Division Laws for Groups

Theorem

Let $G$ be a group.

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

Then:

• $a x = b \iff x = a^{-1} b$
• $x a = b \iff x = b a^{-1}$

Proof

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

Proof of First Result

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle a x\)	\(=\)	\(\displaystyle b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle a^{-1} a x\)	\(=\)	\(\displaystyle a^{-1} b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle e x\)	\(=\)	\(\displaystyle a^{-1} b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle a^{-1} b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

and the converse:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle a^{-1} b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle a x\)	\(=\)	\(\displaystyle a a^{-1} b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle a x\)	\(=\)	\(\displaystyle e b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle a x\)	\(=\)	\(\displaystyle b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Proof of Second Result

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x a\)	\(=\)	\(\displaystyle b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x a a^{-1}\)	\(=\)	\(\displaystyle b a^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x e\)	\(=\)	\(\displaystyle b a^{-1} b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle b a^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

and the converse:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle b a^{-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x a\)	\(=\)	\(\displaystyle b a^{-1} a\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x a\)	\(=\)	\(\displaystyle b e\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x a\)	\(=\)	\(\displaystyle b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Sources

• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 35.2$