# Inverse of Group Inverse/Proof 2

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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

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

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 4.6$. Elementary theorems on groups: Theorem $\text{(iii)}$ - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 28 \ (2)$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 35.5$: Elementary consequences of the group axioms - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $3$: Elementary consequences of the definitions: Corollary $3.5$

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- 1974: Thomas W. Hungerford:
*Algebra*... (previous) ... (next): $\text{I}$: Groups: $\S 1$ Semigroups, Monoids and Groups: Theorem $1.2 \text{(iv)}$