Cancellation Laws

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

$\blacksquare$


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