commutative law is not derivable in group theory?

by Christopher   Last Updated June 13, 2019 12:20 PM

This is the last question of my homework. I solved everything else, but this gives me a hard time. Hope my translation is correct and this is the right place for this kind of question.

Show that the commutative law in group theory is not derivable. A o B = B o A



Answers 1


It's possible that you've been confused by your professor's hint. Remember that Godel's completeness theorem, when phrased as an if-and-only-if, has two parts:

  • If $T$ is consistent, then $T$ has a model.

  • If $T$ has a model, then $T$ is consistent.

The first of these is completeness, and the second is soundness; since soundness is vastly easier to prove - to the point of bordering on trivial - the whole biconditional is often just called "completeness."

But it's soundness that you want here. Specifically, suppose you want to show that $T\not\vdash\varphi$. This is the same as showing that $T\cup\{\neg\varphi\}$ is consistent, which by soundness will follow from the existence of a model of $T\cup\{\neg\varphi\}$. Which is a long-winded way of saying:

It suffices to show that there is such a thing as a non-abelian group.

Noah Schweber
Noah Schweber
June 13, 2019 12:19 PM

Related Questions


Updated June 08, 2015 21:08 PM

Updated September 21, 2017 16:20 PM

Updated December 10, 2017 11:20 AM

Updated April 03, 2018 12:20 PM