External and internal direct product

We proceed now to a precise consideration of the matters described above.

Definition 9.1.1   Let $G_1$, $G_2$, ..., $G_n$ be a finite collection of groups. We form the set $G = G_1 \times G_2 \times ... \times G_n$, the cartesian product of the sets $G_1$, $G_2$, ..., $G_n$. Thus $G$ consists of all $n$-tuples of the form $(a_1, a_2, ..., a_n)$, where $a_i \in G_i$, $i = 1, 2, ..., n$. We introduce an operation which will make $G$ into a group; viz, for any two $n$-tuple of $G$, we define

$$(a_1, a_2, ..., a_n) (b_1, b_2, ..., b_n) = (a_1b_1, a_2b_2, ..., a_nb_n).$$

The group $G$ so constructed is called the (external) direct product of the given groups. We denote this by $G_1 \times G_2 \times ... \times G_n$ (external).

It is understood that each product $a_ib_i$ is performed with the operation of $G_i$. It is now a straight-forward matter to show that, for this operation, the associative law is satisfied. Also the element $(e_1, e_2, ..., e_n)$, where $e_i$ is the identity element of $G_i$, functions as the identity element of $G$. Finally, the inverse of the element $(a_1, a_2, ..., a_n)$ is the element $(a_1^{-1},a_21^{-1}, ..., a_n1^{-1})$ Hence $G$ is a group with respect to the given operation. It is also clear that the following is true.

Proposition 9.1.2   Let $G = G_1 \times G_2 \times ... \times G_n$. (a) If each $G_i$ is finite and $\vert G_i\vert = r_i$, then

$$\vert G\vert=\prod_{i=1}^n r_i.$$

(b) If each $G_i$ is abelian, then $G$ is abelian.

Now we consider the following situation; which will subsequently be shown to be related.

Definition 9.1.3   Let $G$ be a given group and let $G_1, G_2, ..., G_n$ be normal subgroups of $G$ such that

$$G = G_1G_2...G_n$$

(usual product of sets in a group) and $G_i \cap G_1...G_{i- 1}G_{i+1}...G_n = \{e\}$, for every $i = 1, 2, ..., n$. In this situation, we say that $G$ is decomposed into the (internal) direct product of the subgroups $G_1, G_2, ..., G_n$, and we shall write $G = G_1 \times G_2 \times ... \times G_n$ (internal).

For the time being, we shall write in parenthesis after an expression of the form $G_1 \times G_2 \times ... \times G_n$ either ``external'' or ``internal'' to distinguish which of the two situations defined above actually prevails. After we have seen the inter-connection between these two concepts, it will be clear that we can drop this accompanying label without fear of confusion. Our first theorem related to direct products is concerned with the latter situation, i.e., where $G$ is the internal direct product of its subgroups, $G_1, G_2, ..., G_n$. We have

Theorem 9.1.4   $G = G_1 \times G_2 \times ... \times G_n$ (internal) if and only if (1) $a_ia_j = a_ja_i$ for any $a_i \in G_i$ and any $a_j \in G_j$ where $i \not= j$, and (2) Every element of $G$ can be written uniquely in the form $a_1a_2 ... a_n$ where $a_i \in G_i$.

Proof: Suppose first that $G = G_1 \times G_2 \times ... \times G_n$ (internal), and let $a_i \in G_i$ and $a_j \in G_j$ where $i \not= j$. Then the commutator $[a_i,a_j]=a_ia_ja_i^{-1}a_j^{-1}\in G_j$ since $a_ia_ja_i^{-1}\in G_j$ (since $G_j \lhd G$), and $a_j \in G_j$. However, $a_i^{-1}\in G_i$ and $a_ja_ia_j^{-1}\in G_i$ (since $G_i\lhd G$). Therefore $[a_i, a_j] \in G_i$, but $G_i \cap G_j = \{e\}$ (WHY?). Hence

$$[a_i,a_j]=a_ia_ja_i^{-1}a_j^{-1}=e,$$

which implies that $a_ia_j = a_ja_i$ and proves part (1). Next since $G = G_1G_2...G_n$, any $a \in G$ can be written in the form $a = a_1a_2...a_n$ where $a_i \in G_i$. If also $a = b_1b_2...b_n$ where $b_i \in G_i$, then

$$a = a_1a_2...a_n = b_1b_2...b_n$$

and using the commutativity of elements in different $G_i$'s, we get

$$b_ia_i^{-1} = b_1^{-1}a_1 ... b_{i-1}^{-1}a_{i-1} b_{i+1}^{-1}a_{i+1} ... b_{n}^{-1}a_{n}.$$

