Direct Product of Groups

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Products of Groups

We know how to define products in an arbitrary category \(\mathcal{A}\). ([Category Theory] §Limits, ⁋Example 6) The following lemma shows that arbitrary products always exist in the category \(\Grp\).

Lemma 1 \(\Grp\) has arbitrary products; in particular, it is a cartesian monoidal category. ([Category Theory] §Monoidal Categories)

Proof

First, the product set \(\prod_{i\in I} G_i\) satisfying the universal property of products in \(\Set\) was already defined in [Set Theory] §Product of Sets, ⁋Definition 1. For notational convenience, we write an element \(f:I\rightarrow \bigcup G_i\) of \(\prod_{i\in I}G_i\) as a tuple \((a_i)_{i\in I}\).

Now, for any two elements \(x=(x_i)_{i\in I}\) and \(y=(y_i)_{i\in I}\) of the set \(\prod_{i\in I}G_i\), define

\[xy=(x_i)_{i\in I}(y_i)_{i\in I}=(x_iy_i)_{i\in I}.\]

Then \(\prod_{i\in I}G_i\) becomes a group under this operation; its identity element is \((e_i)_{i\in I}\), and the inverse of \(x=(x_i)_{i\in I}\) is \((x_i^{-1})_{i\in I}\). Moreover, for any \(j\in I\),

\[\pr_j(xy)=\pr_j(x_iy_i)_{i\in I}=x_jy_j=\pr_j(x)\pr_j(y),\]

so each \(\pr_j\) is a group homomorphism.

We now prove that \((G=\prod_{i\in I}G_i,(\pr_i)_{i\in I})\) satisfies the universal property. For this, it suffices to show that the map \(f:H\rightarrow G\) furnished by the universal property of the product set is a group homomorphism. For any \(x,y\in H\) and any \(i\in I\),

\[f(xy)=(f_i(xy))_{i\in I}=(f_i(x)f_i(y))_{i\in I}=(f_i(x))_{i\in I}(f_i(y))_{i\in I}=f(x)f(y),\]

so \(f\) is a group homomorphism; therefore \((G=\prod_{i\in I}G_i,(\pr_i)_{i\in I})\) satisfies the universal property.

The following corollaries are also immediate consequences of the universal property of products.

Corollary 2 For a family of groups \((G_i)\), the product of this family is uniquely determined up to a unique isomorphism.

Proof

A terminal object in any category is uniquely determined up to a unique isomorphism.

Corollary 3 Let \((G_i)\) and \((H_i)\) be families of groups indexed by the same set \(I\), and suppose a group homomorphism \(f_i:G_i\rightarrow H_i\) is given for each \(i\). Then there exists a unique group homomorphism \(f:\prod G_i\rightarrow\prod H_i\) making the following diagram

commute. In this case, \(\ker f=\prod\ker f_i\) and \(\im f=\prod\im f_i\).

Proof

\(\prod H_i\) is the terminal object of the collection of cones satisfying the given condition. ([Category Theory] §Limits, §§Universal Property of Limits) From the commutative diagram thus defined,

\[x\in\ker f\iff f(x)=e\iff \forall i(\pr_i^H(f(x))=e_i)\iff \forall i((f_i\circ \pr_i^G)(x)=e_i)\iff \forall i(\pr_i^G(x)\in\ker f_i),\]

so \(\ker f=\prod\ker f_i\).

Similarly, for \(y\in\prod H_i\), the condition \(y\in\im f\) is equivalent to the existence of \(x\in\prod G_i\) with \(y=f(x)\), and for such \(x\),

\[\pr_i^H(y)=\pr_i^H(f(x))=f_i(\pr_i^G(x))\in\im f_i,\]

so \(\im f\subseteq\prod\im f_i\). Conversely, given \(y\in\prod\im f_i\), for each \(i\in I\) we can choose \(x_i\in G_i\) such that \(f_i(x_i)=\pr_i^H(y)\) ([Set Theory] §Axiom of Choice, ⁋The Axiom of Choice.); setting \(x=(x_i)_{i\in I}\), we obtain \(f(x)=y\), so \(\im f=\prod\im f_i\) also holds.

Corollary 4 Let \((G_i)_{i\in I}\) be a family of groups. If for each \(i\in I\) the \(H_i\) are normal subgroups of \(G_i\), then \(\prod H_i\) is also a normal subgroup of \(\prod G_i\), and its quotient group is isomorphic to \(\prod (G_i/H_i)\).

Proof

Apply Corollary 3 to the canonical homomorphisms \(p_i:G_i\rightarrow G_i/H_i\).

Each \(p_i\circ\pr_i\) is a composition of surjective homomorphisms, hence surjective, and therefore by the preceding corollary \(\im p\) equals \(\prod(G_i/H_i)\). Also, the kernel of each \(p_i\) is \(H_i\). Thus, by the first isomorphism theorem,

\[\biggl(\prod_{i\in I} G_i\biggr)\bigg/\biggl(\prod_{i\in I}H_i\biggr)\cong\prod_{i\in I} (G_i/H_i)\]

holds.

Of course, even if the \(H_i\) are not normal subgroups of the \(G_i\), the product \(\prod H_i\) is still a subgroup of \(\prod G_i\).

Corollary 5 Let \((G_i)_{i\in I}\) be a family of groups. If \(H_i\leq G_i\) for each \(i\in I\), then \(\prod H_i\) is a subgroup of \(\prod G_i\).

Proof

Applying Corollary 3 to the inclusion homomorphisms \(\iota_i:H_i\hookrightarrow G_i\), the map \(\iota\) is injective and \(\prod H_i\) is exactly the image of \(\iota\), hence a subgroup of \(\prod G_i\).

Partial Products

The above corollaries are especially useful in the following situation.

Let \((G_i)_{i\in I}\) be a family of groups, and consider a subset \(J\subseteq I\). Then the product \(\prod_{j\in J}G_j\) is well defined. On the other hand, define a family of groups \((G_i')\) by

\[G_i'=\begin{cases} G_i&i\in J\\ \{e\}&i\not\in J\end{cases}\]

and consider the group homomorphisms \(f_i:G_i'\rightarrow G_i\) given by

\[f_i=\begin{cases} \id_{G_i}&i\in J\\ \iota_i&i\not\in J\end{cases}.\]

Then one easily shows that \(\prod_{i\in I}G_i'\cong\prod_{j\in J}G_j\), and thus by Corollary 4 one verifies that

\[\biggl(\prod_{i\in I}G_i\biggr)\bigg/\biggl(\prod_{j\in J}G_j\biggr)\cong\prod_{i\in I\setminus J} G_i\]

holds.

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References

[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.

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