Inverse of Group Inverse

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Theorem

Let $\struct {G, \circ}$ be a group.

Let $g \in G$, with inverse $g^{-1}$.


Then:

$\paren {g^{-1} }^{-1} = g$


Proof 1

Let $g \in G$.

Then:

\(\displaystyle g\) \(\in\) \(\displaystyle G\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle e\) \(=\) \(\displaystyle g^{-1} \circ g\) Definition of Inverse Element
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {g^{-1} }^{-1} \circ e\) \(=\) \(\displaystyle \paren {g^{-1} }^{-1} \circ \paren {g^{-1} \circ g}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {g^{-1} }^{-1} \circ e\) \(=\) \(\displaystyle \paren {\paren {g^{-1} }^{-1} \circ g^{-1} } \circ g\) Definition of Associative Operation
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {g^{-1} }^{-1} \circ e\) \(=\) \(\displaystyle e \circ g\) Definition of Inverse Element
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {g^{-1} }^{-1}\) \(=\) \(\displaystyle g\) Definition of Identity Element

$\blacksquare$


Proof 2

Let $g \in G$.

Then:

\(\displaystyle g g^{-1}\) \(=\) \(\displaystyle e\) Definition of Inverse Element
\(\displaystyle \leadsto \ \ \) \(\displaystyle g\) \(=\) \(\displaystyle \paren {g^{-1} }^{-1}\) Group Product Identity therefore Inverses

$\blacksquare$


Sources