How to prove that a set with an associative binary operation and identity is a group?

You are correct — the group axioms can be weakened. Here is the proof. **Theorem:** A set $G$ with an associative binary operation $\cdot$ such that: - There exists $e \in G$ with $e \cdot a = a$ for all $a$ (left identity) - For each $a$, there exists $a' \in G$ with $a' \cdot a = e$ (left inverse) Then $G$ is a group. **Proof:** First, show $a \cdot a' = e$ (left inverse is also right inverse). Let $a'$ be the left inverse of $a$, and $a''$ be the left inverse of $a'$: \begin{align*} a \cdot a' &= e \cdot (a \cdot a') \\ &= (a'' \cdot a') \cdot (a \cdot a') \\ &= a'' \cdot ((a' \cdot a) \cdot a') \\ &= a'' \cdot (e \cdot a') \\ &= a'' \cdot a' = e \end{align*} Now show $a \cdot e = a$ (left identity is also right identity): $$a \cdot e = a \cdot (a' \cdot a) = (a \cdot a') \cdot a = e \cdot a = a$$ Thus all four axioms hold. **Hierarchy of Algebraic Structures:** | Structure | Closure | Associative | Identity | Inverses | |---|---|---|---|---| | **Magma** | ✓ | | | | | **Semigroup** | ✓ | ✓ | | | | **Monoid** | ✓ | ✓ | ✓ | | | **Group** | ✓ | ✓ | ✓ | ✓ | | **Abelian Group** | ✓ | ✓ | ✓ | ✓ + commutative | **Examples:** - $(\mathbb{N}, +)$ is a semigroup but not a monoid (no identity for $0$ if we exclude it) - $(\mathbb{N}_0, +)$ is a monoid but not a group (no negative numbers) - $(\mathbb{Z}, +)$ is an abelian group