Abelian Groups

We have not paid much attention to the category \(\Ab\) so far, but in this post we examine abelian groups.

Sums of Abelian Groups

First, the universal property of the weak direct product shown in §Restricted Sum, ⁋Theorem 2 applies particularly well when the group \(H\) is abelian.

Theorem 1 Given a family \((G_i)\) of abelian groups, 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 satisfying \(f_i=f\circ\iota_i\), namely \(f:\prod^wG_i\rightarrow H\).

Thus, at least among abelian groups, the weak direct product \(\prod^w G_i\) becomes a coproduct. We call this as follows.

Definition 2 Given a family \((G_i)\) of abelian groups, their weak direct product \(\prod^w G_i\), and the inclusion maps \(\iota_i\), we collectively call \(\prod^w G_i\) and the \(\iota_i\) the direct sum of the \(G_i\), and denote it by \(\bigoplus G_i\).

By 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, we call the subset satisfying \(x_i\neq 0\) in \(I\) the support of \(x\) and write it as \(\supp(x)\), and when the above condition holds, we call the family \((x_i)\) finitely supported.

Abelianization

Now we define the following.

Definition 3 For any group \(G\) and two subgroups of \(G\), \(H_1,H_2\), we define \([H_1,H_2]\) as the subgroup 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 \(G\).

What we are particularly interested in for this post is the case \(H_1=H_2=G\). If \(G\) were an abelian group, then for all \(x,y\in G\) we have \(x^{-1}y^{-1}xy=e\), so \([G,G]=\{e\}\). Thus we may think of \([G,G]\) as representing how far \(G\) is from being an abelian group.

Meanwhile, 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 \(g\in G\),

\[g(x^{-1}y^{-1}xy)g^{-1}=(gx^{-1}g^{-1})(gy^{-1}g^{-1})(gxg^{-1})(gyg^{-1})=(gxg^{-1})^{-1}(gyg^{-1})^{-1}(gxg^{-1})(gyg^{-1})\in [G,G]\]

so it is trivial.

Therefore, \(G/[G,G]\) is well-defined. Since this amounts to treating all elements of the form \(x^{-1}y^{-1}xy\) as \(e\), the quotient \(G/[G,G]\) becomes an abelian group. According to our convention, the operation of an abelian group should properly be written as \(+\), but since \(G/[G,G]\) simultaneously inherits its operation from \(G\), doing so could cause confusion. Therefore, 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 \(x,y\in G\) we have

\[e=f(x)^{-1}f(y)^{-1}f(x)f(y)=f(x^{-1}y^{-1}xy)\]

so \([G,G]\leq\ker f\). Now 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 a unique group homomorphism satisfying \(f=\bar{f}\circ p\), namely \(\bar{f}:G/[G,G]\rightarrow H\).

In particular, suppose a group homomorphism \(f:G\rightarrow H\) is given. Then by the composition \(G\rightarrow H\rightarrow H/[H,H]\), we obtain 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 \(G\), we call the quotient group \(G/[G,G]\) the abelianization of \(G\), and denote it by \(G^\ab\).

Then the preceding argument shows that this defines a functor \(\ab:\Grp\rightarrow\Ab\). That is,

\[\ab(\id_G)=\id_{G^\ab},\qquad \ab(g\circ f)=\ab(g)\circ\ab(f)\]

hold, and these two equalities follow from uniqueness once we show that the right-hand side satisfies the universal property that the left-hand side must satisfy. 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 bijection \(\Hom_\Ab(G^\ab,H)\cong \Hom_\Grp(G,U(H))\) for this claim follows immediately from the correspondence \(\bar{f}\mapsto\bar{f}\circ p\) of Proposition 5, and the only remaining thing to show is naturality in each factor; but this too follows by an argument similar to the uniqueness argument examined above.

Free Abelian Groups

Meanwhile, at the end of the previous post we were able to interpret the free group \(F(X)\) as the free product

\[{\prod_{x\in X}}^\ast \mathbb{Z}\]

Since the direct sum replaces the free product in \(\Ab\), for a set \(X\) we put

\[F_\Ab(X)=\bigoplus_{x\in X} \mathbb{Z}\]

and call this the free abelian group defined by \(X\). Since a function \(u:X\rightarrow Y\) induces a homomorphism \(F_\Ab(u)\) sending the \(x\)-th generator to the \(u(x)\)-th generator, \(F_\Ab\) becomes a functor \(\Set\rightarrow\Ab\).

Now, applying Theorem 1 and the fact that a group homomorphism out of \(\mathbb{Z}\) is determined by the image of \(1\) in turn, we obtain for any abelian group \(H\) the following isomorphism

\[\Hom_\Ab\biggl(\bigoplus_{x\in X}\mathbb{Z},H\biggr)\cong\prod_{x\in X}\Hom_\Ab(\mathbb{Z},H)\cong\prod_{x\in X}U(H)\cong\Hom_\Set(X,U(H))\]

Here each correspondence is given only by composition of homomorphisms and composition of functions, so it is natural in both \(X\) and \(H\), and thus we obtain the following. ([Category Theory] §Adjoint Functors, ⁋Definition 1)

Proposition 8 There exists a left adjoint \(F_\Ab:\Set \rightarrow\Ab\) of the forgetful functor \(U:\Ab \rightarrow \Set\).

