Definition 2.12.1. \(\Hom(G,H)\).
Given groups \(G\) and \(H\) we denote by \(\Hom(G,H)\) the set of all group homomorphisms from \(G\) to \(H\text{.}\)
Section 2.12 Free Abelian groups
Although the main focus of this section is free abelian groups, a significant underlying theme is the concept of universal mapping properties that characterize certain group constructions. As an example, and partly as review, recall the the universal mapping property of a quotient group.
Theorem 2.12.2. Universal property of quotient groups.
Let \(N\) be a normal subgroup of \(G\text{,}\) denoted \(N\trianglelefteq G\text{,}\) and let \(q\colon G\rightarrow G/N\) be the quotient homomorphism defined as \(q(g)=gN\text{.}\) The map
\begin{align*}
\Hom(G/N, H) \amp \rightarrow \Hom(G,H)\\
\overline{\phi} \amp \mapsto \phi=\overline{\phi}\circ q
\end{align*}
is a bijection onto the set
\begin{equation*}
\{\phi\in \Hom(G,H)\colon N\subseteq \ker\phi\}\text{.}
\end{equation*}
In other words given any \(\phi\in \Hom(G,H)\) satisfying \(N\leq \ker\phi\text{,}\) there is a unique map \(\overline{\phi}\in \Hom(G/N, H)\) satisfying \(\overline{\phi}\circ q=\phi\text{.}\) 
In fact, we must have \(\overline{\phi}(gN)=\phi(g)\text{.}\)
Definition 2.12.3. Direct product of groups.
Given a family \(\{G_\alpha\}_{\alpha\in I}\) of groups, its direct product \(\prod_{\alpha\in I}G_\alpha\) is the group with underlying set
\begin{equation*}
\prod_{\alpha\in I}G_{\alpha}=\{(g_\alpha)_{\alpha\in I}\colon g_\alpha\in G_\alpha\}
\end{equation*}
and group operation
\begin{equation*}
(g_\alpha)\cdot (h_\alpha)=(g_\alpha\cdot h_{\alpha})\text{.}
\end{equation*}
Theorem 2.12.4. Universal property of the direct product.
Let \(\{G_\alpha\}_{\alpha\in I}\) be a family of groups, let \(G=\prod G_\alpha\text{.}\)
-
Projections.For each \(\alpha\in I\) the projection map \(\pi_\alpha\colon \prod G_\alpha\rightarrow G_\alpha\) is a surjective group homomorphism.
-
Universal mapping property.Given a group \(H\) the map\begin{align*} \Hom(H, \prod G_\alpha) \amp\longleftrightarrow \prod\Hom(H, G_\alpha) \\ \phi \amp \mapsto (\pi_\alpha\circ \phi)_{\alpha\in I} \end{align*}is a bijection. In other words, given any family of group homomorphisms \(\phi_\alpha\colon H\rightarrow G_\alpha\text{,}\) there is a unique homomorphism \(\phi\colon H\rightarrow G\) satisfying \(\phi_\alpha=\pi_\alpha\circ\phi\) for all \(\alpha\in I\text{.}\)In fact, in this case we must have
\begin{equation} \phi(h)=(\phi_\alpha(h))_{\alpha\in I}\text{.}\tag{2.12.1} \end{equation} - The universal mapping property characterizes \(\prod G_\alpha\) up to isomorphism: i.e., if \(G\) is a group equipped with group homomorphisms \(\pi_\alpha'\colon G\rightarrow G_\alpha\) for which the recipe\begin{align*} \Hom(H, G) \amp\longleftrightarrow \prod\Hom(H,G_\alpha) \\ \phi \amp \mapsto (\pi_\alpha'\circ \phi)_{\alpha\in I} \end{align*}is a bijection, then \(G\cong \prod G_\alpha\text{.}\)
Proof.
The proof of this theorem is very similar to that of Theorem 2.12.7. You are encouraged to adapt that proof.
Definition 2.12.5. Direct sum of abelian groups.
