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A cocomplete abelian category is finitary in the directed-union sense if every cocone under a diagram indexed by a nonempty directed poset, with all its legs monomorphisms, induces a monomorphism from the colimit to the cocone vertex. The category of abelian groups has this property: two colimit representatives mapped to the same vertex become equal at a common upper index because that cocone leg is injective. This use of finitary concerns directed unions, rather than the preservation-of-filtered-colimits meaning used for finitary monads.

A Lawvere theory is a small category with finite products generated by one object $[1]$, so every object is a finite power $[n]$. A model in a finite-product category $\mathcal{C}$ is a finite-product-preserving functor from the theory to $\mathcal{C}$. For a finitary monad $T$ on sets, take the opposite of the full subcategory of its Eilenberg-Moore category on the finitely generated free algebras; the resulting finite-power category is its Lawvere theory, and its set-valued models are $T$-algebras.

If ${\mathbf{Set}}^T$ is additive, its full subcategory of finite free algebras is additive, and passing to the opposite preserves biproducts and abelian-group enrichment. Thus the associated theory is an additive category. A product-preserving model sends the abelian-group object $[1]$ and its addition, zero, and inverse maps to an internal abelian group in $\mathcal{C}$.

In an additive theory the product $[n]=[1{]}^n$ is also a biproduct. If ${\iota }_i:[1]\to [n]$ and ${\pi }_i:[n]\to [1]$ are its injections and projections, every $n$-ary operation decomposes uniquely aswhich in a model reads $\alpha (x_1,\dots ,x_n)={\sum }_i{\alpha }_i(x_i)$. Put $R=\mathrm{End}([1])$, with addition from the enrichment and multiplication from composition. The unary operations give every model an $R$-module structure, and the displayed decomposition says that all operations are exactly $R$-linear combinations. Conversely every $R$-module interprets them this way. Hence ${\mathbf{Set}}^T\simeq R\text{-}\mathbf{Mod}$, up to the conventional choice of left versus right modules.