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Intro to Abstract Algebra Fall 2025

Blake Farman

Chapter 9 Finitely Generated Abelian Groups

Section 9.1 Products of Groups

Definition 9.1.

Assume \(I\) is a set and \(\{G_i\}_{i \in I}\) is a family of groups. The product over this family is the usual Cartesian product set
\begin{equation*} \prod_{i \in I} G = \left\{(g_i)_{i \in I} \;\middle\vert\; i \in I\right\} \end{equation*}
equipped with the binary operation
\begin{equation*} (g_i)_{i \in I} (h_i)_{i \in I} = (g_i h_i)_{i \in I}\text{.} \end{equation*}
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For each \(i \in I\text{,}\) the product is also equipped with a morphism of groups called the projection onto the \(i^\text{th}\) factor, \(G_i\) that outputs the \(i^\text{th}\) element of a tuple
\begin{align*} \pi_i \colon \prod_{i \in I} G_i \amp\to G_i\\ (g_j)_{j \in I} \amp\mapsto g_i \end{align*}
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When the indexing set \(I\) can be placed in bijection with the set \(\{1,2\ldots,n\} \subseteq \N\text{,}\) we will sometimes write \(G_1 \times G_2 \times \cdots \times G_n\) instead of \(\prod_{i \in I} G_i\text{.}\)
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Proof.

Assume \(I\) is a set and \(\{G_i\}_{i \in I}\) is a family of groups indexed by \(I\text{.}\) For associativity, let \((x_i)_{i \in I}\text{,}\) \((y_i)_{i \in I}\text{,}\) and \((z_i)_{i \in I}\) be given. Since the group operation is defined component-wise and each component is a group, we observe that
\begin{align*} (x_i)_{i \in I}\left[(y_i)_{i \in I}(z_i)_{i \in I}\right] \amp=(x_i)_{i \in I}(y_iz_i)_{i \in I}\\ \amp= \left(x_i\left[y_iz_i\right]\right)_{i \in I}\\ \amp= \left(\left[x_iy_i\right]z_i\right)_{i \in I}\\ \amp= \left[(x_i)_{i \in I} (y_i)_{i \in I}\right](z_i)_{i \in i}\text{.} \end{align*}
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The identity element for the group operation is the tuple with the identity for \(G_i\) in the \(i^\text{th}\) component. Indeed, for every \((x_i)_{i \in I}\text{,}\)
\begin{align*} (1_{G_i})_{i \in I} (x_i)_{i \in I} \amp= (1_{G_i} x_i)_{i \in I}\\ \amp= (x_i)_{i \in I}\\ \amp= (x_i 1_{G_i})_{i \in I}\\ \amp= (x_i)_{i \in I} (1_{G_i})_{i \in I}\text{.} \end{align*}
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Finally, for inverses, we note that the inverse of \((x_i)_{i \in I}\) is \(\left(x_i^{-1}\right)_{i \in I}\) because
\begin{align*} (x_i)_{i \in I}\left(x_i^{-1}\right)_{i \in I} \amp= \left(x_ix_i^{-1}\right)_{i \in I}\\ \amp= (1_{G_i})_{i \in I}\\ \amp= \left(x_i^{-1}x_i\right)_{i \in I}\\ \amp= \left(x_i^{-1}\right)_{i \in I} (x_i)_{i \in I}\text{.} \end{align*}
Therefore \(\prod_{i \in I} G_i\) is a group.
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Example 9.3.

The abelian group \(\R^n = \prod_{i=1}^n \R\) is the underlying abelian group of the real vector space of dimension \(n\text{.}\) The elements are \(n\)-tuples of real numbers and the group operation is the usual pointwise addition of vectors:
\begin{equation*} (x_1, x_2, \ldots, x_n) + (y_1, y_2, \ldots, y_n) = (x_1 + y_1, x_2 + y_2, \ldots, x_n + y_n)\text{.} \end{equation*}
We can regard the projections as the familiar projection onto the \(n^\text{th}\) coordinate,
\begin{equation*} \pi_i(x_1, x_2, \ldots, x_n) = x_i\text{.} \end{equation*}
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The same construction works for any of the other number systems we have discussed. For example, \(\Z^n \subseteq \R^n\) consists of the points in \(\R^n\) with integer coordinates and \(\Q^n \subseteq \R^n\) consists of the points in \(\R^n\) with rational coordinates.
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The following theorem tells us about how products interact with other groups. In particular, every morphism of groups into a product is uniquely determined by the projections.
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Proof.

