First order logic: statement is or isn't a semantic consequence of group theory?

Question

Where Group theory is defined by:

• Signature $\Sigma=\{e, ⋅\}$

• Axioms:

  1. Closure: $\forall x\forall y\exists z(x⋅y=z)$
  2. Associativity: $∀x∀y∀z ( (x⋅y)⋅z = x⋅(y⋅z) )$
  3. Identity element: $∀x ( (x⋅e = x) ∧ (e⋅x = x) )$
  4. Inverse element: $∀x∃y ( (x⋅y = e) ∧ (y⋅x = e) )$

Prove/disprove:

• $∀x (x · x = e → (x = e))$ is a semantic consequence of group theory.

I'm kind of sure that it ISN'T a semantic consequence, because there could be several identity elements. How can I show this formally? With propositional logic it's easy, but this one gets fuzzy. Thank you!

Answer 1

Suppose $e$ and $f$ are identity elements. Then \begin{align} e = ef & \text{ since $f$ is an identity element} \\ ef = f & \text{ since $e$ is an identity element.} \end{align} So $e=f$ and there cannot be more than one identity element.

But now look at the group with two elements and observe that $x^2=e$ is true of both of them: the one non-identity element must be its own inverse.

And there are many groups with non-identity elements that are their own inverses. For example, in the group of all permutations on $n$ elements, every product of disjoint $2$-cycles is its own inverse. In the integers modulo $2n$, the element $n$ is its own inverse. In any direct product of two groups or more groups, if each of the factors has an element that is its own inverse, then every way of picking one of those from each of the factors, and the identity element from the other factors yields an element that is its own inverse. In Euclidean $3$-space, every line is the axis of an $180^\circ$ rotation, and each of those is its own inverse in the group of all isometries of $\mathbb R^3.$ And for each point $p\in\mathbb R^3,$ the mapping $x\mapsto 2p-x$ is its own inverse.

Answer 2

Hint

We can show (by contradiction) that:

$∀x(x⋅x=x → (x=e))$

is a consequence of the axioms for group theory above.

Assume that the sought conclusion is not implied by the axioms, i.e.

$\exists z \ (z \cdot z= z \land \lnot (z=e))$.

Thus: $z \cdot z = z$, and multiply by inverse $z^{-1}$ : $z^{-1} \cdot (z \cdot z) = z^{-1} \cdot z = e$.

By associativity : $z^{-1} \cdot (z \cdot z) = (z^{-1} \cdot z) \cdot z = e \cdot z$, and thus:

$e \cdot z = e$.

Now multiply by $e^{-1}$ : $e^{-1} \cdot (e \cdot z) = e^{-1} \cdot e = e$.

Again by associativity: $(e^{-1} \cdot e) \cdot z = e \cdot z = z$, and thus:

$z = e$

contradicting the assumption that: $\lnot (z=e)$.