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 $(1)$

\(\ds a x\) \(=\) \(\ds b\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds a^{-1} \paren{a x }\) \(=\) \(\ds a^{-1} b\)
\(\ds \leadsto \ \ \) \(\ds \paren{a^{-1} a } x\) \(=\) \(\ds a^{-1} b\) Group Axiom $\text G 1$: Associativity
\(\ds \leadsto \ \ \) \(\ds e x\) \(=\) \(\ds a^{-1} b\) Definition of Inverse Element
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds a^{-1} b\) Definition of Identity Element


and the converse:

\(\ds x\) \(=\) \(\ds a^{-1} b\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds a x\) \(=\) \(\ds a \paren{a^{-1} b }\)
\(\ds \leadsto \ \ \) \(\ds a x\) \(=\) \(\ds \paren{a a^{-1} } b\) Group Axiom $\text G 1$: Associativity
\(\ds \leadsto \ \ \) \(\ds a x\) \(=\) \(\ds e b\) Definition of Inverse Element
\(\ds \leadsto \ \ \) \(\ds a x\) \(=\) \(\ds b\) Definition of Identity Element

$\blacksquare$


Proof of $(2)$

\(\ds x a\) \(=\) \(\ds b\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds \paren{x a } a^{-1}\) \(=\) \(\ds b a^{-1}\)
\(\ds \leadsto \ \ \) \(\ds x \paren{a a^{-1} }\) \(=\) \(\ds b a^{-1}\) Group Axiom $\text G 1$: Associativity
\(\ds \leadsto \ \ \) \(\ds x e\) \(=\) \(\ds b a^{-1} b\) Definition of Inverse Element
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds b a^{-1}\) Definition of Identity Element


and the converse:

\(\ds x\) \(=\) \(\ds b a^{-1}\) by hypothesis
\(\ds \leadsto \ \ \) \(\ds x a\) \(=\) \(\ds \paren{b a^{-1} } a\)
\(\ds \leadsto \ \ \) \(\ds x a\) \(=\) \(\ds b \paren{a^{-1} a }\) Group Axiom $\text G 1$: Associativity
\(\ds \leadsto \ \ \) \(\ds x a\) \(=\) \(\ds b e\) Definition of Inverse Element
\(\ds \leadsto \ \ \) \(\ds x a\) \(=\) \(\ds b\) Definition of Identity Element

$\blacksquare$


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