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.
We have not paid much attention to the category \(\Ab\) so far; in this post we study 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 abelian.
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 it 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. Then the pair of \(\prod^w G_i\) and the \(\iota_i\) is called the direct sum of the \(G_i\), and is denoted by \(\bigoplus G_i\).
With a slight abuse of notation, if we identify \(\iota_i(G_i)\) with \(G_i\), then 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 those \(i\) with \(x_i\neq 0\) is called the support of \(x\) and denoted by \(\supp(x)\), and when the above condition holds we say that the family \((x_i)\) is finitely supported.
Abelianization
Now we define the following.
Definition 3 For any group \(G\) and any two subgroups \(H_1,H_2\) of \(G\), define \([H_1,H_2]\) to be the subgroup of \(G\) generated by the following commutators
\[[h_1,h_2]=h_1^{-1}h_2^{-1}h_1h_2,\qquad h_1\in H_1,h_2\in H_2\]In particular, what interests us in this post is the case \(H_1=H_2=G\). If \(G\) were an abelian group, then \(x^{-1}y^{-1}xy=e\) for all \(x,y\in G\), so \([G,G]=\{e\}\). Hence one may think of \([G,G]\) as measuring how far \(G\) is from being abelian.
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 obvious.
Therefore, \(G/[G,G]\) is well-defined. This amounts to declaring every element of the form \(x^{-1}y^{-1}xy\) to be \(e\), so \(G/[G,G]\) becomes an abelian group. According to our convention, the operation of an abelian group should be written as \(+\), but since \(G/[G,G]\) simultaneously inherits its operation from \(G\), doing so might cause confusion. Hence we agree to 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 any \(x,y\in G\) the following equation
\[e=f(x)^{-1}f(y)^{-1}f(x)f(y)=f(x^{-1}y^{-1}xy)\]holds, so \([G,G]\leq\ker f\). Now by §Isomorphism Theorems, ⁋Proposition 3 we obtain the following.
Proposition 5 Let \(G\) be any group and let \(p:G\rightarrow G/[G,G]\) be the quotient homomorphism. Then for any abelian group \(H\) and any 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 by composing \(G\rightarrow H\rightarrow H/[H,H]\) we obtain a group homomorphism from \(G\) 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 is denoted by \(G^\ab\).
Then the preceding 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 \(\ab:\Grp \rightarrow \Ab\), there exists an adjunction \(\ab\dashv U\).
The proof of this has already been completed above.
Free Abelian Groups
At the end of the previous post we were able to interpret the free group \(F(X)\) as the free product
\[{\prod_{x\in X}}^\ast \mathbb{Z}\]However, since we already know that \(\Ab\) has coproducts \(\bigoplus\), by the same argument we can obtain the left adjoint \(F_\Ab:\Set\rightarrow \Ab\) of 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 resulting \(F_\Ab(X)\) is defined to be the free abelian group. That is, the following proposition holds.
Proposition 8 The forgetful functor \(U:\Ab \rightarrow \Set\) has a left adjoint \(F_\Ab:\Set \rightarrow\Ab\).
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 \(H\). But this set has an interesting property: namely, \(\Hom_\Ab(G,H)\) 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 it can actually be regarded as a bifunctor to \(\Ab\). That is, one may think of \(\Hom_\Ab(-,-)\) as something similar to an internal \(\Hom\). However, with only 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 \(\Hom\) for this structure. That is,
\[\Hom_\Ab(G\times H, A)\cong \Hom_\Ab(G,\Hom_\Ab(H,A))\]does not hold in general. For example, if \(G=\mathbb{Z}\), then the above 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 almost all cases. (For instance, try plugging in \(H=\{e\}\).)
Therefore, in order to think of \(\Hom_\Ab(-,-)\) as an internal \(\Hom\), we must endow \(\Ab\) with a new symmetric monoidal category structure. Moreover, from equation (1) above one can also guess that \(\mathbb{Z}\) should behave like the unit for this monoidal product.
Tensor Products
The fundamental reason why the equation in Example 10 cannot hold is quite simple. The reason the above isomorphism held in \(\Set\) was that for any function \(f:A\times B \rightarrow C\), once an element of \(A\) or an element of \(B\) was fixed, the remaining map became a function from \(B\) or from \(A\) to \(C\).
On the other hand, the only maps \(f:G\times H \rightarrow A\) for which fixing the first or second component yields a group homomorphism are the zero maps. For if \(f(x, -)\) is a group homomorphism for arbitrary \(x\in G\), then \(f(x,0)=0\), and similarly \(f(0,y)=0\) for arbitrary \(y\in H\); substituting this into the condition that \(f\) is a group homomorphism,
\[f(x+0,0+y)=f(x,0)+f(0,y)\]yields that \(f(x,y)=0\) must hold for all \((x,y)\in G\times H\).
Looking at the above argument, requiring that the function obtained by fixing one component of the source of \(f\) be a group homomorphism seems quite natural.
Definition 11 For two abelian groups \(G,H\), a function \(f:G\times H \rightarrow A\) is called bilinear if the following 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, if we replace the left-hand side of the first equation in Example 10 with \(\Bilin(G,H;A)\), then one can verify that we obtain the isomorphism
\[\Bilin(G,H;A)\cong \Hom_\Ab(G,\Hom_\Ab(H,A))\]Moreover, one can verify that \(\Bilin(G,H;-)\) becomes a representable functor from \(\Ab\) to \(\Set\).
Theorem 12 The functor \(\Bilin(G,H;-)\) 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 the free abelian group, for any function \(f:G\times H \rightarrow A\) 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 \(\Bilin(G,H;A)\cong\Hom_\Ab(F_\Ab(G\times H)/S,A)\) still needs to be shown, but it is a simple computation so we omit it.
Definition 13 The representing object of Theorem 12 is called the tensor product of \(G\) and \(H\), and is denoted by \(A\otimes B\).
One can see that the elements of \(A\otimes B\) are represented by finite sums of elements of the form \(a\otimes b\). Then one 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 the symmetor \(\sigma\) are obtained from the universal property of \(\otimes\). That \(\mathbb{Z}\) is the tensor unit follows from the fact that the following 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 \(f:A \rightarrow A'\), \(g:B \rightarrow B'\), 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 \(f(a)\otimes g(b)\). Occasionally one also considers 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 point of view, it is conventional to define \(A^{\otimes 0}\) to be \(\mathbb{Z}\).
On the other hand, once we regard \((\Ab,\otimes, \mathbb{Z})\) as a symmetric monoidal category, we have already verified that the following holds.
Theorem 15 (\(\otimes\dashv\Hom\)) There exists an adjunction
\[\Hom_\Ab(G\otimes H, A)\cong\Hom_\Ab(G,\Hom_\Ab(H, A))\cong\Hom_\Ab(H,\Hom_\Ab(G, A))\]Hence, one can regard \(\Hom_\Ab(-,-)\) 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 especially 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.
For now, for a commutative monoid \(I\), there is no difference between taking the direct sum regarding \(I\) as a set and the graded abelian group defined above, so at present this definition is merely a new name for an existing concept. The reason for defining it is to later impose a relation between a new operation defined on an abelian group and the addition of \(I\).
References
[nLab] Tensor product of abelian groups. Link
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