whether the operation was called addition or multiplication turns out not to matter.
The kernel \(\ker\varphi = \{g : \varphi(g) = e\}\) is a subgroup, and \(\varphi\) is injective exactly when \(\ker\varphi = \{e\}\). That is the group version of the fact that a linear map is injective exactly when its null space is trivial — the same argument in a different setting.
A subgroup \(N\) with \(gNg^{-1} = N\) for every \(g\) — equivalently, its left and right cosets coincide, \(gN = Ng\). In an abelian group every subgroup is normal, so the condition only has content when the group is not commutative.
In \(S_3\), the two-element subgroup \(\{e, (01)\}\) is not normal: its left and right cosets differ. The alternating group \(A_3\), by contrast, is normal — as every kernel is.
If \(k \in \ker\varphi\) then \(\varphi(gkg^{-1}) = \varphi(g)\,e\,\varphi(g)^{-1} = e\), so \(gkg^{-1}\) is in the kernel too.
When \(N\) is normal, its cosets form a group under \((aN)(bN) = (ab)N\), written \(G/N\), with \(|G/N| = |G|/|N|\). Normality is exactly what makes this well-defined; without it the answer would depend on which representatives you chose.
\(\mathbb{Z}/n\mathbb{Z}\) is the integers modulo the subgroup \(n\mathbb{Z}\). Its elements are cosets — residue classes — and modular arithmetic is the quotient operation. The notation was already written in the form the general definition uses.
Collapse everything the map sends to the identity, and what is left is a copy of what the map hits. On the sign map: \(S_3 / A_3 \cong \{+1, -1\}\), so \(A_3\) has exactly two cosets — the even and the odd permutations — and \(|A_n| = n!/2\) follows.
\(S_n\) is the group of all permutations of \(n\) objects under composition, with \(|S_n| = n!\).
Swapping positions \(0\) and \(1\) and then \(1\) and \(2\) does not give the same arrangement as doing them the other way round. At order \(6\), \(S_3\) is the smallest non-abelian group — and it is not cyclic either, despite containing elements of order \(3\).
Every group is isomorphic to a subgroup of some symmetric group. Each element \(g\) permutes the group by \(x \mapsto gx\), and that assignment is an injective homomorphism. So permutation groups are not one example among many — every group is one of them.
Go back over the number theory with the new vocabulary, and five separate results become three arguments that work in every group at once:
| Number theory result | As group theory |
|---|---|
| Orders divide \(\varphi(n)\) | Lagrange applied to \(\langle a \rangle\) |
| Euler’s theorem | \(a^{|G|} = e\), from Lagrange |
| Fermat’s little theorem | The same, with \(|G| = p - 1\) |
| \(\varphi(p-1)\) primitive roots mod \(p\) | A cyclic group of order \(n\) has \(\varphi(n)\) generators |
| Wilson’s theorem | Pair each unit with its inverse |
In \((\mathbb{Z}/p\mathbb{Z})^{\times}\), pair each element with its inverse; each pair multiplies to \(1\). What is left over are the elements equal to their own inverse, and mod a prime \(x^2 \equiv 1\) forces \(x \equiv \pm 1\). So the product of all units is \(1 \cdot (p - 1) \equiv -1\), which is \((p-1)! \equiv -1 \pmod p\).
Checked for every prime up to \(47\). This is the same inverse-pairing proof that Euler’s Criterion and Gauss’s Lemma gives as a lemma. What the group language adds is why it works: the units mod \(p\) form a group, so every element has an inverse to pair with, and the only question is which elements are their own inverse. The converse of Wilson’s theorem — that a composite \(n\) never gives \(-1\) — is proved separately, and together they make \((n-1)! \equiv -1\) a test for primality.
Mod \(8\), every unit — \(1, 3, 5, 7\) — squares to \(1\), so nothing pairs off, and the product is \(1\) rather than \(-1\). The argument fails at precisely the step it relied on, which is what a correct proof should do when its hypothesis is removed. And it is the same fact that made \((\mathbb{Z}/8\mathbb{Z})^{\times}\) fail to be cyclic.
This opens Tier 5, which the plan marks optional. Abstract algebra was chosen first because it is the natural successor to the number theory and linear algebra units: most of what this page says, those units had already proved in concrete form.
| Concept | Statement | Already on this site as |
|---|---|---|
| Group | closure, associativity, identity, inverses | units mod \(n\), permutations, invertible matrices |
| \(|(\mathbb{Z}/n\mathbb{Z})^{\times}|\) | \(\varphi(n)\) | Euler’s totient |
| Order of \(a\) | smallest \(k\) with \(a^k = e\) | multiplicative order |
| Lagrange | \(|H|\) divides \(|G|\) | orders divide the totient |
| \(a^{|G|} = e\) | corollary of Lagrange | Euler and Fermat |
| Cyclic group | \(G = \langle g \rangle\) | a primitive root exists |
| Generators | \(\varphi(n)\) of them | \(\varphi(p-1)\) primitive roots |
| Homomorphism | \(\varphi(ab) = \varphi(a)\varphi(b)\) | \(\det\) and \(\operatorname{sgn}\) |
| Quotient | \(G/N\), \(N\) normal | \(\mathbb{Z}/n\mathbb{Z}\) |
| First Isomorphism Theorem | \(G/\ker\varphi \cong \operatorname{im}\varphi\) | \(|A_n| = n!/2\) |