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:

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

$\blacksquare$


Proof 2

Let $g \in G$.

Then:

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

$\blacksquare$


Sources


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