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One class of algebraic structures that are typically studied are those given by a set $X$ and a set of $n$-ary operations defined on $X$ for each $n\in \mathbb{N}$. Perhaps most studied are those binary operations that make up monoids, groups, rings, etc; to specify these structures further, there are identities put on the varying operations defined on the underlying set.

For example, for a semi-group $(X,\ast)$, there is binary operation $\ast: X\times X\rightarrow X$ such that the identity $a \ast (b\ast c)=(a\ast b) \ast c$ (where it is assumed that there are universal quantifiers for each of the free variables $a,b,c$) holds, i.e. that $\ast$ is associative. Another example includes commutativite magmas, which has the identity $a \ast b=b\ast a$.

I was wondering if it were possible (in certain circumstances) to reduce a set of identities to a smaller set. In particular:

  • Is there an identity/equational law such that the algebras satisfying it are exactly the commutative semi-groups? Thus, the equational law captures both associativity and commutativity.
  • Is there an identity/equational law such that the algebras satisfying it are exactly the monoids? Thus, the equational law captures both the existence of an identity and associativity.
  • Is there an identity/equational law such that the algebras satisfying it are exactly the groups? Thus, the equational law captures the existence of an identity, inverses, and associativity.
  • Is there an identity/equational law such that the algebras satisfying it are exactly the commutative semi-rings? Thus, the equational law captures the existence of an additive and multiplicative identity, associativity of both operations, commutativity of both operations, and distributivity.

I'm inclined to think not, because performing several different operations in tandem can be confounding, but can't think of a way to show this precisely aside from counterexamples for each attempt to create such an equational law.

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  • $\begingroup$ I'm wondering if it can even be done for monoids - semigroups with an identity. It seems really hard without the ability to cancel to reduce any equation to something much shorter to deduce that $1\star x = x$. $\endgroup$ – Thomas Andrews Aug 4 '14 at 19:20
  • $\begingroup$ Essentially, if any of your axioms are of the form $x=P(x,\dots)$ then your single axiom also has to have one term on one side, or you are never able to reduce any equation to one term. $\endgroup$ – Thomas Andrews Aug 4 '14 at 19:57
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    $\begingroup$ Once we have unit, the axiom $(ab)(cd)=(ac)(bd)$ is equivalent to commutativity and associativity. $\endgroup$ – Berci Aug 4 '14 at 20:12
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If your single axiom is of the form $P(x,\dots)=Q(x,\dots)$ then you can never reduce an expression in your algebra to fewer terms than the minimum number of terms in $P$ and $Q$.

So if your original axioms include $x=1\star x$ then, if you can get a single axiom, it must be of the form $x=Q(x,\dots)$.

Now, in a free monoid on more than $2$ generators, the only relation of the form:

$$x=Q(x,\dots)$$

consists only of one $x$ and otherwise all $1$ terms.

But with such a $Q$, if you have any non-associative binary operation with a 2-sided identity, you'd have $x=Q(x)$ true for all $x$, but associativity would not true.

So there is no single identity that defines monoids.

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  • $\begingroup$ Very interesting; your answer makes it very clear that the identity ruins the ability to create a single axiom. Berci's comment also shows that if an identity exists, then we can combine both associativity and commutativity together. Is there any way to remove the dependence on the identity in this respect? $\endgroup$ – Hayden Aug 4 '14 at 22:43
  • $\begingroup$ There isn't a way to write the identity in a general monoid other that $1$ - there is no expression that is always equal to $1$. $\endgroup$ – Thomas Andrews Aug 4 '14 at 22:51
  • $\begingroup$ I think I might has mis-worded my first comment. I was asking more along the lines of whether there was a single axiom for a general commutative semi-group, rather than the two axioms of $1\cdot x=x$ and $(xy)(zw)=(xz)(yw)$ that Berci gave when it was moreover a commutative monoid, and not about whether there was another way to represent the axiom for the identity. Nevertheless, I'll make sure to keep in mind that there is no other way to do that. $\endgroup$ – Hayden Aug 4 '14 at 23:01
  • $\begingroup$ I guess the question I asked above is actually null and void, and your first sentence gets that point across; if we have $P(x,\ldots)=Q(x,\ldots)$ that gives both associativity and commutativity, then because it gives commutativity both sides can have at most 2 terms, but because associativity has three terms on each side, it could not be that it would be derivable. Thanks for the wonderful insight! $\endgroup$ – Hayden Aug 4 '14 at 23:36

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