Suppose \(Z\) is a group equipped with morphisms \(\{\zeta_i \colon Z \to G_i\}_{i \in I}\text{.}\) For existence, we do the only thing we can: given \(z \in Z\text{,}\) form the tuple \((\zeta_i(z))_{i \in I}\text{.}\) A priori, this defines a morphism of sets \(\zeta \colon Z \to \prod_{i \in I} G_i\text{.}\) To see it is actually a morphism of groups, suppose \(z_1, z_2 \in G\text{.}\) Since each \(\zeta_i\) is itself a morphism of groups, we observe that
\begin{align*} \zeta(z_1z_2) \amp= \left(\zeta_i(z_1z_2)\right)_{i \in I}\\ \amp= \left(\zeta_i(z_1) \zeta_i(z_2)\right)_{i \in I}\\ \amp= \left(\zeta_i(z_1)\right)_{i \in I} \left(\zeta_i(z_2)\right)_{i \in I}\\ \amp= \zeta(z_1) \zeta(z_2) \end{align*}
by the way we defined the binary operation in the product. Moreover, by definition of the projections, we can see that
\begin{equation*} \pi_i \circ \zeta(z) = \pi_i\left(\zeta_j(z)\right)_{j \in I} = \zeta_i(z)\text{.} \end{equation*}
This establishes the existence of a morphism.
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For uniqueness, assume \(\eta \colon Z \to \prod_{i \in I} G_i\) satisfies the condition that for all \(i \in I\text{,}\) \(\pi_i \circ \eta = \zeta_i\text{.}\) Let \(z \in Z\) be given. Write \(\eta(z) = (x_i)_{i \in I}\text{.}\) By assumption,
\begin{equation*} x_i = \pi_i(x_i)_{i \in I} = \pi_i \circ \eta(z) = \zeta_i(z)\text{.} \end{equation*}
Hence
\begin{equation*} \zeta(z) = (\zeta_i(z))_{i \in I} = (x_i)_{i \in I} = \eta(z) \end{equation*}
Therefore the morphism \(\zeta\) is the unique morphism with the property that for all \(i \in I\text{,}\)
\begin{equation*} \pi_i \circ \zeta(z) = \zeta_i\text{.} \end{equation*}
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Remark 9.6.

When dealing with the product of two groups, \(G_1\) and \(G_2\text{,}\) it is common to see the morphism, \(\zeta\text{,}\) induced by \(\zeta_1 \colon Z \to G_1\) and \(\zeta_2 \colon Z \to G_2\) written as
\begin{equation*} \zeta_1 \times \zeta_2 \colon Z \to G_1 \times G_2\text{.} \end{equation*}
The proof justifies this notation because
\begin{equation*} \zeta_1 \times \zeta_2(z) = \left(\zeta_1(z), \zeta_2(z)\right)\text{.} \end{equation*}
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When writing the product explicitly becomes too cumbersome, it is common to see the induced morphism, \(\zeta\text{,}\) written as
\begin{equation*} \prod_{i \in I} \zeta_i \colon Z \to \prod_{i \in I} G_i \end{equation*}
to remind the reader that the induced map satisfies the property
\begin{equation*} \zeta(z) = \left(\zeta_i(z)\right)_{i \in I}\text{.} \end{equation*}
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In general, universal properties are an extraordinarily useful tool for producing maps into or out of some algebraic construction. However, the real power is in knowing that a construction defined by a universal property is unique (up to unique isomorphism).
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Proof.

