# Group/Examples/Self-Inverse and Cancellable Elements

## 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.