Group/Examples/Self-Inverse and Cancellable Elements

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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