For ease of notation, write \(G = \prod_{i \in I} G_i\text{.}\) Assume \(Z\) is a group equipped with morphisms \(\zeta_i \colon Z \to G_i\) that enjoys the same universal property as the product. First, we observe that if we can use the universal property to produce our isomorphism, then it will be unique. This tells us how to get started: use the universal property for each element to produce a morphism between the two groups. Once we have the two morphisms, we just need to prove they are inverses to one another.
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Using the universal property for \(G\text{,}\) we can produce a map \(\zeta \colon Z \to G\) that satisfies the condition for all \(i \in I\text{,}\) \(\pi_i \circ \zeta = \zeta_i\text{.}\) Similarly, using the universal property for \(Z\text{,}\) we can produce a map \(\pi \colon G \to Z\) that satisfies the condition for all \(i \in I\text{,}\) \(\zeta_i \circ \pi = \pi_i\text{.}\) It remains only to show that these maps compose to the appropriate identity morphisms.
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For every \(z \in Z\text{,}\) consider \(\pi \circ \zeta(z) \in Z\text{.}\) By the construction of the product, this element is completely determined by the projections, \(\zeta_i \colon Z \to G_i\text{:}\)
\begin{equation*} \zeta_i \circ \pi \circ \zeta(z) = \pi_i \circ \zeta(z) = \zeta_i(z) \end{equation*}
That is, \(\pi \circ \zeta(z)\) and \(z\) are the same element because they have the same components.
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The reverse direction is similar. Let \(x \in G\) be given and consider \(\zeta \circ \pi(x)\text{.}\) This element is completely determined by the projections, \(\pi_i \colon G \to G_i\text{:}\)
\begin{equation*} \pi_i \circ \zeta \circ \pi(x) = \zeta_i \circ \pi(x) = \pi_i(x)\text{.} \end{equation*}
As above, we conclude that \(\zeta \circ \pi(x)\) and \(x\) are the same because they have the same components.
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Example 9.8.

To illustrate a typical example of how to use the corollary, we prove that for any two groups \(G\) and \(H\text{,}\) \(G \times H \cong H \times G\text{.}\)
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Solution.
The corollary tells us that if \(G \times H\) satisfies the universal property for \(H \times G\text{,}\) then there is a unique isomorphism between these two groups. We note that the order of the projections is important here. The group \(G \times H\) is equipped with the projections \(\pi_1 \colon G \times H \to G\) and \(\pi_2 \colon G \times H \to H\text{.}\) The content of Corollary 9.7 is that it suffices to prove that if \(Z\) is any group equipped with maps \(\zeta_1 \colon Z \to H\) and \(\zeta_2 \colon Z \to G\text{,}\) then there exists a unique morphism of groups \(\zeta \colon Z \to G \times H\) that satisfies
\begin{equation*} \pi_1 \circ \zeta = \zeta_2 \end{equation*}
and \(\pi_2 \circ \zeta = \zeta_1\text{.}\) Of course, this map \(\zeta\) is obtained from the universal property for \(G \times H\text{.}\) Therefore there exists a unique isomorphism \(G \times H \cong H \times G\) by Corollary 9.7.
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Remark 9.9.

One refers to this as a universal property because it can be used to characterize the product in any setting. In order to define the product in another setting, we need only specify the type of object (e.g. groups) and the type of morphisms (e.g. morphisms of groups) involved. Everything else remains the same.
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For example, if we wanted to work with sets, then we can define the product of a family of sets as follows: If \(\{X_i\}_{i \in I}\) is a family of sets indexed by \(I\text{,}\) then the product \(X = \prod_{i \in I} X_i\) is the unique set (up to bijection) equipped with morphisms of sets (also called projections)
\begin{equation*} \pi_i \colon X \to X_i \end{equation*}
such that for every set \(Z\) equipped with morphisms of sets \(\zeta_i \colon Z \to X_i\text{,}\) there exists a unique morphism of sets \(\zeta \colon Z \to X\) such that for all \(i \in I\text{,}\)
\begin{equation*} \pi_i \circ \zeta = \zeta_i\text{.} \end{equation*}
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It is a good exercise to check that the usual Cartesian product satisfies this property.
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Example 9.10.

