Group Product Identity therefore Inverses/Part 1

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Theorem

Let $g$ and $h$ be elements of a group $G$ whose identity element is $e$.


Then:

$g h = e \implies h = g^{-1}$ and $g = h^{-1}$


Proof 1

From the Division Laws for Groups:

$g h = e \implies g = e h^{-1} = h^{-1}$

Also by the Division Laws for Groups:

$g h = e \implies h = g^{-1} e = g^{-1}$

$\blacksquare$


Proof 2

Let $g h = e$.

Then:

\(\ds h\) \(=\) \(\ds e h\) Group Axiom $\text G 2$: Existence of Identity Element
\(\ds \) \(=\) \(\ds \paren {g^{-1} g} h\) Group Axiom $\text G 3$: Existence of Inverse Element
\(\ds \) \(=\) \(\ds g^{-1} \paren {g h}\) Group Axiom $\text G 1$: Associativity
\(\ds \) \(=\) \(\ds g^{-1} e\) by hypothesis
\(\ds \) \(=\) \(\ds g^{-1}\) Group Axiom $\text G 2$: Existence of Identity Element


and:

\(\ds g\) \(=\) \(\ds g e\) Group Axiom $\text G 2$: Existence of Identity Element
\(\ds \) \(=\) \(\ds g \paren {h h^{-1} }\) Group Axiom $\text G 3$: Existence of Inverse Element
\(\ds \) \(=\) \(\ds \paren {g h} h^{-1}\) Group Axiom $\text G 1$: Associativity
\(\ds \) \(=\) \(\ds e h^{-1}\) by hypothesis
\(\ds \) \(=\) \(\ds h^{-1}\) Group Axiom $\text G 2$: Existence of Identity Element

$\blacksquare$