We have not paid particularly close attention to the category \(\Ab\) until now; in this article, we examine abelian groups.
Sums of Abelian Groups
First, the universal property of the weak direct product shown in §Restricted Sums, ⁋Theorem 2 applies particularly well when the group \(H\) is an abelian group.
Theorem 1 Let a family \((G_i)\) of abelian groups be given, and consider \(\prod^w G_i\) and the inclusion maps \(\iota_i\). Then for any abelian group \(H\) and group homomorphisms \(f_i:G_i\rightarrow H\), there exists a unique group homomorphism \(f:\prod^wG_i\rightarrow H\) such that \(f_i=f\circ\iota_i\).
Thus, at least among abelian groups, the weak direct product \(\prod^w G_i\) becomes a coproduct. We call this as follows.
Definition 2 Let a family \((G_i)\) of abelian groups, their weak direct product \(\prod^w G_i\), and the inclusion maps \(\iota_i\) be given. We call \(\prod^w G_i\) together with the \(\iota_i\) the direct sum of the \(G_i\), and denote it by \(\bigoplus G_i\).
With a slight abuse of notation, regarding \(\iota_i(G_i)\) and \(G_i\) as the same, any element of \(\bigoplus G_i\) can be written as
\[x=\sum_{i\in I} x_i,\qquad\text{$x_i\in G_i$, $x_i=0$ for all but finitely many $i$}\]In this situation, the subset of \(I\) consisting of indices where \(x_i\neq 0\) is called the support of \(x\) and denoted by \(\supp(x)\), and when this condition holds, the family \((x_i)\) is called finitely supported.
Abelianization
Now we make the following definition.
Definition 3 For any group \(G\) and two subgroups \(H_1,H_2\) of \(G\), define \([H_1,H_2]\) as the subgroup of \(G\) generated by the commutators
\[[h_1,h_2]=h_1^{-1}h_2^{-1}h_1h_2,\qquad h_1\in H_1,h_2\in H_2\]We are particularly interested in the case \(H_1=H_2=G\). If \(G\) is an abelian group, then for all \(x,y\in G\), we have \(x^{-1}y^{-1}xy=e\), so \([G,G]=\{e\}\). Thus, \([G,G]\) can be thought of as measuring how far \(G\) is from being an abelian group.
On the other hand, the following holds.
Proposition 4 For any group \(G\), the commutator subgroup \([G,G]\) is a normal subgroup of \(G\).
Proof
For any \(x,y\in G\) and \(g\in G\),
\[g(x^{-1}y^{-1}xy)g^{-1}=(gx^{-1}g^{-1})(gy^{-1}g^{-1})(gxg^{-1})(gyg^{-1})=(gxg^{-1})^{-1}(gyg^{-1})^{-1}(gxg^{-1})(gyg^{-1})\in [G,G]\]which is clear.
Thus, \(G/[G,G]\) is well-defined. This amounts to treating all elements of the form \(x^{-1}y^{-1}xy\) as \(e\), so \(G/[G,G]\) becomes an abelian group. According to our convention, the operation of an abelian group should be written as \(+\), but since \(G/[G,G]\) inherits its operation from \(G\), writing it this way might cause confusion. Therefore, we write the operation of \(G/[G,G]\) as multiplication rather than \(+\).
On the other hand, for any abelian group \(H\), if a group homomorphism \(f:G\rightarrow H\) is given, then for all \(x,y\in G\),
\[e=f(x)^{-1}f(y)^{-1}f(x)f(y)=f(x^{-1}y^{-1}xy)\]holds, so \([G,G]\leq\ker f\). By §Group Isomorphisms, ⁋Proposition 3, we obtain the following.
Proposition 5 Consider any group \(G\) and the quotient homomorphism \(p:G\rightarrow G/[G,G]\). Then for any abelian group \(H\) and group homomorphism \(f:G \rightarrow H\), there exists \(\bar{f}:G/[G,G]\rightarrow H\) satisfying \(f=\bar{f}\circ p\).
In particular, let any group homomorphism \(f:G\rightarrow H\) be given. Then the composition \(G\rightarrow H\rightarrow H/[H,H]\) gives a group homomorphism from the group \(G\) to the abelian group \(H/[H,H]\), and by Proposition 5, this induces a group homomorphism from \(G/[G,G]\) to \(H/[H,H]\).
Definition 6 For any group \(G\), the quotient group \(G/[G,G]\) is called the abelianization of \(G\) and denoted by \(G^\ab\).
The above argument shows that this defines a functor \(\ab:\Grp\rightarrow\Ab\). Moreover, the following holds.
