Cancellation laws

Remarks

The first law is left cancellation and the second law is right cancellation. To prove the first law, multiply $ab = ac$ on the left by $a^{-1}$. A consequence is that in the Cayley table of a finite group, no element occurs two times in one row or in one column. The laws can fail in a monoid: in $(\mathbb{Z}, \cdot)$, $0 \cdot 1 = 0 \cdot 2$ but $1 \neq 2$.

Going deeper

For a student

The cancellation laws let us remove a common factor from both sides of an equation. The factor must be on the same side in both products.

The proof of left cancellation has three steps:

1. Suppose that $ab = ac$.
2. Multiply both sides on the left by $a^{-1}$.
3. Associativity and $a^{-1}a = e$ give $b = c$.

A worked example in $S_3$, with composition from right to left:

1. Let $a = (1\,2)$. Suppose that $ax = (1\,2\,3)$ and $ay = (1\,2\,3)$.
2. Then $ax = ay$, so $x = y$ by left cancellation.
3. Thus $x = a^{-1}(1\,2\,3) = (1\,2)(1\,2\,3) = (2\,3)$ is the only solution.

This shows a useful consequence. For fixed $a$ and $c$, the equation $ax = c$ has exactly one solution, $x = a^{-1}c$. Thus each row of the Cayley table of a finite group contains each element exactly one time. The same is true for each column.

A common error is to cancel across sides. From $ab = ca$ we cannot conclude $b = c$. In $S_3$, let $a = (1\,2)$, $b = (1\,3)$ and $c = (2\,3)$. Then $ab = (1\,3\,2)$ and $ca = (1\,3\,2)$, but $b \ne c$.

The laws need inverses. In the monoid $(\mathbb{Z}, \cdot)$, $0 \cdot 1 = 0 \cdot 2$, but $1 \ne 2$, because $0$ has no inverse.