Cancellation Laws

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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

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