Given a family \(\{A_\alpha\}_{\alpha\in I}\) of abelian groups, its direct sum \(\oplus_{\alpha\in I}A_\alpha\) is the subgroup of \(\prod_{\alpha\in I}A_\alpha\) defined as
\begin{equation*}
\bigoplus\limits_{\alpha\in I}A_\alpha=\{(g_\alpha)\in \prod_{\alpha\in I}A_\alpha\colon g_{\alpha}=0_\alpha \text{ for all but finitely many } \alpha\}\text{.}
\end{equation*}
Remark 2.12.6.
The construction for the direct sum does not require that the groups \(A_\alpha\) be abelian. So why do we include this condition? The reason, briefly, is that the term “direct sum” describes not the construction of \(\bigoplus A_\alpha\) but rather the universal mapping property it satisfies among abelian groups, as described in Theorem 2.12.7. This property fails if the abelian condition on groups is removed. (Put another way, the notion of a direct sum is really a category theory concept.)Theorem 2.12.7. Universal property of direct sums.
Let \(\{A_\alpha\}_{\alpha\in I}\) be a family of abelian groups.
-
Injections.For each \(\alpha\in I\) the map \(i_\alpha\colon A_\alpha\rightarrow \bigoplus A_\alpha\) that maps \(g_\alpha\) to the tuple \(g\in \bigoplus A_\alpha\) satisfying \(\pi_\alpha(g)=g_\alpha\) and \(\pi_{\alpha'}(g)=0_{\alpha'}\) for all \(\alpha'\ne \alpha\) is an injective group homomorphism.
-
Universal mapping property.Given an abelian group \(A\) the map\begin{align*} \Hom(\bigoplus A_\alpha,A) \amp\longleftrightarrow \prod\Hom( A_\alpha,A) \\ \phi \amp \mapsto (\phi\circ i_\alpha)_{\alpha\in I} \end{align*}is a bijection. In other words, given any abelian group \(A\) and family of group homomorphisms \(\phi_\alpha\colon A_\alpha\rightarrow A\text{,}\) there is a unique homomorphism \(\phi\colon \bigoplus A_\alpha\rightarrow A\) satisfying \(\phi_\alpha=\phi\circ i_\alpha\) for all \(\alpha\in I\text{.}\)In fact, in this case we must have
\begin{equation} \phi((g_\alpha)_{\alpha\in I})=\sum_{\alpha\in I}\phi_\alpha(g_\alpha)\text{.}\tag{2.12.2} \end{equation} - The universal mapping property characterizes \(\bigoplus A_\alpha\) among abelian groups up to isomorphism: i.e., if \(G\) is an abelian group equipped with group homomorphisms \(i_\alpha'\colon A_\alpha\rightarrow G\) for which the recipe\begin{align*} \Hom(G,A) \amp\longleftrightarrow \prod\Hom(A_\alpha,A) \\ \phi \amp \mapsto (\phi\circ i_\alpha')_{\alpha\in I} \end{align*}is a bijection for any abelian group \(A\text{,}\) then \(G\cong \bigoplus A_\alpha\text{.}\)
Proof.
- This is easy to see.
