Division Laws for Groups

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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 $(1)$

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


and the converse:

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

$\blacksquare$


Proof of $(2)$

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


and the converse:

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

$\blacksquare$


Sources