On page 66 of Lang’s Algebra he states the following:

Let $G$ = ∏ $G_i$ be a direct product of groups. Then each $G_j$ admits an injective homomorphism into the product, on the $j$-th component, namely the map $λ_j$ : $G_j$ → ∏ $G_i$ such that for $x$ in $G_j$, the $i$-th component of $λ_j(x)$ is the is the unit element of $G_i$ if $i ≠ j$, and is equal to $x$ itself if $i = j$. This embedding will be called the canonical one. But we still don’t have a coproduct of the family, because the factors commute with each other.

I understand each point until the final sentence. What is meant by “the factors commute with each other” and why does this imply that the direct product cannot be a coproduct of a family?

  • $\begingroup$ What is Lang's definition of coproduct? $\endgroup$
    – coffeemath
    Aug 14, 2021 at 1:30
  • $\begingroup$ I would imagine it means that for $i \neq j$ and $g \in G_i$, $h \in G_j$ then $\lambda_i(g)\ \cdot \lambda_j(h) = \lambda_j(h) \cdot \lambda_i(g)$. (This would not happen if the $\lambda$-maps where category-theoretic co-projection maps.) $\endgroup$ Aug 14, 2021 at 1:31
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    $\begingroup$ If this were a coproduct, then any map from the $G_i$ to a group $H$ would necessarily yield a map from the product to $H$ commuting with the embeddings $\lambda_j$. But it is easy to concoct maps in which the image of $G_i$ in $H$ does not commute with the image of $G_j$ in $H$, for $i\neq j$, and then you cannot possibly obtain this as a map from the product, because in the product $\lambda_i(g)$ always commutes with $\lambda_j(g')$ for any $g\in G_i$, $g'\in G_j$, if $i\neq j$. $\endgroup$ Aug 14, 2021 at 1:34

1 Answer 1


A coproduct of the family $\{G_i\}_{i\in I}$ is a group $C$ together with maps (embeddings in fact, though that can be deduced from the universal property) $\lambda_i\colon G_i\to C$ with the following universal property:

Given any group $H$ and a family of morphisms $f_i\colon G_i\to H$, there exists a unique morphism $F\colon C\to H$ such that $f_i=F\circ\lambda_i$ for each $i\in I$.

Lang is noting that you can embedd each $G_i$ into the product using the functions $\lambda_i$. He also notes that if $i\neq j$, then for every $x\in G_i$, $y\in G_j$, you have that $\lambda_i(x)$ and $\lambda_j(y)$ commute. That means that the image of $\lambda_i(x)$ and the image of $\lambda_j(y)$ will commute under any homomorphism $F\colon\prod G_i\to H$.

But this is a problem: if you can find a group $H$ and morphisms $f_i\colon G_i\to H$ with the property that there are $i$ and $j$, $i\neq j$, such that not every element of $f_i(G_i)$ commutes with every element of $f_j(G_j)$, then there is no way to find a morphism $F\colon \prod G_i\to H$ that will satisfy $F\circ\lambda_i=f_i$. And such groups and maps abound, which means that the fact that images of the $G_i$ inside the product commute with one another doom the product from being able to be the coproduct in the category of all groups.

(The product does have the relevant universal property relative to any family of morphisms $f_i\colon G_i\to H$ that have the additional property that every element of $f_i(G_i)$ commutes with every element of $f_j(G_j)$ whenever $i\neq j$; and of course you can see why, at least in the finite index case, the product will work as a coproduct for abelian groups, that is, if every group in sight is abelian.)

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    $\begingroup$ A simple example is with all the $G_i=G$ and $H=G,$ where $G$ is non-abelian and all $f_i$ are identities. $\endgroup$ Aug 14, 2021 at 2:26
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    $\begingroup$ @ThomasAndrews: Yes, but you want more than that. You'd want to show that for any family of nontrivial groups there is always a choice of $H$ and $f_i$ that will prevent the product from being the coproduct. (The only time it "works" is if at most one of the $G_i$ is nontrivial). That is, show not only that it may fail to be the coproduct in some cases, but that it is never the coproduct (except in a very special case). $\endgroup$ Aug 14, 2021 at 2:53

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