Binary Operations and the Group Axioms

Group theory is the mathematical language of symmetry, and binary operations satisfying these four axioms appear everywhere structure is preserved under combination. In physics, the symmetries of physical laws (rotations, Lorentz transformations, gauge symmetries of the Standard Model) form groups, and Noether's theorem ties every continuous symmetry group to a conservation law. In chemistry and crystallography, point groups and space groups classify molecular and crystal symmetries. In cryptography, the difficulty of certain group-theoretic problems (discrete logarithms in $(\mathbb{Z}/p\mathbb{Z})^\times$ or in elliptic curve groups) underlies widely deployed public-key systems such as Diffie–Hellman key exchange and elliptic curve cryptography. In computer science, group-based hashing, error-correcting codes over finite groups, and the structure of permutation groups underlie algorithms for sorting networks and combinatorial search. In pure mathematics, groups are the entry point to ring theory, field theory, and Galois theory (Cyclic Groups and Modular Arithmetic gives the first concrete family beyond the abstract definition), and the basic consequences proved here — cancellation, uniqueness of identity and inverses — are used silently in essentially every subsequent argument about groups, as previewed in Basic Consequences of the Group Axioms.

Ask the tutor: Why does associativity guarantee that any parenthesization of a long product gives the same answer, and how would you prove that rigorously by induction?

Think of a binary operation as a machine: feed it two inputs from $S$, and it outputs one element, also required to lie in $S$. This "stay inside the set" requirement is why closure is not listed as a separate group axiom in the definition above — it is baked into the very meaning of "a binary operation on $S$." If someone hands you a rule that sometimes produces an output outside $S$, that rule is simply not a binary operation on $S$ in the first place. Associativity is the axiom that makes long expressions meaningful. Without it, a string like $a * b * c * d$ would be ambiguous — you would need to specify a full parenthesization, and different choices could give different answers. With associativity, one proves (by induction, foreshadowed in Basic Consequences of the Group Axioms) that any parenthesization of a fixed sequence of elements, in a fixed order, gives the same result. This is why we can write $abcd$ unambiguously. The identity element is the "do-nothing" element — combining it with anything leaves that thing unchanged, whether the identity sits on the left or the right. Inverses are what make equations solvable: given $a * x = b$, having an inverse for $a$ lets us solve for $x$ by combining both sides with $a^{-1}$. A convenient way to picture a small group is its Cayley table, a grid recording every product; associativity, identity, and inverses are all visible structural patterns in that grid.

Example 1: The nonzero rationals under multiplication. Let $G = \mathbb{Q}^{\times} = \mathbb{Q} \setminus \{0\}$, and let $*$ be ordinary multiplication of rational numbers. We verify every axiom explicitly. Binary operation / closure. If $a = p/q$ and $b = r/s$ are nonzero rationals (with $p,q,r,s$ integers, $q,s \neq 0$, and $p,r\neq 0$ since $a,b \neq 0$), then

Since $p, r \neq 0$ we have $pr \neq 0$, so $a \cdot b \neq 0$; and $qs \neq 0$ since $q, s \neq 0$. Hence $a \cdot b \in \mathbb{Q}^\times$. So $\cdot : \mathbb{Q}^\times \times \mathbb{Q}^\times \to \mathbb{Q}^\times$ is genuinely a binary operation on $\mathbb{Q}^\times$ — note this required checking $ab \ne 0$, which is why we excluded $0$ from $G$ in the first place: $0 \cdot 1 = 0 \notin \mathbb{Q}^\times$ would have broken closure had we included $0$. Associativity. For rationals $a = p/q$, $b=r/s$, $c=t/u$,

Both equal $\frac{prt}{qsu}$ (using associativity and commutativity of multiplication of integers, which we take as known arithmetic), so $(a\cdot b)\cdot c = a\cdot(b\cdot c)$. Identity. Take $e = 1$. For any $a = p/q \in \mathbb{Q}^\times$, $1 \cdot a = a \cdot 1 = p/q = a$. So $1$ is a two-sided identity. Inverses. For $a = p/q$ with $p \neq 0$, set $a^{-1} = q/p$. This is a nonzero rational since $q \neq 0$. Check:

So every element has a two-sided inverse in $G$. All four axioms hold, so $(\mathbb{Q}^\times, \cdot)$ is a group. It is abelian, since rational multiplication commutes. Note that $\mathbb{Q}$ itself under multiplication is not a group (Axiom 3 fails: $0$ has no multiplicative inverse), which is exactly why $0$ was excluded.

Example 2: Invertible $2\times 2$ matrices over $\mathbb{R}$ — a genuinely non-abelian group. Let $G = GL_2(\mathbb{R})$, the set of $2\times 2$ real matrices $A$ with $\det A \neq 0$, and let $*$ be matrix multiplication. We check the axioms with a specific asymmetric pair to make the non-commutativity concrete. Closure. If $A, B \in GL_2(\mathbb{R})$ then $\det(AB) = \det(A)\det(B) \neq 0$ (a standard fact about determinants), so $AB \in GL_2(\mathbb{R})$. Thus matrix multiplication is a binary operation on $G$. Associativity. Matrix multiplication is associative: for any three $2\times2$ matrices $A,B,C$ (invertible or not), entrywise expansion of $(AB)C$ and $A(BC)$ both give the same sum $\sum_k \sum_j A_{ik}B_{kj}C_{jl}$ for each entry, by the distributive law in $\mathbb{R}$; this is a general fact about matrix multiplication that we use here without reproof, as it holds for all matrices over any field, invertible or not. Identity. Take $e = I = \begin{pmatrix}1&0\\0&1\end{pmatrix}$, which is invertible ($\det I = 1 \neq 0$) and satisfies $IA = AI = A$ for all $2\times2$ matrices $A$. Inverses. By definition of $GL_2(\mathbb{R})$, every $A \in G$ has $\det A \neq 0$, and the classical adjugate formula

gives a matrix in $G$ (since $\det(A^{-1}) = 1/\det A \neq 0$) satisfying $A A^{-1} = A^{-1}A = I$; this can be checked by direct entrywise multiplication. So $(GL_2(\mathbb{R}), \cdot)$ is a group. Now take the concrete, deliberately asymmetric elements

Both have determinant $1 \neq 0$, so both lie in $G$. Compute:

Since $AB = \begin{pmatrix}7&2\\3&1\end{pmatrix} \neq \begin{pmatrix}1&2\\3&7\end{pmatrix} = BA$, this group is non-abelian: the commutativity axiom is not part of the group definition, and here we exhibit a group where it genuinely fails. This is the first indication that groups are strictly more general than the familiar commutative number systems.