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 operation is cancellable.
Let $e$ be the identity element of $G$.
Then:
Corollary 1
- $g h = g \implies h = e$
Corollary 2
- $h g = g \implies h = e$
Proof 1
Let $a, b, c \in G$ and let $a^{-1}$ be the inverse of $a$.
Suppose $b a = c a$.
Then:
\(\ds \paren {b a} a^{-1}\) | \(=\) | \(\ds \paren {c a} a^{-1}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds b \paren {a a^{-1} }\) | \(=\) | \(\ds c \paren {a a^{-1} }\) | Definition of Associative Operation | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds b e\) | \(=\) | \(\ds c e\) | Definition of Inverse Element | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds b\) | \(=\) | \(\ds c\) | Definition of Identity Element |
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 $5$