Group Product Identity therefore Inverses/Part 1/Proof 2
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Theorem
- $g h = e \implies h = g^{-1}$ and $g = h^{-1}$
Proof
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 |
$\blacksquare$
Sources
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: The Group Property: Theorem $1 \ \text{(iii)}$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 28 \ (1)$