# Group/Examples/Self-Inverse and Cancellable Elements

## Contents

## Theorem

Let $S$ be a set with an operation which assigns to each $\tuple {a, b} \in S \times S$ an element $a \ast b \in S$ such that:

- $(1): \quad \exists e \in S: a \ast b = e \iff a = b$
- $(2): \quad \forall a, b, c \in S: \paren {a \ast c} \ast \paren {b \ast c} = a \ast b$

Then $\struct {S, \circ}$ is a group, where $\circ$ is defined as $a \circ b = a \ast \paren {e \ast b}$.

## Proof

We verify the group axioms, in the following order (for convenience):

### G0: Closure

Let $a, b \in S$. Then from the definition of $\ast$, we have $a \ast b \in S$, and hence also $a \circ b = a \ast \paren {e \ast b} \in S$.

This proves closure of $\circ$.

$\Box$

### G2: Identity

We assert that $e$ is the identity with respect to $\circ$.

We verify this as follows (let $a \in S$):

\(\displaystyle a \circ e\) | \(=\) | \(\displaystyle a \ast \paren {e \ast e}\) | Definition of $\circ$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \ast e\) | Property $(1)$ of $\ast$ | ||||||||||

\(\displaystyle e \circ a\) | \(=\) | \(\displaystyle e \ast \paren {e \ast a}\) | Definition of $\circ$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a \ast a} \ast \paren {e \ast a}\) | Property $(1)$ of $\ast$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \ast e\) | Property $(2)$ of $\ast$ |

It follows that $a \circ e = e \circ a = a \ast e$.

It remains to prove the following identity:

- $(3): \quad a \ast e = a$

By property $(1)$, it suffices to prove $a \ast \paren {a \ast e} = e$:

\(\displaystyle a \ast \paren {a \ast e}\) | \(=\) | \(\displaystyle \paren {a \ast e} \ast \paren {\paren {a \ast e} \ast e}\) | Property $(2)$ of $\ast$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a \ast e} \ast \paren {\paren {a \ast e} \ast \paren {e \ast e} }\) | Property $(1)$ of $\ast$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a \ast e} \ast \paren {a \ast e}\) | Property $(2)$ of $\ast$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e\) | Property $(1)$ of $\ast$ |

So indeed $e$ is the identity for $\circ$.

$\Box$

### G3: Inverses

For any element $a \in S$, we claim that $e \ast a$ is the inverse for $a$.

This is verified as follows:

\(\displaystyle \paren {e \ast a} \circ a\) | \(=\) | \(\displaystyle \paren {e \ast a} \ast \paren {e \ast a}\) | Definition of $\circ$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e\) | Property $(1)$ of $\ast$ | ||||||||||

\(\displaystyle a \circ \paren {e \ast a}\) | \(=\) | \(\displaystyle a \ast \paren {e \ast \paren {e \ast a} }\) | Definition of $\circ$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \ast \paren {e \circ a}\) | Definition of $\circ$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \ast a\) | $e$ is the identity element for $\circ$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e\) | Property $(1)$ of $\ast$ |

We conclude that $e \ast a$ is indeed the inverse for $a$.

$\Box$

### G1: Associativity

Lastly, we prove that $\circ$ is associative, i.e.:

- $a \circ \paren {b \circ c} = \paren {a \circ b} \circ c$

To this end, we observe the following property of $\ast$:

\((4):\quad\) | \(\displaystyle e \ast \paren {a \ast b}\) | \(=\) | \(\displaystyle \paren {b \ast b} \ast \paren {a \ast b}\) | Property $(1)$ of $\ast$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle b \ast a\) | Property $(2)$ of $\ast$ |

Subsequently, we compute:

\(\displaystyle \paren {a \circ b} \circ c\) | \(=\) | \(\displaystyle \paren {a \ast \paren {e \ast b} } \ast \paren {e \ast c}\) | Definition of $\circ$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a \ast \paren {e \ast b} } \ast \paren {\paren {e \ast c} \ast e}\) | Property $(3)$ of $\ast$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {a \ast \paren {e \ast b} } \ast \paren {\paren {\paren {e \ast c} \ast b} \ast \paren {e \ast b} }\) | Property $(2)$ of $\ast$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \ast \paren {\paren {e \ast c} \ast b}\) | Property $(2)$ of $\ast$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \ast \paren {e \ast \paren {b \ast \paren {e \ast c} } }\) | Property $(4)$ of $\ast$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle a \circ \paren {b \circ c}\) | Definition of $\circ$ |

This somewhat cumbersome calculation shows that we indeed have associativity.

$\Box$

It follows that $\struct {S, \circ}$ is indeed a group.

$\blacksquare$

## Also see

- $B$-Algebra, where it can be seen that the operation $*$ as defined here is related to that in the definition of a $B$-Algebra.

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 28 \delta$

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Exercise $7.7$