- Suppose we are given an abelian group \(A\) and homomorphisms \(\phi_\alpha\colon A_\alpha \rightarrow A\text{.}\) First we show there is at most one such homomorphism \(\phi\colon \bigoplus A_\alpha\rightarrow A\text{.}\) Indeed, since by assumption \(\phi\) satisfies \(\phi_\alpha=\phi\circ i_\alpha\) for all \(\alpha\text{,}\) we must have\begin{align*} \phi\left( (g_\alpha)_{\alpha\in I}\right) \amp = \phi\left(\sum_{\alpha\in I}i_\alpha(g_\alpha) \right) \\ \amp = \sum_{\alpha\in I}\phi(i_\alpha(g_\alpha)) \amp (\phi \text{ is homo.})\\ \amp =\sum_{\alpha\in I}\phi_\alpha(g_\alpha) \amp (\phi_\alpha=\phi\circ i_\alpha)\text{.} \end{align*}
- Suppose \(G\) is an abelian group equipped with maps \(i_\alpha'\colon A_\alpha\rightarrow G\) satisfying the given property. By the universal mapping property of \(\oplus A_\alpha\) there is a homomorphism \(\phi\colon \bigoplus A_\alpha\rightarrow G\) satisfying\begin{equation*} i_\alpha'=\phi\circ i_\alpha \end{equation*}for all \(\alpha\in I\text{.}\) Since the group \(G\) and maps \(i_{\alpha}'\) also satisfies the universal mapping property, the maps \(i_\alpha\colon A_\alpha\rightarrow \bigoplus A_{\alpha}\) gives rise to a homomorphism \(\psi\colon G\rightarrow \bigoplus A_{\alpha}\) satisfying\begin{equation*} i_\alpha=\psi\circ i_\alpha' \end{equation*}for all \(\alpha\in I\text{.}\) We claim that \(\phi\circ\psi=\id_G\) and \(\psi\circ\phi=\id_{\oplus A_{\alpha}}\text{,}\) from whence it follows that \(\phi\) (and \(\psi\)) is an isomorphism. To see why these two equalities hold, we use the uniqueness claim in the universal mapping property. For example, we have\begin{align*} (\phi\circ \psi)\circ i_\alpha' \amp =\phi(\psi\circ i_\alpha')\\ \amp= \phi\circ i_\alpha \amp (\psi\circ i_{\alpha'}=i_\alpha)\\ \amp = i_\alpha' \amp (\phi\circ i_\alpha=i_{\alpha'}) \end{align*}for all \(\alpha\text{.}\) Since \(\id_G\) also satisfies \(\id_G\circ i_{\alpha}'=i_\alpha'\) for all \(\alpha\in I\text{,}\) we conclude that \(\phi\circ \psi=\id_G\text{.}\) The argument for \(\psi\circ \phi\) is exactly similar.
Example 2.12.8.
Let \(G=\bigoplus\limits_{\alpha\in I} A_\alpha\)
- Consider the system of maps \(i_\alpha\colon A_\alpha\rightarrow G\text{.}\) According to the universal mapping property, there is a unique map \(\phi\colon G\rightarrow G\) satisfying \(i_\alpha=\phi\circ i_\alpha\) for all \(\alpha\text{.}\) What is \(\phi\text{?}\)
- Fix \(\alpha_0\in I\) and consider the system of maps \(j_\alpha\colon A_\alpha\rightarrow A_{\alpha_0}\) defined as \(j_{\alpha}=0_{A_\alpha}\) for \(\alpha\ne \alpha_0\) (the zero map), and \(j_{\alpha_0}=\id_{A_{\alpha_0}}\text{.}\) According to the universal mapping property, there is a unique map \(\phi\colon G\rightarrow A_{\alpha_0}\) satisfying \(j_\alpha=\phi\circ i_\alpha\) for all \(\alpha\text{.}\) What is \(\phi\text{?}\)
Definition 2.12.9. Direct sum of subgroups.
Let \(\{A_\alpha\}_{\alpha\in I}\) be a family of subgroups of the abelian group \(G\text{.}\) The inclusion maps \(j_\alpha\colon A_\alpha\rightarrow G\) give rise to a unique group homomorphism \(\phi\colon \bigoplus A_\alpha\rightarrow G\text{.}\)
-
Sum of subgroups.The group \(G\) is the of the \(A_\alpha\text{,}\) denoted \(G=\sum A_\alpha\text{,}\) if \(\phi\) is surjective.
-
Direct sum of subgroups.The group \(G\) is the of the \(A_\alpha\text{,}\) denoted \(G=\bigoplus A_\alpha\text{,}\) if \(\phi\) is an isomorphism.
Theorem 2.12.10. Direct sum equivalence.
Let \(\{A_\alpha\}_{\alpha\in I}\) be a family of subgroups of the abelian group \(G\text{.}\)
- We have \(G=\sum A_\alpha\) if and only if for all \(g\in G\) we can write \(g=\sum_{k=1}^ng_{\alpha_k}\) for some elements \(g_{\alpha_k}\in A_{\alpha_k}\text{.}\)
- We have \(G\cong \bigoplus A_\alpha\) if and only if for all \(g\in G\) we can write \(g=\sum_{k=1}^ng_{\alpha_k}\) for some elements \(g_{\alpha_k}\in A_{\alpha_k}\) in a unique way.
