Group is Abelian iff it has Middle Cancellation Property
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Theorem
Let $G$ be a group.
Then the following are equivalent:
- $(1): \quad G$ is abelian
- $(2): \quad G$ satisfies the middle cancellation property
Proof
Let us suppress the operation of $G$ for brevity.
$(2) \implies (1)$
Suppose that $G$ satisfies the middle cancellation property.
Then, for all $g, h \in G$:
\(\ds e h\) | \(=\) | \(\ds h e\) | Definition of Identity Element | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds g g^{-1} h\) | \(=\) | \(\ds hg^{-1}g\) | Definition of Inverse Element | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds g h\) | \(=\) | \(\ds h g\) | Definition of Middle Cancellation Property |
Thus $G$ is abelian.
$\Box$
$(1) \implies (2)$
Conversely, suppose $G$ is abelian.
Then, for all $a, b, c, d, x \in G$:
\(\ds a x b\) | \(=\) | \(\ds c x d\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a b x\) | \(=\) | \(\ds c d x\) | Definition of Abelian Group | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a b\) | \(=\) | \(\ds c d\) | Right Cancellation Law |
Thus the middle cancellation property holds in $G$.
$\blacksquare$