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:

\(\displaystyle \left({b a}\right) a^{-1}\) \(=\) \(\displaystyle \left({c a}\right) a^{-1}\)                    
\(\displaystyle \implies\) \(\displaystyle b \left({a a^{-1} }\right)\) \(=\) \(\displaystyle c \left({a a^{-1} }\right)\)          Definition of Associativity          
\(\displaystyle \implies\) \(\displaystyle b e\) \(=\) \(\displaystyle c e\)          Definition of Inverse          
\(\displaystyle \implies\) \(\displaystyle b\) \(=\) \(\displaystyle c\)          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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