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I have no idea how to do this because in my notes there are no examples and HSP doesn't show up until the next section. How do I prove something is a variety without HSP? I think the only thing that I have is the definiton of a variety:

A variety $V$ of algebras is defined by a language $L$ and a set $I$ of identities over $L$. An algebra $A$ over $L$ belongs to a variety $V$ if all identities $I$ hold in $A$. The full details of the question are let $L$ =('\') be a language consisting of a single binary operator '\'. For any group $G$ consider the algebra over $L$ by defining '\' to be the operation $x$\ $y = x^{-1}y$

Also, if anybody knows some good resources that deal with varieties of algebras it would be greatly appreciated. All I have are my class notes and they are pretty unclear. I can't find any youtube lectures on the subject, notes that maybe have a few more examples, or even examples on stack exchange so I am really struggling

Vinny Chase
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  • (1) You must have some definition of variety at your disposal; otherwise, there's nothing you can do about it. (2) You can't define $x\backslash y$ that way without having defined $x^{-1}y$, which seems to involve two operations: product and inverse; is $G$ a group? (3) A good resource about Universal Algebra which is freely available is A Course in Universal Algebra by Burris and Sankappanavar. – amrsa May 18 '18 at 07:39
  • @amrsa Thanks. The full definition of variety I havev ia variety $V$ of algebras is defined by a language $L$ and a set $I$ of identities over $L$. An algebra $A$ over $L$ belongs to a variety $V$ if all identities $I$ hold in $A$. The full details of the question are let $L$ =('') be a language consisting of a single binary operator ''. For any group $G$ consider the algebra over $L$ by defining '' to be the operation I've mentioned above – Vinny Chase May 18 '18 at 07:43
  • @VinnyChase You should add the details from your comment to the question so potential answerers can find it easier. Also, you never officially defined $G^\backslash$. Also also, formatting your question into separate paragraphs will make it easier to read. Downvote from me until you fix some of these. – Eran May 18 '18 at 19:50
  • @Eran fixed. Please let me know if you have any input – Vinny Chase May 18 '18 at 20:59
  • One big gotcha with this problem is that every group is nonempty, whereas any variety with only a single binary operation in its signature is automatically going to contain the empty set as an algebra. – Daniel Schepler May 18 '18 at 21:47
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    @DanielSchepler I suppose that can happen with certain definitions. For the definition of algebra in the book in the link I gave above, an algebra is always a non-empty set together with a set of finitary operations. I'm not sure about the definition the OP is using... – amrsa May 18 '18 at 21:50
  • These questions are related: (https://math.stackexchange.com/questions/399638/what-is-a-simple-axiomatisation-of-groups-using-division) and (https://math.stackexchange.com/questions/1225445/abelian-groups-axioms-with-minus-in-place-of-plus?rq=1) – Eran May 19 '18 at 16:39

1 Answers1

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If you don't want to use HSP, then the best bet would be to go to the definition, and explicitly find a set of identities involving only $\backslash$ that together imply the algebra comes from a group. Now, observe that given $\backslash$ coming from a group, you can recover the group structure: $e = x \backslash x$; $x^{-1} = x \backslash e$; and then $x \cdot y = x^{-1} \backslash y$. The only gotcha is that the first requires the algebra to be nonempty to get started - so it will depend on your definition whether or not the empty set is considered to be an algebra over a signature with no nullary operations. If it is, then you are pretty much doomed, since every group is nonempty; if the empty set is never considered to be an algebra under your definition, then you can use this to bootstrap $e$.

Now, you could go through and expand all the group identities in terms of $\backslash$; for example, associativity of multiplication would expand to:

$$ [ [(x \backslash (x \backslash x)) \backslash y] \backslash (x \backslash x)] \backslash z = [x \backslash (x \backslash x)] \backslash [[y \backslash (x \backslash x)] \backslash z].$$

You will probably also need a relation specifying that $e$ is actually a constant:

$$ x \backslash x = y \backslash y. $$

Then automatically, any algebra of the form $G^{\backslash}$ for $G$ a group will satisfy these relations. Conversely, suppose you start with any algebra satisfying these relations, and define group operations by $e := x_0 \backslash x_0$ (choosing $x_0$ in the underlying set by using the assumption that it must be nonempty), $x^{-1} := x \backslash e$, $x \cdot y := x^{-1} \backslash y$. Then this will give a group, pretty much by construction. Furthermore, using this definition you will have $x^{-1} \cdot y = [(x^{-1})^{-1}] \backslash y = x \backslash y$, so applying the transformation given in the problem to this group will recover the original given $\backslash$ operation.

It actually turns out that in this case, you can come up with a similar set of relations by hand which is equivalent, but "prettier": $x \backslash x = y \backslash y$; $(x \backslash x) \backslash x = x$; and $(x \backslash y) \backslash (x \backslash z) = y \backslash z$.

Daniel Schepler
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  • In the question, $x\backslash y$ was $x^{-1}y$; you seem to interpret it as $xy^{-1}$. Apart from that notational issue, this answer is, as far as I can see, the best approach. – Andreas Blass May 19 '18 at 00:07
  • @DanielSchepler How did you come up with that identity? I worked through it and it works, but I'm trying to come up with the other identities but I can't seem to – Vinny Chase May 19 '18 at 17:31
  • Which identity were you asking about? If it's the associativity of multiplication one: for example, $x \cdot y = x^{-1} \backslash y = (x \backslash e) \backslash y = [x \backslash (x \backslash x)] \backslash y$; and then $(x \cdot y) \cdot z = (x \cdot y)^{-1} \backslash z = [(x \cdot y) \backslash e] \backslash z = [(x \cdot y) \backslash (x \backslash x)] \backslash z$, then substitute $x \cdot y$. And similarly for $x \cdot (y \cdot z)$. Which looks like I need to make another correction... – Daniel Schepler May 20 '18 at 00:43
  • For a simpler one, let's take the left identity relation $e \cdot x = x$. Then $e \cdot x$ expands to $e^{-1} \backslash x = (e \backslash e) \backslash x = [(x \backslash x) \backslash (x \backslash x)] \backslash x$, so the identity would expand to $[(x \backslash x) \backslash (x \backslash x)] \backslash x = x$. – Daniel Schepler May 20 '18 at 00:52