Group/Examples/Self-Inverse and Cancellable Elements

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


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

Then $\left({S, \circ}\right)$ is a group, where $\circ$ is defined as $a \circ b = a \ast \left({e \ast b}\right)$.

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 $e \ast b \in S$, and hence also $a \circ b = a \ast \left({e \ast b}\right) \in S$.

This proves closure of $\circ$.

G2: Identity
We assert that $e$ is the identity with respect to $\circ$.

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

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 $a \ast \left({a \ast e}\right) = e$:

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

G3: Inverses
For any element $a \in S$, we claim that $e \ast a$ is the inverse for $a$.

This is verified as follows:

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

G1: Associativity
Lastly, we prove that $\circ$ is associative, i.e.:


 * $a \circ \left({b \circ c}\right) = \left({a \circ b}\right) \circ c$

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

Subsequently, we compute:

This somewhat cumbersome calculation shows that we indeed have associativity.

It follows that $\left({S, \circ}\right)$ is indeed a group.