Direct Products

Section 7.1 Direct Products

We now discuss how to build new groups from old ones.

Subsection Direct Products of Groups

“Evil is the product of the ability of humans to make abstract that which is concrete.”
―John-Paul Sartre

Definition 7.1. Direct Product.

Let \(G_\alpha\) be a group for all \(\alpha\) in an index set \(J\text{.}\) The direct product of the groups \(G_\alpha\) is the Cartesian product \(\prod_{\alpha\in J} G_\alpha\) with multiplication defined by

\begin{equation*} (g_a)_{\alpha\in J}(h_\alpha)_{\alpha\in J} = (g_\alpha h_\alpha)_{\alpha\in J}. \end{equation*}

The direct sum of the groups \(G_\alpha\) is the subset \(\bigoplus_{\alpha\in J} G_\alpha\) of the direct product \(\prod_{\alpha\in J} G_\alpha\) given by

\begin{equation*} \bigoplus_{\alpha\in J} G_\alpha= \{(g_\alpha)_{\alpha\in J} \mid g_\alpha= e_{G_\alpha} \text{ for all but finitely many } \alpha\}, \end{equation*}

with the same multiplication as the direct product.

Remark 7.2.

A direct sum is the same thing as a finite direct product.

Theorem 7.3. Direct Product of Groups is a Group.

The direct product of a collection of groups is a group, and the direct sum of the collection is a subgroup of the direct product.

Remark 7.4.

The direct sum notation will not be seen or heard from again in group theory. It was fun while it lasted.

Theorem 7.5. Properties of Direct Products.

The direct product of groups is abelian if and only if every factor of the product is abelian.
Let \(G\) be a direct product of groups. If \(H\) is a direct product of subgroups of \(G\text{,}\) then \(H\leq G\text{.}\)
If \(G_1\cong G_2\) and \(H_1\cong H_2\text{,}\) then \(G_1\times H_1\cong G_2\times H_2\text{.}\)
Let \((g_1,\cdots,g_n)\in G_1\times\cdots\times G_n\text{.}\) Then \(|(g_1,\cdots,g_n)|=\lcm(|g_1|,\dots,|g_n|)\text{.}\)

Theorem 7.6. Direct Products and Homomorphisms.

Let \(G\) and \(H\) be groups. Then the projection map \(\pi_G:G\times H\to G\) is a surjective homomorphism of groups.
Let \(G\) and \(H\) be groups. Then the inclusion map \(\iota_G:G\to G\times H\) is an injective homomorphism of groups.

Example 7.7. Direct Product of Cyclic Groups.

If \(\operatorname{gcd}(m,n)=1\) then \(\mathbb{Z}/m\times \mathbb{Z}/n\cong \mathbb{Z}/mn\text{.}\) Indeed consider the elements \(x=(1,0)\) and \(y=(0,1)\) in \(\mathbb{Z}/m\times \mathbb{Z}/n\text{.}\) Then \(|x|=m, |y|=n\) and \(x+y=y+x=(1,1)\text{.}\) Therefore \(|xy|=\operatorname{lcm}(|x|,|y|)=mn\text{.}\) Since \(\langle x+y \rangle \subseteq \mathbb{Z}/m\times \mathbb{Z}/n\) and both of these sets have cardinality \(mn\) it must be the case that \(\mathbb{Z}/m\times \mathbb{Z}/n=\langle x+y \rangle=\langle (1,1) \rangle\text{.}\) Since \(\langle x+y \rangle\) and \(\mathbb{Z}/mn\) are both cyclic groups of order \(mn\) they are isomorphic. Thus

\begin{equation*} \mathbb{Z}/m\times \mathbb{Z}/n\cong \mathbb{Z}/mn. \end{equation*}

Exercise 7.8. No Cancellation in Products.

Find an example of groups \(G,H\) such that there is an isomorphism \(G\times H\cong G\) but \(H\) is not trivial. Note: since \(G\times H\cong G\) can be rewritten as \(G\times H\cong G\times \{e\}\text{,}\) the above shows that in general one cannot cancel groups in isomorphisms between direct products.

Subsection Internal and External Direct Products

“Don’t worry when you are not recognized, but strive to be worthy of recognition.”
―Abraham Lincoln

Theorem 7.9. Recognition Theorem for Direct Products.

Suppose \(G\) is a group with normal subgroups \(H\mathrel{\unlhd}G\) and \(K\mathrel{\unlhd}G\) such that \(H\cap K=\{e\}\text{.}\) Then the following hold:

\(HK\cong H\times K\) via the isomorphism of groups \(\theta: H \times K \to HK\) defined by \(\theta(h,k) = hk\text{.}\)
\(\displaystyle H\cong \theta^{-1}(H)=\{(h,e)\mid h\in H\}\leq H\times K\)
\(K\cong \theta^{-1}(K)=\{(e,k)\mid k\in K\}\leq H\times K\text{.}\)

Proof.

Notice that the hypothesis implies \(HK\leq G\text{.}\) Furthermore \(H\mathrel{\unlhd}G, K\mathrel{\unlhd}G\) and \(H\cap K=\{e\}\) imply that the elements of \(H\) commute with the elements of \(K\text{.}\) Indeed, consider \(h\in H,k\in K\text{.}\) Then since \(H\mathrel{\unlhd}G\text{,}\) \(khk^{-1} \in H\text{,}\) so also \([k,h]=khk^{-1}h^{-1}\in H\text{.}\) Similarly it follows that \([k,h]\in K\text{,}\) but since \(H \cap K = \{e\}\) it follows that \([k,h]= e\text{,}\) i.e. \(hk = kh\) for any \(h\in H, k\in K\text{.}\)

Using the above we have

\begin{equation*} \begin{aligned} \theta((h_1, k_1) (h_2, k_2)) & = \theta(h_1h_2, k_1k_2) \\ & = h_1h_2k_1k_2 \\ & = h_1k_1h_2k_2 = \theta(h_1, k_1) \theta(h_2, k_2) \end{aligned} \end{equation*}

and thus \(\theta\) is a homomorphism. It’s kernel is \(\{(k,h) \mid k = h^{-1} \}\text{,}\) which is just \(\{e\}\) since \(H \cap K = \{e\}\text{.}\) The image of \(\theta\) is clearly \(HK\text{.}\) This proves \(\theta\) is an isomorphism.

Definition 7.10. Internal and External Direct Products.

If \(H\mathrel{\unlhd}G\) and \(K\mathrel{\unlhd}G\) are such that \(H\cap K=\{e\}\) then we call \(HK\) the internal direct product of \(H\) and \(K\) and \(H\times K\) the external direct product of \(H\) and \(K\text{.}\)

Summary

The Direct Product of groups is again a group.
¹
See: Theorem 7.3.
\(\mathbb{Z}/m\times \mathbb{Z}/n\cong \mathbb{Z}/mn\text{.}\)
We define Internal and External Direct Products and identify them with the Recognition Theorem for Direct Products, the requirement being \(H,K\nsg G\) and \(H\cap K=\{e\}\text{.}\)

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