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 |
$\blacksquare$