The Abelian Group \(\Hom_\Ab(G,H)\)

For any abelian groups \(G,H\), \(\Hom_\Ab(G,H)\) 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 \(G,H\), \(\Hom_\Ab(G,H)\) is an abelian group.

Proof

For any \(f,g:G \rightarrow H\), it suffices to define \(f+g\) by

\[(f+g)(x)=f(x)+g(x)\qquad\text{for all $x\in G$}\]

That \(f+g\) is again a group homomorphism is due to \(H\) being abelian: swapping the middle two terms of \((f+g)(x+y)=f(x)+f(y)+g(x)+g(y)\) yields \((f+g)(x)+(f+g)(y)\). The identity element is the zero map, the inverse of \(f\) is \((-f)(x)=-f(x)\), and the commutativity of \(\Hom_\Ab(G,H)\) also follows from the commutativity of \(H\).

\(\Hom_\Ab(-,-)\) was originally defined as a bifunctor from \(\Ab^\op\times \Ab\) to \(\Set\), but by this proposition we can in fact view it as a bifunctor to \(\Ab\). That is, we may think of \(\Hom_\Ab(-,-)\) as something similar to the internal \(\Hom\) of [Category Theory] §Adjoint Functors, ⁋Definition 8. However, using only the language we have so far, this is impossible.

Example 10 \(\Ab\) is a cartesian monoidal category with respect to \(\times\), as immediately after [Category Theory] §Monoidal Categories, ⁋Proposition 4. 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 equation would become

\[\Hom_\Ab(\mathbb{Z}\times H,A)\cong \Hom_\Ab(\mathbb{Z},\Hom_\Ab(H,A))\cong \Hom_\Ab(H,A)\tag{1}\]

and this equation will almost always be false. (For instance, one can try plugging in \(H=\{e\}\).)

Therefore, in order to regard \(\Hom_\Ab(-,-)\) as an internal \(\Hom\), we must endow \(\Ab\) with a new symmetric monoidal category structure. Also, taking a hint from equation (1) above, we can guess that \(\mathbb{Z}\) should behave like the unit of this monoidal product.

Tensor Product

The fundamental reason why the equation in Example 10 cannot hold is quite simple. The reason why the above isomorphism held in \(\Set\) was that for any function \(f:A\times B \rightarrow C\), fixing an element of \(A\) or an element of \(B\) left a function from \(B\) or \(A\) to \(C\).

On the other hand, among group homomorphisms \(f:G\times H \rightarrow A\), the only \(f\) such that fixing each component yields a group homomorphism is the zero map. This is because for any \(x\in G\), if \(f(x, -)\) is a group homomorphism, then we must have \(f(x,0)=0\), and similarly for any \(y\in H\), we must have \(f(0,y)=0\), so 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 three abelian groups \(G,H,A\), 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 \(\Bilin(G,H;A)\), we can verify that we obtain the isomorphism

\[\Bilin(G,H;A)\cong \Hom_\Ab(G,\Hom_\Ab(H,A))\]

Furthermore, we can verify that \(\Bilin(G,H;-)\) is a representable functor from \(\Ab\) to \(\Set\).

Theorem 12 \(\Bilin(G,H;-)\) is representable.

Proof

In the free abelian group \(F_\Ab(G\times H)\), define the subgroup \(S\) 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, 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 \(A\).

Although the naturality of the isomorphism \(\Bilin(G,H;A)\cong\Hom_\Ab(F_\Ab(G\times H)/S,A)\) additionally needs to be shown, it is a simple computation, so we omit it.

Definition 13 We call the representation of Theorem 12 the tensor product of \(G\) and \(H\), and denote it by \(G\otimes H\).

We can see that elements of \(G\otimes H\) are represented as finite sums of elements of the form \(x\otimes y\). 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 is obtained 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 \(f:A \rightarrow A'\), \(g:B \rightarrow B'\), 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 we also 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, by convention \(A^{\otimes 0}\) is defined to be \(\mathbb{Z}\).

Meanwhile, once we regard \((\Ab,\otimes, \mathbb{Z})\) as a symmetric monoidal category in this way, we have already verified that the following holds.

Theorem 15 (\(\otimes\dashv\Hom\)) An adjunction

\[\Hom_\Ab(G\otimes H, A)\cong\Hom_\Ab(G,\Hom_\Ab(H, A))\cong\Hom_\Ab(H,\Hom_\Ab(G, A))\]

exists. Therefore, we may 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 \(I\). Then we call the direct sum \(\bigoplus G_i\) of this family 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 merely gives a new name to an existing concept. The reason for defining this is to impose a relation between this operation and the addition of \(I\) when a new operation is later defined on the abelian group.

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References

[nLab] Tensor product of abelian groups. Link