This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Previously, we verified in §Direct Products of Groups that arbitrary products exist in \(\Grp\), and in §Group Homomorphisms, ⁋Proposition 2 that every morphism in \(\Grp\) has an equalizer. Therefore, by the argument following [Category Theory] §Limits, ⁋Example 7, \(\Grp\) is a complete category.
On the other hand, every morphism in \(\Grp\) has a coequalizer. (§Isomorphism Theorems, ⁋Proposition 8) Thus, if \(\Grp\) has arbitrary coproducts, then \(\Grp\) would be a cocomplete category and hence a bicomplete category.
However, as in §Direct Products of Groups, ⁋Lemma 1, it seems difficult to find an obvious way to put a group structure on the coproduct \(\coprod G_i\) in \(\Set\). ([Set Theory], §Sum of Sets, ⁋Proposition 5)
In this post, we first show that the category of abelian groups has coproducts. In the next post, we will show, through a different approach from this one, that for arbitrary groups there also exists a group satisfying the universal property of coproducts.
Restricted sum
Let a family of groups \((G_i)\) and their product be given. Then each \(G_i\) can be viewed as a subgroup of \(\prod G_i\) via \(\iota_i\). Naturally, one may consider whether the following formula
\[\prod_{i\in I} G_i=\left\langle\bigcup \iota_i(G_i)\right\rangle\]holds. This formula almost never holds if \(I\) is an infinite set. As the simplest example, let \(I=\mathbb{N}\) and \(G_i=\mathbb{Z}/2\mathbb{Z}=\{\bar{0}, \bar{1}\}\). Then, for instance, the left-hand side contains the element
\[(\bar{1},\bar{1},\cdots)\]but the right-hand side contains only elements obtained by finite operations on the \(\iota_i(\bar{1})\), so it cannot contain the above element.
Definition 1 Let a family of groups \((G_i)\) be given, and fix subgroups \(H_i\) of the \(G_i\). Then the subgroup consisting of those \(x\) such that \(\pr_ix\in H_i\) for all but finitely many \(i\) is called the restricted sum of the \(G_i\) with respect to the \(H_i\), and is denoted by \(\prod^H G_i\).
In the special case where \(H_i=\{e\}\) for all \(i\), it is called the weak direct product of the \(G_i\), and is denoted simply by
\[{\prod_{i\in I}}^w G_i\]The notation \(\prod^H\) is not particularly good, but fortunately we are only interested in the weak direct product, so we will not use this notation again.
By definition,
\[\left\langle\bigcup \iota_i(G_i)\right\rangle={\prod_{i\in I}}^w G_i\]holds. Also, if \(I\) is a finite set, then the weak direct product coincides with the ordinary direct product.
Then \(\prod^wG_i\) has the following universal property.
Theorem 2 Let a family of groups \((G_i)\) and their weak direct product \(\prod^w G_i\) be given. For another group \(H\), if group homomorphisms \(f_i:G_i\rightarrow H\) satisfy the following condition:
For any \(i\neq j\), if \(x\in G_i\) and \(y\in G_j\), then \(f_i(x)f_j(y)=f_j(y)f_i(x)\)
then there exists a unique group homomorphism \(f:\prod^w G_i\rightarrow H\) such that \(f_i=f\circ\iota_i\) holds for every \(i\).
Proof
Let us first show uniqueness. If \(f, f'\) satisfy the above equation, they must take the same values on \(\bigcup\iota_i(G_i)\), and hence also on \(\prod^w G_i\); therefore \(f=f'\).
Now we must show the existence of \(f\). For any \(x\in \prod^w G_i\), define \(f(x)\) by the formula
\[f(x)=\prod_{i\in I} f_i(\pr_ix)\]Here \(\prod\) denotes the product of ordinary elements. Since \(x\) is an element of \(\prod^w G_i\), all but finitely many of the \(f_i(\pr_ix)\) on the right-hand side are the identity, so this product is well-defined.
That the equation \(f_i=f\circ\iota_i\) holds is obvious, and that \(f\) is a group homomorphism follows because for any \(x,y\in\prod^wG_i\),
\[f(xy)=\prod_{i\in I}f_i(\pr_i(xy))=\prod_{i\in I}f_i(\pr_ix)f_i(\pr_iy)\]holds, so picking only the finitely many indices for which \(\pr_i(xy)\) is not \(e_i\) and calling these indices \(1,\ldots, n\), we obtain
\[f_1(\pr_1x)f_1(\pr_1y)f_2(\pr_2x)f_2(\pr_2y)\cdots f_n(\pr_nx)f_n(\pr_ny)\]and since \(f_i(\pr_ix)\) and \(f_j(\pr_jy)\) always commute when \(i\neq j\), this expression can be rewritten as
\[f_1(\pr_1x)f_2(\pr_2x)\cdots f_n(\pr_nx)f_1(\pr_1y)f_2(\pr_2y)\cdots f_n(\pr_ny)\]Therefore \(f(xy)=f(x)f(y)\) and \(f\) is a group homomorphism. That \(f_i=f\circ\iota_i\) is obvious.
