Inverse of an Inverse
Contents |
Theorem
General Algebraic Structures
Let $\left({S, \circ}\right)$ be an algebraic structure.
Let $x \in S$ be invertible, and let $y$ be an inverse of $x$.
Then $x$ is also an inverse of $y$.
Monoids
Let $\left({S, \circ}\right)$ be a monoid.
Let $x \in S$ be invertible, and let its inverse be $x^{-1}$.
Then $x^{-1}$ is also invertible, and:
- $\left({x^{-1}}\right)^{-1} = x$
Proof
Algebraic Structure
Let $x \in S$ be invertible, where $y$ is an inverse of $x$.
Then:
- $x \circ y = e = y \circ x$
by definition.
Proof for Monoid
If $\left({S, \circ}\right)$ is a monoid then by definition $\circ$ is associative.
So any inverse of $x$ is unique, and can be denoted $x^{-1}$.
From the result for algebraic structures, $x^{-1}$ is also invertible and its inverse is $x$.
Thus we see that $\left({x^{-1}}\right)^{-1} = x$.
$\blacksquare$
Proof for Group
For use when $G$ is a group.
Let $g \in G$.
Then:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle g\) | \(\in\) | \(\displaystyle G\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle e\) | \(=\) | \(\displaystyle g^{-1}g\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of inverse | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle (g^{-1})^{-1}e\) | \(=\) | \(\displaystyle ((g^{-1})^{-1}(g^{-1}g)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle (g^{-1})^{-1}e\) | \(=\) | \(\displaystyle ((g^{-1})^{-1}g^{-1})g\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Associativity | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle (g^{-1})^{-1}e\) | \(=\) | \(\displaystyle eg\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of inverse | ||
| \(\displaystyle \) | \(\displaystyle \implies\) | \(\displaystyle \) | \(\displaystyle (g^{-1})^{-1}\) | \(=\) | \(\displaystyle g\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of identity |
$\blacksquare$
Sources
- J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 4.6$: Theorem $\text{(iii)}$
- Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 4$: Theorem $4.3$
- George McCarty: Topology: An Introduction with Application to Topological Groups (1967): Chapter $\text{II}$
- C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 1.4$: Theorem $2 \ \text{(i)}$
- Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 28 \ (2)$
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 35.5$
- John F. Humphreys: A Course in Group Theory (1996): $\S 3$: Corollary $3.5$