Group is Abelian iff it has Middle Cancellation Property

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 $G$.

$\blacksquare$