Abelian Groups

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

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 Direct 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 \(f:\prod^wG_i\rightarrow H\) 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 introduce the following terminology.

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 call the pair consisting of \(\prod^w G_i\) 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 of \(I\) consisting of those \(i\) with \(x_i\neq 0\) the support of \(x\), denoted \(\supp(x)\), and when the above condition holds we say the family \((x_i)\) is finitely supported.

Abelianization

Now we make the following definition.

Definition 3 For any group \(G\) and any two subgroups \(H_1,H_2\) of \(G\), we define \([H_1,H_2]\) to be the subgroup of \(G\) generated by the commutators

\[[h_1,h_2]=h_1^{-1}h_2^{-1}h_1h_2,\qquad h_1\in H_1,h_2\in H_2\]

What we are particularly interested in for 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 \([G,G]=\{e\}\). Thus we may regard \([G,G]\) as measuring how far \(G\) is from being abelian.

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 \(g\in G\),

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

which is immediate.

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 \(G/[G,G]\) is 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, not addition.

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 Homomorphisms, ⁋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 a unique \(\bar{f}:G/[G,G]\rightarrow H\) satisfying \(f=\bar{f}\circ p\).

In particular, suppose a group homomorphism \(f:G\rightarrow H\) is 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 \(G\), 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 together with the fact that a group homomorphism out of \(\mathbb{Z}\) is determined by the image of \(1\), 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 \(G,H\), 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

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

That \(f+g\) is again a group homomorphism is due to \(H\) being abelian: in \((f+g)(x+y)=f(x)+f(y)+g(x)+g(y)\) we may swap the middle two terms to obtain \((f+g)(x)+(f+g)(y)\). The identity 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 \(H\).

Although \(\Hom_\Ab(-,-)\) was originally defined as a bifunctor from \(\Ab^\op\times \Ab\) to \(\Set\), 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, with only the language we have so far this is impossible.

Example 10 \(\Ab\) is a cartesian monoidal category with respect to \(\times\), just as in [Category Theory] §Monoidal Categories, ⁋Proposition 4 and immediately after. 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 for almost all cases. (For instance, try \(H=\{e\}\).)

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

Tensor Products

The fundamental reason why the equation in Example 10 cannot hold is quite simple. The reason the analogous 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 from \(A\) to \(C\).

On the other hand, the only group homomorphisms \(f:G\times H \rightarrow A\) 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 \(y\in H\), so substituting this into the condition that \(f\) be a group homomorphism

\[f(x+0,0+y)=f(x,0)+f(0,y)\]

gives \(f(x,y)=0\) 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 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 of Example 10 with \(\Bilin(G,H;A)\), then we can verify that we obtain the 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 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 \(A\).

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

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

We know that elements of \(G\otimes H\) are represented by finite sums of elements of the form \(x\otimes y\). Then we can verify that \(\otimes\) is a monoidal product with \(\mathbb{Z}\) as 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 following isomorphism is natural:

\[\Hom_\Ab(\mathbb{Z}\otimes G, H)\cong\Bilin(\mathbb{Z},G;H)\cong\Hom_\Ab(\mathbb{Z},\Hom_\Ab(G,H))\cong\Hom_\Ab(G,H)\]

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 \(\otimes\) we have

\[A^{\otimes m}\otimes A^{\otimes n}\cong A^{\otimes(m+n)}\]

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, we have already verified that the following holds.

Theorem 15 (\(\otimes\dashv\Hom\)) There exists an adjunction

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

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 introducing it is to later impose a relation between a new operation defined on the abelian group and the addition of \(I\).

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References

[nLab] Tensor product of abelian groups. Link