Proposition 7 For the forgetful functor \(U:\Ab \rightarrow \Grp\) and the abelianization functor \(\ab:\Grp \rightarrow \Ab\), an adjunction \(\ab\dashv U\) exists.
The proof has already been completed above.
Free Abelian Groups
At the end of the previous article, we could interpret the free group \(F(X)\) as the free product
\[{\prod_{x\in X}}^\ast \mathbb{Z}\]Since we already know that \(\Ab\) has the coproduct \(\bigoplus\), by the same argument, we can obtain the left adjoint \(F_\Ab:\Set\rightarrow \Ab\) to the forgetful functor \(U:\Ab \rightarrow \Set\) through the formula
\[F_\Ab(X)=F_\Ab\left(\coprod_{x\in X} \{x\}\right)\cong \coprod_{x\in X} F_\Ab(\ast)=\bigoplus_{x\in X} \mathbb{Z}\]The \(F_\Ab(X)\) thus obtained is defined as the free abelian group. Thus the following proposition holds.
Proposition 8 A left adjoint \(F_\Ab:\Set \rightarrow\Ab\) to the forgetful functor \(U:\Ab \rightarrow \Set\) exists.
The Abelian Group \(\Hom_\Ab(G,H)\)
For any abelian groups \(G,H\), the set \(\Hom_\Ab(G,H)\) is the set of group homomorphisms from \(G\) to \(H\). This set has an interesting property: \(\Hom_\Ab(G,H)\) is already an abelian group. This is a result that does not hold in \(\Grp\).
Proposition 9 For any abelian groups \(G,H\), the set \(\Hom_\Ab(G,H)\) is an abelian group.
Proof
For any \(f,g:G \rightarrow H\), define \(f+g\) by
\[(f+g)(x)=f(x)+g(x)\qquad\text{for all $x\in G$}\]\(\Hom_\Ab(-,-)\) was originally defined as a bifunctor from \(\Ab^\op\times \Ab\) to \(\Set\), but by this proposition, we can actually regard it as a bifunctor to \(\Ab\). That is, we can think of \(\Hom_\Ab(-,-)\) as something similar to an internal \(\Hom\). However, in terms of the language we have so far, this is impossible.
Example 10 \(\Ab\) is a cartesian monoidal category with respect to \(\times\). However, \(\Hom_\Ab(-,-)\) cannot be regarded as an internal \(\Hom\) with respect to this structure. That is,
\[\Hom_\Ab(G\times H, A)\cong \Hom_\Ab(G,\Hom_\Ab(H,A))\]does not generally hold. For example, if \(G=\mathbb{Z}\), then the above equation becomes
\[\Hom_\Ab(\mathbb{Z}\times H,A)\cong \Hom_\Ab(\mathbb{Z},\Hom_\Ab(H,A))\cong \Hom_\Ab(H,A)\tag{1}\]which will be false in most cases. (For instance, try \(H=\{e\}\).)
Thus, to regard \(\Hom_\Ab(-,-)\) as an internal \(\Hom\), we need to give \(\Ab\) a new symmetric monoidal category structure. Also, taking a hint from equation (1) above, we can guess that \(\mathbb{Z}\) should act like the unit of this monoidal product.
Tensor Products
The fundamental reason why the equation in Example 10 cannot hold is quite simple. In \(\Set\), the above isomorphism held because for any function \(f:A\times B \rightarrow C\), if we fix one element of \(A\) or \(B\), what remains is a function from \(B\) or \(A\) to \(C\).
On the other hand, for a group homomorphism \(f:G\times H \rightarrow A\), the only \(f\) for which fixing the first or second component gives a group homomorphism is the zero map. For any \(x\in G\), if \(f(x, -)\) is a group homomorphism, then \(f(x,0)=0\) must hold, and similarly for any \(y\in H\), \(f(0,y)=0\) must hold, so substituting this into the condition that \(f\) is a group homomorphism,
\[f(x+0,0+y)=f(x,0)+f(0,y)\]we see that \(f(x,y)=0\) must hold for all \((x,y)\in G\times H\).
Looking at the above argument, it seems quite natural to require that the function obtained by fixing one component of the source of \(f\) be a group homomorphism.
Definition 11 For two abelian groups \(G,H\), a function \(f:G\times H \rightarrow A\) is called bilinear if the two equations
\[f(x,y_1+y_2)=f(x,y_1)+f(x,y_2),\qquad f(x_1+x_2,y)=f(x_1,y)+f(x_2,y)\]always hold.
