Elementary Group Theory - Alternative Axioms

Alternative Axioms

The pair of axioms A3 and A4 may be replaced either by the pair:

  • A3’, left neutral. There exists an such that for all, .
  • A4’, left inverse. For each, there exists an element such that .

or by the pair:

  • A3”, right neutral. There exists an such that for all, .
  • A4”, right inverse. For each, there exists an element such that .

These evidently weaker axiom pairs are trivial consequences of A3 and A4. We will now show that the nontrivial converse is also true. Given a left neutral element and for any given then A4’ says there exists an such that .

Theorem 1.2:

Proof. Let be an inverse of Then:


\begin{align}
e & = y \perp (a \perp x) &\quad (1) \\ & = y \perp (a \perp (e \perp x)) &\quad (A3') \\ & = y \perp (a \perp ((x \perp a) \perp x)) &\quad (A4') \\ & = y \perp (a \perp (x \perp (a \perp x))) &\quad (A2) \\ & = y \perp ((a \perp x) \perp (a \perp x)) &\quad (A2) \\ & = (y \perp (a \perp x)) \perp (a \perp x) &\quad (A2) \\ & = e \perp (a \perp x) &\quad (1) \\ & = a \perp x &\quad (A3') \\
\end{align}

This establishes A4 (and hence A4”).

Theorem 1.2a:

Proof.


\begin{align}
a \perp e & = a \perp (x \perp a) &\quad (A4') \\ & = (a \perp x) \perp a &\quad (A2) \\ & = e \perp a &\quad (A4) \\ & = a &\quad (A3') \\
\end{align}

This establishes A3 (and hence A3”).

Theorem: Given A1 and A2, A3’ and A4’ imply A3 and A4.

Proof. Theorems 1.2 and 1.2a.

Theorem: Given A1 and A2, A3” and A4” imply A3 and A4.

Proof. Similar to the above.

Read more about this topic:  Elementary Group Theory

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