Inverse in Group is Unique/Proof 3

Theorem

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


Then every element $x \in G$ has exactly one inverse:

$\forall x \in G: \exists_1 x^{-1} \in G: x \circ x^{-1} = e^{-1} = x^{-1} \circ x$

where $e$ is the identity element of $\struct {G, \circ}$.


Proof

Let $x, y \in G$.

We already have, from the definition of inverse element, that:

$\forall x \in G: \exists x^{-1} \in G: x \circ x^{-1} = e = x^{-1} \circ x$

By Group has Latin Square Property, there exists exactly one $a \in G$ such that $a \circ x = y$.

Similarly, there exists exactly one $b \in G$ such that $x \circ b = y$.

Substituting $e$ for $y$, it follows that $x^{-1}$ as defined above is unique.

$\blacksquare$


Sources

  • 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory
  • 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 4.6$. Elementary theorems on groups: Theorem $\text{(i)}$
  • 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $3$: Elementary consequences of the definitions: Proposition $3.3$: Remark $3$
This article is issued from ProofWiki. The text is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Additional terms may apply for the media files.