Now, for fixed \(G,H\in\obj(\Ab)\), define the set \(\Bilin(G,H;A)\) by
\[\Bilin(G,H;A)=\{\text{bilinear maps from $G\times H$ to $A$}\}\]By the above argument, we can verify that if we replace the left side of the first equation in Example 10 with \(\Bilin(G,H;A)\), we obtain an isomorphism
\[\Bilin(G,H;A)\cong \Hom_\Ab(G,\Hom_\Ab(H,A))\]Moreover, we can verify that \(\Bilin(G,H;-)\) becomes a representable functor from \(\Ab\) to \(\Set\).
Theorem 12 \(\Bilin(G,H;-)\) is representable.
Proof
Define the subgroup \(S\) of the free abelian group \(F_\Ab(G\times H)\) by
\[S=\left\langle (x, y_1+y_2)-(x,y_1)-(x,y_2), (x_1+x_2,y)-(x_1,y)-(x_2,y)\mathop{\big\vert}x,x_1,x_2\in G, y,y_1,y_2\in H\right\rangle\]Then by the universal property of free abelian groups, whenever a function \(f:G\times H \rightarrow A\) is given, there exists a group homomorphism \(\hat{f}:F_\Ab(G\times H)\rightarrow A\), and if \(f\) is bilinear, then the kernel of this \(\hat{f}\) contains \(S\), so \(\hat{f}\) defines a group homomorphism from \(F_\Ab(G\times H)/S\) to \(A\).
The naturality of the isomorphism \(\Bilin(G,H;A)\cong\Hom_\Ab(F_\Ab(G\times H)/S,A)\) requires additional verification, but this is a straightforward computation, so we omit it.
Definition 13 The representation in Theorem 12 is called the tensor product of \(G\) and \(H\), and is denoted by \(A\otimes B\).
We can see that elements of \(A\otimes B\) can be expressed as finite sums of elements of the form \(a\otimes b\). Then we can verify that \(\otimes\) is a monoidal product with \(\mathbb{Z}\) as the tensor unit.
Theorem 14 \((\Ab,\otimes, \mathbb{Z})\) is a symmetric monoidal category.
Proof
The associator \(\alpha\) and symmetor \(\sigma\) are obtained from the universal property of \(\otimes\). That \(\mathbb{Z}\) is the tensor unit follows from the fact that the isomorphism
\[\Hom_\Ab(\mathbb{Z}\otimes G, H)\cong\Bilin(\mathbb{Z},G;H)\cong\Hom_\Ab(\mathbb{Z},\Hom_\Ab(G,H))\cong\Hom_\Ab(G,H)\]is natural.
In particular, for any \(f:A \rightarrow A'\) and \(g:B \rightarrow B'\), since \(\otimes\) is a bifunctor, there exists a morphism \(f\otimes g:A\otimes B \rightarrow A'\otimes B'\). This is the group homomorphism determined by sending elements of the form \(a\otimes b\) to \(f(a)\otimes g(b)\). Sometimes we need to consider the \(n\)-fold tensor product of an abelian group \(A\), in which case we write
\[A^{\otimes n}=\underbrace{A\otimes\cdots\otimes A}_\text{$n$ times}\]Then by the associativity of \(\otimes\),
\[A^{\otimes m}\otimes A^{\otimes n}\cong A^{\otimes(m+n)}\]holds. From this perspective, it is conventional to define \(A^{\otimes 0}\) as \(\mathbb{Z}\).
Thus, regarding \((\Ab,\otimes, \mathbb{Z})\) as a symmetric monoidal category, we have already verified the following.
Theorem 15 (\(\otimes\dashv\Hom\)) There exists an adjunction
\[\Hom_\Ab(G\otimes H, A)\cong\Hom_\Ab(G,\Hom_\Ab(H, A))\cong\Hom_\Ab(H,\Hom_\Ab(G, A))\]Thus, \(\Hom_\Ab(-,-)\) can be regarded as the internal \(\Hom\) of \((\Ab,\otimes,\mathbb{Z})\).
Graded Abelian Groups
For a family \((G_i)\) of abelian groups, the direct sum \(\bigoplus G_i\) is well-defined. The following definition is particularly useful in other algebraic structures.
Definition 16 For a commutative monoid \(I\), consider a family \((G_i)_{i\in I}\) of abelian groups indexed by \(I\). This is called a graded abelian group.
Currently, for a commutative monoid \(I\), there is no difference between regarding \(I\) as a set and taking the direct sum versus the graded abelian group defined above, so this definition is merely giving a new name to an existing concept. The reason for defining this is to later specify the relationship between a new operation defined on abelian groups and the addition of \(I\).
References
[nLab] Tensor product of abelian groups. Link
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