Proof.
According to (2.12.2) the map \(\phi\circ \oplus A_\alpha\rightarrow G\) coming from the inclusions \(j\colon A_\alpha\hookrightarrow G_\alpha\) has formula
\begin{align*}
\phi((g_\alpha)) \amp = \sum_{\alpha\in I}j_\alpha(g_\alpha)\\
\amp = \sum_{\alpha\in I}g_\alpha \amp (j_\alpha=\id\vert_{A_\alpha})\text{.}
\end{align*}
From this it is clear that \(\phi\) is surjective if and only if every element \(g\in G\) can be so written, and bijective if and only if this expression is unique. The result follows.
Definition 2.12.11. Free abelian group.
A group \(G\) is a free abelian group if \(G\cong \bigoplus\limits_{\alpha\in I}\Z\) for some set \(I\text{.}\)
Theorem 2.12.12. Free abelian groups.
Let \(G\) be an abelian group. The following statements are equivalent.
- \(G\) is free abelian.
- There is a set \(I\) and tuple \((g_\alpha)_{\alpha\in I}\in \prod_{\alpha\in I}G\) such that the map\begin{align*} \bigoplus_{\alpha\in I}\Z \amp \rightarrow G\\ (m_{\alpha}) \amp \mapsto \sum_{\alpha\in I} m_\alpha g_\alpha \end{align*}is an isomorphism.
- There is a set \(I\) and tuple \((g_\alpha)_{\alpha\in I}\in \prod_{\alpha\in I}G\) such that for all \(g\in G\) we can write \(g=\sum_{\alpha\in I}m_{\alpha}g_\alpha\text{,}\) where \(m_\alpha=0\) for all but finitely many \(\alpha\in I\text{,}\) in a unique way.
- There is a set \(I\) and tuple \((g_\alpha)_{\alpha\in I}\in \prod_{\alpha\in I}G\) such that for any abelian group \(A\) the map\begin{align*} \Hom(G,A) \amp \rightarrow \prod_{\alpha\in I} A \\ \phi \amp \mapsto (\phi(g_\alpha))_{\alpha\in I} \end{align*}is a bijection. In other words, there is a tuple \((g_\alpha)_{\alpha\in I}\) such that given any abelian group \(A\) and any elements \(h_\alpha\in A\text{,}\) there is a unique homomorphism \(\phi\colon G\rightarrow A\) satisfying \(\phi(g_\alpha)=h_\alpha\text{.}\)
Proof.
\((1)\implies (2)\text{.}\) If \(G\) is free abelian, there is an isomorphism \(\phi\colon \bigoplus_{\alpha\in I}\Z\rightarrow G\text{.}\) Let \(\phi_\alpha=\phi\circ i_\alpha\) and let \(g_{\alpha}=\phi_\alpha(1)\text{.}\) From (2.12.2) we have
\begin{align*}
\phi((m_\alpha))\amp = \sum_{\alpha\in I} \phi_\alpha(m_\alpha) \\
\amp = \sum_{\alpha\in I}m_\alpha\phi_\alpha(1) \amp (\phi_\alpha \text{ a homo.})\\
\amp =\sum_{\alpha\in I}m_\alpha g_\alpha\text{.}
\end{align*}
\((2)\implies (3)\text{.}\) This is clear.
\((3)\implies (4)\text{.}\) Since for any \(g\) we can write \(g=\sum_{\alpha\in I}m_\alpha g_\alpha\text{,}\) given a homomorphism \(\phi\colon G\rightarrow A\) we have
\begin{align*}
\phi(g) \amp =\phi\left( \sum_{\alpha\in I}m_\alpha g_\alpha\right)\\
\amp = \sum_{\alpha\in I}m_\alpha\phi(g_\alpha)\text{.}
\end{align*}
Thus \(\phi\) is completely determined by its values \((\phi(g_\alpha))_{\alpha\in I}\text{.}\) Furthermore, given any tuple \((a_\alpha)_{\alpha\in I}\) it is easy to see that the function \(\phi\colon G\rightarrow A\) defined as
\begin{equation*}
\phi(g)=\phi \left( \sum_{\alpha\in I}m_\alpha g_\alpha\right)=\sum_{\alpha\in I}m_\alpha a_\alpha
\end{equation*}
is well-defined and a group homomorphism.