The universal property also provides us with a convenient mechanism for defining subgroups of the product. Assume \(G, H\) are groups, \(G^\prime \leq G\) and \(H^\prime \leq G\text{.}\) Since \(G^\prime\) and \(H^\prime\) are groups, the inclusions \(\mu \colon G^\prime \to G\) and \(\nu \colon H^\prime \to H\) are injective morphisms of groups. Composing with the projections \(\pi_{G^\prime} \colon G^\prime \times H^\prime \to H^\prime\) and \(\pi_{H^\prime} \colon G^\prime \times H^\prime \to H^\prime\text{,}\) we obtain two morphisms of groups \(\mu \circ \pi_{G^\prime} \colon G^\prime \times H^\prime \to G\) and \(\nu \circ \pi_{H^\prime} \colon G^\prime \times H^\prime \to H\text{.}\) These morphisms induce a unique morphism of groups \(\phi \colon G^\prime \times H^\prime \to G \times H\) via the universal property.
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We can check \(\phi\) is injective. Indeed, if \(g_1,g_2 \in G^\prime\) and \(h_1,h_2 \in H^\prime\) satisfy \(\phi(g_1,h_1) = \phi(g_2,h_2)\text{,}\) then
\begin{align*} g_1 \amp= \mu \circ \pi_{G^\prime}(g_1,g_2)\\ \amp= \pi_G \circ \phi (g_1,h_1) \\ \amp= \pi_G \circ \phi (g_2,h_2)\\ \amp= \mu \circ \pi_{G^\prime}(g_1,g_2)\\ \amp= g_2 \end{align*}
Similarly,
\begin{align*} h_1 \amp= \nu \circ \pi_{H^\prime}(g_1,g_2)\\ \amp= \pi_H \circ \phi (g_1,h_1) \\ \amp= \pi_H \circ \phi (g_2,h_2)\\ \amp= \nu \circ \pi_{H^\prime}(g_1,g_2)\\ \amp= h_2 \end{align*}
Since \(\phi\) is surjective onto its image, we see that
\begin{equation*} G^\prime \times H^\prime \cong \left\{(g,h) \;\middle\vert\; g \in G^\prime, h \in H^\prime\right\} \leq G \times H \end{equation*}
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Remark 9.11.

It should be pointed out that not all subgroups of a product group arise in this manner. Two important examples follow.
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Example 9.12. Diagonal Subgroup.

Assume \(G\) is a non-trivial group. Consider the function
\begin{align*} \Delta \colon G \amp\to G \times G\\ x \amp\mapsto (g,g) \end{align*}
For all \(a,b \in G\text{,}\)
\begin{equation*} \Delta(ab) = (ab,ab) = (a,a)(b,b) = \Delta(a)\Delta(b) \end{equation*}
implies \(\Delta\) is a morphism of groups. Moreover,
\begin{equation*} \Delta(a) = (a,a) = (1,1) \iff a = 1 \end{equation*}
implies \(\Delta\) is injective. Hence \(\Delta\) is an isomorphism onto its image
\begin{equation*} \Delta(G) = \left\{(g,g) \;\middle\vert\; g \in G\right\} \subseteq G \times G\text{,} \end{equation*}
called the diagonal of \(G\).
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The diagonal is not a product of subgroups of \(G\text{.}\) Suppose to the contrary that there exist \(H,K \leq G\) such that \(\Delta(G) = H \times K\text{.}\) For all \(g \in G\text{,}\) we observe that \((g,g) \in H \times K\) and thus
\begin{equation*} \pi_H(g,g) = g \in H \quad\text{and}\quad \pi_K(g,g) = g \in K\text{.} \end{equation*}
This implies \(G \leq H\) and \(G \leq K\text{.}\) Hence \(H = K = G\) and \(\Delta(G) = G \times G\text{.}\) However, since \(G\) was assumed to be non-trivial, there exists \(g \in G\) such that \(g \neq 1\text{,}\) but \((g,1) \not\in \Delta(G) = G \times G\text{,}\) a contradiction. Therefore \(\Delta(G)\) is subgroup of \(G \times G\) that is not the product of subgroups of \(G\text{.}\)
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Definition 9.13. Direct Sum.

Assume \(I\) is a set and \(\{G_i\}_{i \in I}\) is a family of groups. The direct sum of \(\{G_i\}_{i \in I}\) is the subset of the product \(\prod_{i \in I} G_i\text{,}\)
\begin{equation*} \bigoplus_{i \in I} G_i = \left\{(x_i)_{i \in I} \in \prod_{i \in I} G_i \;\middle\vert\; \text{for only finitely many}\ i \in I, x_i \neq 1_{G_i}\right\}\text{.} \end{equation*}
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Proof.

