# Cancellation Laws

## Theorem

Let $G$ be a group.

Let $a, b, c \in G$.

Then:

- $b a = c a \implies b = c$
- $a b = a c \implies b = c$

These are respectively called the **right and left cancellation laws**.

That is, the group product is cancellable.

## Proof 1

Let $a, b, c \in G$ and let $a^{-1}$ be the inverse of $a$.

Suppose $b a = c a$.

Then:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left({b a}\right) a^{-1}\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \left({c a}\right) a^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle b \left({a a^{-1} }\right)\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle c \left({a a^{-1} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | by associativity | ||

\(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle b e\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle c e\) | \(\displaystyle \) | \(\displaystyle \) | by the definition of inverse | ||

\(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle b\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle c\) | \(\displaystyle \) | \(\displaystyle \) | by the definition of identity |

Thus, the right cancellation law holds. The proof of the left cancellation law is analogous.

$\blacksquare$

## Proof 2

From its definition, a group is a monoid, all of whose elements have inverses and thus are invertible.

From Invertible Element of Monoid is Cancellable, it follows that all its elements are therefore cancellable.

$\blacksquare$

## Proof 3

Suppose $x = b a = c a$.

By Group has Latin Square Property, there exists exactly one $y \in G$ such that $x = y a$.

That is, $x = b a = c a \implies b = c$.

Similarly, suppose $x = a b = a c$.

Again by Group has Latin Square Property, there exists exactly one $y \in G$ such that $x = a y$.

That is, $a b = a c \implies b = c$.

$\blacksquare$

## Sources

- Seth Warner:
*Modern Algebra*(1965)... (previous)... (next): $\S 7$: Theorem $7.1$ Corollary - Thomas W. Hungerford:
*Algebra*(1974)... (previous)... (next): $\text{I}$: Groups: $\S 1$ Semigroups, Monoids and Groups: Theorem $1.2 \text{(ii)}$ - Paul Halmos and Steven Givant:
*Introduction to Boolean Algebras*(2008)... (previous)... (next): $\S 1$: Exercise $5$