A group is the simplest structure in which every equation $a \ast x = b$ can be solved, the way you solve $3 + x = 7$ or $3x = 7$. Solving needs exactly three tools besides a binary operation: regrouping, a neutral element, and a way to undo.
Definition. A group is a set $G$ with a binary operation $\ast$ such that
- (associativity) $(a \ast b) \ast c = a \ast (b \ast c)$ for all $a, b, c \in G$;
- (identity) there is $e \in G$ with $e \ast a = a \ast e = a$ for all $a \in G$;
- (inverses) for each $a \in G$ there is $a^{-1} \in G$ with $a \ast a^{-1} = a^{-1} \ast a = e$.
Closure is built into "binary operation on $G$", but when you test an example, check it first. A group is abelian if also $a \ast b = b \ast a$ always. The number of elements $|G|$ is the order of $G$. We usually write $ab$ for $a \ast b$; for abelian groups like $\mathbb{Z}$ we often write $a + b$, $0$ and $-a$ instead.
Why it works
| Set and operation | Group? | Reason |
|---|---|---|
| $\mathbb{Z}$, $+$ | yes | identity $0$, inverse $-a$ |
| $\mathbb{Q} \setminus \{0\}$, $\times$ | yes | identity $1$, inverse $1/a$ |
| invertible $2 \times 2$ real matrices, $\times$ | yes (not abelian) | $\det(AB) = \det A \det B \ne 0$; inverses |
| $\mathbb{Z}$, $-$ | no | $(5 - 3) - 1 \ne 5 - (3 - 1)$ |
| $\mathbb{N}$, $+$ | no | $1$ has no inverse |
| $\mathbb{Z}_{6}$, $\times$ mod $6$ | no | $0$ and $2$ have no inverse |
Worked example: A finite group from a table
- Closure: no product of two of these is a multiple of $5$, so every product lies in the set.
- Associativity is inherited from multiplication of integers.
- Identity: $1$. Inverses: $1 \cdot 1 = 1$, $2 \cdot 3 = 6 \equiv 1$, $4 \cdot 4 = 16 \equiv 1$.
Answer: Yes. It is the group $U(5)$ of the next lesson.
Worked example: A strange-looking group
- Associativity: both $(a \ast b) \ast c$ and $a \ast (b \ast c)$ equal $a + b + c + 2$.
- Identity: $a \ast u = a + u + 1 = a$ forces $u = -1$. Check: $(-1) \ast a = a$ too.
- Inverse: $a \ast y = a + y + 1 = -1$ gives $y = -a - 2$.
- Check with $a = 4$: $4 \ast (-6) = 4 - 6 + 1 = -1$, the identity.
Answer: $e = -1$ and $a^{-1} = -a - 2$.
Common mistake
Key idea