Group Product Identity therefore Inverses/Part 2/Proof 1
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Theorem
- $g h = e \implies h = g^{-1}$ and $g = h^{-1}$
Proof
From the Division Laws for Groups:
- $h g = e \implies g = e h^{-1} = h^{-1}$
Also by the Division Laws for Groups:
- $h g = e \implies h = g^{-1} e = g^{-1}$
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 35.3$: Elementary consequences of the group axioms