\((4)\implies (1)\text{.}\) Given such a tuple \((g_\alpha)_{\alpha\in I}\text{,}\) we define group homomorphisms \(\phi_\alpha\colon \Z\rightarrow G\) as \(\phi_{\alpha}(m)=mg_\alpha\text{.}\) This gives rise to a group homomorphism \(\phi\colon \oplus_{\alpha\in I}\Z\rightarrow G\) satisfying
\begin{equation*}
\phi_\alpha=\phi\circ i_\alpha
\end{equation*}
for all \(\alpha\in I\text{.}\) Next, for all \(\alpha\in I\) let \(\bolde_\alpha=i_\alpha(1)\text{:}\) this is the tuple in \(\bigoplus_{\alpha\in I}\Z\) whose \(\alpha\)-th component is \(1\text{,}\) and whose every other component is 0. The property described in (4) guarantees that there is a unique homomorphism \(\psi\colon G\rightarrow \bigoplus_{\alpha\in I}\Z\) sending \(g_\alpha\) to \(\psi(g_\alpha)=\bolde_\alpha\text{.}\) We will show that \(\phi\) and \(\psi\) are inverses of each other, and hence that \(G\cong \bigoplus_{\alpha\in I}\Z\text{.}\)
\(\psi\circ\phi=\id\text{.}\) We have \(\psi\circ\phi\colon \bigoplus \Z\rightarrow \bigoplus \Z\text{.}\) Using the universal mapping property of \(\bigoplus_{\alpha\in I} \Z\text{,}\) if we show that \((\psi\circ\phi)\circ i_\alpha=i_\alpha\) for all \(\alpha\in I\text{,}\) we conclude that \(\psi\circ\phi=\id\text{.}\) For any \(m\in \Z\) we have
\begin{align*}
(\psi\circ\phi)\circ i_\alpha (m) \amp=\psi(\phi\circ i_\alpha(m)) \\
\amp =\psi(\phi_\alpha(m))\\
\amp=\psi(mg_\alpha) \\
\amp =m\psi(g_\alpha)\\
\amp =mi_\alpha(1)\\
\amp =i_\alpha(m)\text{.}
\end{align*}
This shows that \((\psi\circ\phi)\circ i_\alpha=i_\alpha\text{,}\) and thus \(\psi\circ\phi=\id\text{,}\) as desired.
\(\phi\circ\psi=\id\text{.}\) Since \(\phi\circ\psi\colon G\rightarrow G\text{,}\) using the uniqueness condition in the property described in (4), we need only show that \(\phi\circ \psi (g_\alpha)=g_\alpha\) for all \(\alpha\in I\text{.}\) We have
\begin{align*}
\phi\circ \psi (g_\alpha) \amp=\phi(\psi (g_\alpha)) \\
\amp =\phi(i_\alpha(1))\\
\amp = g_\alpha\text{,}
\end{align*}
as desired.
Definition 2.12.13. Basis of a free abelian group.
Let \(G\) be an abelian group. A tuple \((g_\alpha)_{\alpha\in I}\) satisfying any of the equivalent conditions of Theorem 2.12.12 is called a basis of \(G\text{.}\)
In this case we define the rank of \(G\text{,}\) denoted \(\rank G\text{,}\) as \(\rank G=\abs{I}\text{.}\)
Proof.
Why is \(\rank G\) well-defined? As you will show in a homework exercise, if \(G\cong\oplus_{\alpha\in I}\Z\text{,}\) then \(G/2G\cong \bigoplus_{k\in I}\Z/2\Z\text{.}\) Thinking of \(\Z/2\Z\) as the finite field \(\F_2\text{,}\) we see that \(\abs{I}\) is just the dimension of \(G/2G\) as a \(\F_2\)-vector space: i.e., \(\rank G=\dim_{\F_2}G/2G\text{.}\)
