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