# Cancellation Laws

## Theorem

Let $G$ be a group.

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

Then the following hold:

- Right cancellation law

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

- Left cancellation law

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

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:

\(\displaystyle \paren {b a} a^{-1}\) | \(=\) | \(\displaystyle \paren {c a} a^{-1}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle b \paren {a a^{-1} }\) | \(=\) | \(\displaystyle c \paren {a a^{-1} }\) | $\quad$ Definition of Associative Operation | $\quad$ | ||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle b e\) | \(=\) | \(\displaystyle c e\) | $\quad$ Definition of Inverse Element | $\quad$ | ||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle b\) | \(=\) | \(\displaystyle c\) | $\quad$ Definition of Identity Element | $\quad$ |

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

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