Two operations: the ring
A group has one operation. But the integers have *two* that interlock — addition and multiplication — and that richer setting is a [[ring|ring]]. A ring is a set with two operations where: addition makes it an [[abelian-group|abelian group]] (so subtraction always works), multiplication is associative, and the two are tied together by the [[distributive-property|distributive law]], a(b + c) = ab + ac. The integers, polynomials with real coefficients, and n×n matrices are all rings.
Notice what a plain ring does *not* promise. Multiplication need not commute (matrix rings don't), there need not be a multiplicative identity 1, and even when there is, most elements have no multiplicative inverse — you can't divide by 2 inside the integers and stay there. Rings are deliberately loose; we tighten them by adding conditions, and each condition we demand carves out a smaller, better-behaved class.
Tightening: integral domains and fields
First refinement: forbid surprises in multiplication. An [[integral-domain|integral domain]] is a commutative ring with a 1, in which a product of two nonzero elements is never zero. That last rule — no zero divisors — is exactly what lets you cancel: if ab = ac with a ≠ 0, you may conclude b = c. The integers are the prototype. The clock ring mod 6 is *not* a domain, because 2·3 = 0 there even though neither factor is 0.
Second refinement: demand division. A [[field|field]] is a commutative ring with 1 in which *every* nonzero element has a multiplicative inverse — so you can add, subtract, multiply, and divide (except by zero) freely. The rationals, the reals, and the complex numbers are fields; the integers are not, since 2 has no integer inverse. Every field is an integral domain, but not conversely — the integers are a domain that just misses being a field.
When is the clock Z/nZ a FIELD? Need every nonzero a to have an inverse: some b with a·b ≡ 1 (mod n). Mod 6 (n = 6, composite): 2·1=2 2·2=4 2·3=0 2·4=2 2·5=4 → 2 NEVER gives 1 also 2·3 ≡ 0 with 2,3 ≠ 0 → zero divisors so Z/6Z is NOT even an integral domain, let alone a field. Mod 5 (n = 5, prime): find each inverse 1·1 ≡ 1 so 1⁻¹ = 1 2·3 = 6 ≡ 1 so 2⁻¹ = 3 (and 3⁻¹ = 2) 4·4 = 16 ≡ 1 so 4⁻¹ = 4 every nonzero element has an inverse → Z/5Z IS a field. Rule: Z/nZ is a field exactly when n is PRIME.
Comparing structures: homomorphisms and isomorphisms
Once you have many structures, you want to compare them. A [[homomorphism|homomorphism]] is a map between two structures that preserves the operation: for a group map, f(a ∗ b) = f(a) ∗ f(b); for a ring map, it also respects multiplication. It need not be one-to-one and need not hit everything — it just carries the *pattern* of combination from one world into another. The map sending each integer to its remainder mod n is a ring homomorphism: it doesn't matter whether you add first and then reduce, or reduce first and then add.
An [[isomorphism|isomorphism]] is a homomorphism that is also a perfect pairing — one-to-one and onto, with an inverse that is itself a homomorphism. When two structures are isomorphic they are the same structure wearing different clothes: relabel the elements of one and you get the other exactly. This is what “(a copy of)” quietly meant in Cayley's theorem — the hidden symmetric group is an isomorphic twin, not a literal equality.