Consider the class of groups in the signature {*}. Is the equational theory of that class axiomatized by the associative law? I asked this on math stack exchange but I didn't receive a satisfactory answer?
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1$\begingroup$ Is this your MSE question? It appears to me that Berci has answered your question in the comments. If you still don't understand, try asking Berci to clarify. (Hint: there is an additional axiom that groups satisfy beyond the associative law. If you look up semigroups, as Berci suggested, you will find examples of semigroups which are not groups. Or maybe you can come up with examples yourself.) $\endgroup$– Tim CampionCommented Aug 31, 2019 at 23:01
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3$\begingroup$ @TimCampion That's not the question - the question is about axiomatizing the equational theory, not capturing the class of structures. $\endgroup$– Noah SchweberCommented Aug 31, 2019 at 23:50
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1$\begingroup$ Ah, I see. I guess I wasn't the only one to read it that way, so I'll leave my comment up. $\endgroup$– Tim CampionCommented Sep 1, 2019 at 4:34
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1$\begingroup$ I guess that the question can be restated (without model theory language, and universal algebra instead) as: among magmas, is the variety of semigroups generated by the class of groups? $\endgroup$– YCorCommented Sep 1, 2019 at 8:58
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$\begingroup$ @Ycor, although you could do that, it obscures the linguistic intent. That is why I added the answer I did, as Bjorn's answer (while correct) does not address the question explicitly. Gerhard "Will Have Nonexperts Reading This" Paseman, 2019.09.01. $\endgroup$– Gerhard PasemanCommented Sep 1, 2019 at 15:32
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2 Answers
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Yes. It suffices to show that any free semigroup embeds in a group.
For this I refer you to MO question 3235:
Let $F$ be a free semigroup (say, $2$-generated) which is embedded in a group $G$, and suppose that $G$ (as a group) is generated by $F$. The most simple such situation would be when $G$ is a free group, but there are lot of groups, besides free ones, which could occur in this situation (for example, $G$ could be solvable).
answered Aug 31, 2019 at 23:06
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The title should read " Is the equational theory of (the class of those semigroups which are groups) axiomatized by the associative law?", and the answer given by Bjorn Kjos-Hanssen makes even more sense (and seems more immediate): Take any free group on a set X of generators, and consider its semigroup reduct. The set X generates a subsemigroup S of this reduct, which is free (in semigroups) on X. (If it weren't, the words in X would be identified in the free group.) So any hope of finding a nontrivial semigroup identity in the equational theory of this class is toast.
Gerhard "Please Pass The Butter Knife" Paseman, 2019.08.31.
answered Sep 1, 2019 at 1:17
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$\begingroup$ A tricky thing here is that the class in question is not a closed class under the Galois connection of $Mod$ and $Id$. Fortunately that did not trip up Bjorn. Gerhard "Also Accepts The Accepted Answer" Paseman, 2019.08.31. $\endgroup$ Commented Sep 1, 2019 at 1:32
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2$\begingroup$ I'm not sure it would make a better title. An important implicit point is that it's among class with signature a single binary law, i.e., the question is among magmas. Your suggestion of title keeps this implicit, only at the cost of making the title more complicated. $\endgroup$– YCorCommented Sep 1, 2019 at 14:07
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$\begingroup$ @Ycor, the signature is important, and I agree that my suggestion does not make it fully explicit. (Among some colleagues in universal algebra, the convention is that semi groups have the same signature as magmas, which is different from the signature for groups, so it seems explicit to me.) How about "(the magmas which are groups)"? Gerhard "Can Go With The Flow" Paseman, 2019.09.01. $\endgroup$ Commented Sep 1, 2019 at 15:21