Cancellation Laws
Contents |
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:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left({b a}\right) a^{-1}\) | \(=\) | \(\displaystyle \) | \(\displaystyle \left({c a}\right) a^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle b \left({a a^{-1} }\right)\) | \(=\) | \(\displaystyle \) | \(\displaystyle c \left({a a^{-1} }\right)\) | \(\displaystyle \) | \(\displaystyle \) | by associativity | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle b e\) | \(=\) | \(\displaystyle \) | \(\displaystyle c e\) | \(\displaystyle \) | \(\displaystyle \) | by the definition of inverse | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle b\) | \(=\) | \(\displaystyle \) | \(\displaystyle c\) | \(\displaystyle \) | \(\displaystyle \) | 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 the 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 the 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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Sources
- Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 7$: Theorem $7.1$ Corollary
- Paul Halmos and Steven Givant: Introduction to Boolean Algebras (2008)... (previous)... (next): $\S 1$: Exercise $5$
