Prove the class of all algebras $G^ \backslash$ is a variety 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
 A: 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 $(x \backslash y) \backslash (x \backslash z) = y \backslash z$.