The condition imposed on the \(f_i\),
For any \(i\neq j\), if \(x\in G_i\) and \(y\in G_j\), then \(f_i(x)f_j(y)=f_j(y)f_i(x)\)
is a condition that necessarily arises, because these are exactly the conditions satisfied by the \(\iota_i\). Because of this, Theorem 6 becomes the answer to our question only for abelian groups.
Using the universal property of the weak direct product, we can prove several properties analogous to those for the direct product. For instance, the following holds.
Proposition 3 Let the \(G_i\) be groups and the \(H_i\) be normal subgroups of the \(G_i\). Then the \(\prod^w H_i\) are also normal subgroups of \(\prod^w G_i\), and the quotient group is \(\prod^w (G_i/H_i)\).
Internal weak product
Let \(G\) be a group and let \((H_i)\) be a family of subgroups of \(G\). If the elements of \(H_i\) commute with the elements of \(H_j\) whenever \(i\neq j\), then the inclusion homomorphisms \(\iota_i:H_i\rightarrow G\) induce a homomorphism \(\iota\) from \(\prod^w H_i\) to \(G\).
Furthermore, we define the following.
Definition 4 In the above situation, if \(\iota\) is an isomorphism, then \(G\) is called the internal weak direct product of the \(H_i\).
Thinking about the form of the homomorphism \(f\) constructed in Theorem 2, one can verify that \(G\) being the internal weak direct product of the \(H_i\) is equivalent to the following condition:
Every \(x\in G\) can be expressed as a product \(\prod y_i\) of finitely supported families \((y_i)_{i\in I}\) with \(y_i\in H_i\).
If the subgroups \(H_i\) are all normal subgroups of \(G\), then \(G\) becomes the internal weak direct product of the \(H_i\) if the following additional conditions are satisfied.
Proposition 5 If the normal subgroups \((H_i)\) of a group \(G\) satisfy the following two conditions:
- \(G=\bigl\langle\bigcup_{i\in I} H_i\bigr\rangle\),
- \[H_k\cap \bigl\langle\bigcup_{i\neq k} H_i\bigr\rangle=\{e\}\]
then \(G\) is the internal weak direct product of the \(H_i\).
Proof
First, condition 2 shows in particular that \(H_i\cap H_j=\{e\}\) holds for every pair \(i\neq j\). Now taking arbitrary \(x_i\in H_i, x_j\in H_j\),
\[x_ix_jx_i^{-1}x_j^{-1}=x_i\bigl(x_jx_i^{-1}x_j^{-1}\bigr)=\bigl(x_ix_jx_i^{-1}\bigr)x_j^{-1}\in H_i\cap H_j=\{e\}\]shows that the elements of \(H_i\) and \(H_j\) commute. Therefore the inclusion homomorphisms \(\iota_i\) induce \(\iota\) well, as in Theorem 2.
To show that \(G\) is the internal weak direct product of the \(H_i\), we must show that this induced \(\iota\) is an isomorphism. First, by condition 1, any \(a\in G\) is obtained by finite operations on the \(\bigcup H_i\). Also, since the \(H_i\) commute with each other, we can write \(a\) as
\[a=\prod_{i\in I} h_i=\prod_{i\in I}\iota_i(h_i),\qquad\text{$\supp(h_i)$ finite and $h_i\in H_i$}\]Letting \(h=\prod_{i\in I} \iota_i(h_i)\in\prod^w H_i\),
\[a=\prod_{i\in I}\iota_i(h_i)=\iota\left(\prod_{i\in I}\iota_i(h_i)\right)=\iota(h)\]so \(\iota\) is surjective.
Now suppose \(\iota(a)=e\). Then we can write \(a=(a_i)_{i\in I}\) for a finitely supported family \((a_i)\) with each term in \(H_i\). From the equation
\[\iota(a)=\prod_{i\in I}\iota_i(a_i)=\prod_{i\in I} a_i=e\]if \(\supp(a_i)\) has at least one element and \(i\in\supp(a_i)\), then
\[a_i^{-1}=\prod_{j\in I\setminus\{i\}}a_j\in H_i\cap \left\langle\bigcup_{j\neq i} H_i\right\rangle=\{e\}\]which contradicts the assumption that \(i\in\supp(a_i)\). Therefore \(\supp(a_i)\) is empty and \(a\) is the identity.
References
[Hun] Thomas W. Hungerford, Algebra, Graduate texts in mathematics, Springer, 2003.
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