Division Laws for Groups

Theorem

Let $G$ be a group.

Let $a, b, x \in G$.

Then:

$(1): \quad a x = b \iff x = a^{-1} b$

$(2): \quad x a = b \iff x = b a^{-1}$

Proof

All derivations can be achieved using applications of the group axioms:

Proof of $(1)$

\(\ds a x\)

\(=\)

\(\ds b\)

\(\ds \leadsto \ \ \)

\(\ds a^{-1} a x\)

\(=\)

\(\ds a^{-1} b\)

\(\ds \leadsto \ \ \)

\(\ds e x\)

\(=\)

\(\ds a^{-1} b\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds a^{-1} b\)

and the converse:

\(\ds x\)

\(=\)

\(\ds a^{-1} b\)

\(\ds \leadsto \ \ \)

\(\ds a x\)

\(=\)

\(\ds a a^{-1} b\)

\(\ds \leadsto \ \ \)

\(\ds a x\)

\(=\)

\(\ds e b\)

\(\ds \leadsto \ \ \)

\(\ds a x\)

\(=\)

\(\ds b\)

$\blacksquare$

Proof of $(2)$

\(\ds x a\)

\(=\)

\(\ds b\)

\(\ds \leadsto \ \ \)

\(\ds x a a^{-1}\)

\(=\)

\(\ds b a^{-1}\)

\(\ds \leadsto \ \ \)

\(\ds x e\)

\(=\)

\(\ds b a^{-1} b\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds b a^{-1}\)

and the converse:

\(\ds x\)

\(=\)

\(\ds b a^{-1}\)

\(\ds \leadsto \ \ \)

\(\ds x a\)

\(=\)

\(\ds b a^{-1} a\)

\(\ds \leadsto \ \ \)

\(\ds x a\)

\(=\)

\(\ds b e\)

\(\ds \leadsto \ \ \)

\(\ds x a\)

\(=\)

\(\ds b\)

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 35.2$: Elementary consequences of the group axioms