We observe that the direct sum is not empty because \(1 = (1_{G_i})_{i \in I} \in \bigoplus_{i \in I} G_i\text{.}\) Let \(x = (x_i)_{i \in I}, y = (y_i)_{i \in I} \in \bigoplus_{i \in I} G_i\) be given. By definition, there exist finite subsets \(I_x, I_y \subseteq I\) such that for all \(i \in I_x\text{,}\) \(x_i \neq 1_{G_i}\) and for all \(i \in I_y\text{,}\) \(y_i \neq 1_{G_i}\text{.}\) We observe that for every \(i \in I_x \cup I_y\text{,}\)
\begin{equation*} x_iy_i^{-1} = \begin{cases} 1_{G_i} \amp\text{if}\ x_i = y_i,\\ x_iy_i^{-1} \amp\text{else} \end{cases} \end{equation*}
and for all \(i \in I \setminus (I_x \cup I_y)\text{,}\)
\begin{equation*} x_iy_i^{-1} = 1_{G_i}1_{G_i}^{-1} = 1_{G_i}\text{.} \end{equation*}
That is to say, there exists \(J \subseteq I_x \cup I_y\) such that for all \(i \in J\text{,}\) \(x_iy_i^{-1} \neq 1_{G_i}\) and for all \(i \not \in J\text{,}\) \(x_iy_i^{-1} = 1_{G_i}\text{.}\) Since \(I_x\) and \(I_y\) are both finite, so is their union and also the set \(J\text{.}\) Hence
\begin{equation*} xy^{-1} = \left(x_i y_i^{-1}\right)_{i \in I} \in \bigoplus_{i \in I} G_i\text{.} \end{equation*}
Therefore \(\bigoplus_{i \in I} G_i \leq \prod_{i \in I} G_i\) by The Subgroup Criterion.
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For the final claim, assume \(I\) is a finite set. Trivially, for every \(x = (x_i)_{i \in I} \in \prod_{i \in I} G_i\text{,}\) there exist only finitely many \(i \in I\) such that \(x_i \neq 1_{G_i}\text{.}\) Therefore if \(I\) is a finite set, then \(\bigoplus_{i \in I} G_i = \prod_{i \in I} G_i\text{.}\)
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Remark 9.15.

Generally, one is interested in the direct sum for abelian groups. In this case, we typically write \(+\) for the operation and \(0\) for the identity element. The elements of the direct sum of \(\{A_i\}_{i \in I}\) can then be viewed as formal sums
\begin{equation*} \sum_{i \in I} a_i \end{equation*}
where for all but finitely many \(i \in I\text{,}\) \(a_i = 0\text{.}\)
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If we restrict our attention solely to abelian groups, then the direct sum is the coproduct (of abelian groups). That is to say, the direct sum of abelian groups satisfies a universal proeprty that is dual to the product in the sense that all the maps are reversed. Instead of being equipped with projections onto the components, the direct sum is equipped with inclusions from the components
\begin{align*} \mu_i \colon A_i \amp\mapsto \bigoplus_{i \in I} A_i\\ a \amp\mapsto \sum_{j \in I} a_j \end{align*}
where
\begin{equation*} a_j = \begin{cases} a \amp \text{if}\ j = i\\0 \amp\text{else}\end{cases} \end{equation*}
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The dual mapping property is this: if \(Z\) is any abelian group equipped with morphisms \(\zeta_i \colon A_i \to Z\text{,}\) then there exists a unique morphism \(\zeta \colon \bigoplus_{i \in I} A_i \to Z\) such that for all \(i \in I\text{,}\) the diagram below commutes.
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Figure 9.16. The diagram associated with the universal property for the direct sum.
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As with the product, we typically write \(\zeta = \bigoplus_{i \in I} \zeta_i\text{.}\) That is to say if \(\sum_{i \in I} a_i \in \bigoplus_{i \in I}\text{,}\) then there exists a finite \(J \subseteq I\) such that for all \(i \in J\text{,}\) \(a_i \neq 0\) and so
\begin{equation*} \zeta_i(a_i) = \begin{cases} \zeta_i(a_i) \amp\text{if}\ i \in J\\ 0 \amp \text{else}\end{cases} \end{equation*}
yields a xfinite sum of elements in \(Z\)
\begin{equation*} \zeta\left(\sum_{i \in I} a_i\right) = \sum_{i \in I} \zeta_i(a_i) = \sum_{i \in J} \zeta_i(a_i)\text{.} \end{equation*}
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We leave it as an exercise for the interested reader to state and prove the analogous results from products.
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Warning 9.17.

If we do not require all groups in consideration to be abelian, then the direct sum is not generally a coproduct of groups. When our groups are not necessarily abelian, the coproduct is the free product of groups
 1 
en.wikipedia.org/wiki/Free_product
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