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 Apr 26 awarded Popular Question Jul 13 comment Why Lawvere theories have finite products? and more Andrej, I quote from nLab article on Lawvere theories: "Remark. For T a Lawvere theory, we are to think of the hom-set T(n,1) as the set of n-ary operations defined by the theory. For instance for T the theory of abelian groups, we have T(2,1)={+,−} and T(0,1)={0}." Why is T(2,1) just {+,-},? Wouldn't it include other 'unnamed' operations as well? Jul 13 awarded Scholar Jul 13 accepted Why Lawvere theories have finite products? and more Jul 13 comment Why Lawvere theories have finite products? and more Thanks, after reading Andrej's post, now (3) makes sense. Jul 13 comment Why Lawvere theories have finite products? and more Thank you, very clear. I was misunderstanding it all, not thinking in syntax at all, just thinking about the models. Jul 12 awarded Student Jul 12 comment Why Lawvere theories have finite products? and more But isn't the monoid described just by the functor that tells what's the set ($F T$), the operation ($F *$) and the neuter ($F e$)? Jul 12 comment Why Lawvere theories have finite products? and more I think this is the point where I get confused. Why would I need a product of four things? If model for $T$ would be a functor $F$ from $T$ to $Set$, then the product would be $F * : F T^2 \to F T$. For example if $F T = \mathbb{Z}_n$, then $F * : \mathbb{Z}_n^2 \to \mathbb{Z}_n$ would be the product (+) of my monoid (Z_n, +, 0) in Set, wouldn't it? Jul 12 asked Why Lawvere theories have finite products? and more