# 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

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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
June 13, 2019 12:19 PM

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