Group is B-Algebra Iff All Elements Self-Inverse

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Theorem

Let $\left({G, \circ}\right)$ be a group whose identity element is $e$.


Then $\left({G, \circ}\right)$ is also a $B$-algebra if and only if:

$\forall g \in G: g = g^{-1}$

That is, if and only if all elements of $G$ are self-inverse.


Proof

Let $\left({G, \circ}\right)$ be a group such that $\forall g \in G: g = g^{-1}$.

From Group Induces B-Algebra, the algebraic structure $\left({G, *}\right)$ where $*$ is defined as:

$\forall a, b \in G: a * b := a \circ b^{-1}$

is a $B$-algebra.

But as $b = b^{-1}$, it follows that:

$\forall a, b \in G: a * b := a \circ b$

and so $\left({G, \circ}\right)$ is itself a $B$-algebra.


Now suppose it is not the case that $\forall g \in G: g = g^{-1}$.

Suppose $\left({G, \circ}\right)$ were a $B$-algebra.

By axiom $A2$ of the definition of a $B$-algebra, the identity element $e$ would be the $0$, as:

$\forall a \in G: a \circ 0 = a$

As an Identity is Unique, there can be no other element in $G$ which can satisfy the conditions to be the $0$ of a $B$-algebra.

Now as not every element is self-inverse:

$\exists x \in G: x \circ x \ne e$

Then by axiom $A1$ of the definition of a $B$-algebra it follows that $\left({G, \circ}\right)$ is not a $B$-algebra.

Hence the result.

$\blacksquare$