This in turn implies, since $G$ is the internal direct product of the $G_i$, that $b_ia_i^{-1} = e$, or that $b_i = a_i$. But this can be done for every $i = 1, 2, ..., n$. This establishes part (2). Conversely suppose that (1) and (2) hold. We claim each $G_i\lhd G$. Indeed, for if $g_i \in G_i$ and $a = a_1a_2...a_n$, $a_j \in G_j$ is an arbitrary element of $G$, then

$$ag_ia^{-1} = a_1a_2 ... a_n g_i a_n^{-1} ... a_2^{-1}a_1^{-1} =a_i g_i a_i^{-1}$$

since, by (1), $a_j$ commutes with $g_i$ for all $j > i$ and $a_j$ commutes with $g_i$ for all $j < i$. From (2), we have that $G = G_1G_2...G_n$. Finally we note that a typical element of $G_1...G_{i- 1}G_{i+1}...G_n$ is of the form $a_1a_2...a_{i- 1}a_{i+1}...a_n$, where $a_j \in G_j$. Suppose such an element is also in $G_i$ and, hence, is equal to some $a_i \in G_i$:

$$a_i = a_1a_2...a_{i- 1}a_{i+1}...a_n,$$

or

$$e...ea_ie...e = a_1a_2...a_{i- 1}ea_{i+1}...a_n,$$

where $e$ is the identity of $G$. By the uniqueness part of (2), we now have $a_i = e$, so

$$G_i \cap G_1...G_{i- 1} G_{i+1}...G_n = \{e\}.$$

We now proceed to establish two theorems which will show that, in the future, we need not distinguish between internal and external direct products.

Theorem 9.1.5   Let $G = G_1 \times G_2 \times ... \times G_n$ (internal) and let $G_i\cong H_i$, $i = 1, 2, ..., n$. Form $H = H_1 \times H_2 \times ... \times H_n$ (external); then $G\cong H$.

Proof: Let $f_iH_i\rightarrow G_i$ be an isomorphism of $H_i$ onto $G_i$. We now define a mapping

$$f : H\rightarrow G,$$

by

$$f((a_1, a_2, ..., a_n)) = f_1(a_1)f_2(a_2) ... f_n(a_n).$$

We claim that $f$ is an isomorphism of $H$ onto $G$. First we observe that $f$ is onto $G$: For any element $g$ in $G$, we know is of the form $g = b_1b_2...b_n$, where $b_i \in G_i$. Therefore, since each $f_i$ is onto, there exist $a_i \in G_i$ so that $b_i = f(a_i)$, and so

$$g = b_1b_2...b_n=f(a_1)f(a_2)...f(a_n)=f(a_1,a_2,...,a_n),$$

where $(a_1, a_2, ..., a_n)\in H$. (We note that here we have dropped the double parentheses around the $n$-tuple.) Thus $f$ is onto $G$. Second, we note that $f$ is a homomorphism, i.e., $f$ preserves the group operation. For

$$f((a_1, a_2, ..., a_n)(b_1, b_2, ..., b_n)) =f((a_1b_1, a_2b_2, ..., a_nb_n)) =f(a_1b_1)f(a_2b_2)...f(a_nb_n).$$

Since each $f_i$ is a homomorphism, this is

$$=f(a_1)f(b_1)f(a_2)f(b_2)...f(a_n)f(b_n) =(f(a_1)f(a_2)...f(a_n))(f(b_1)f(b_2)...f(b_n)),$$

since elements from different factors of an internal direct product commute from Theorem 9.1.4 (1). Finally, we have from the definition of f that the above is equal to

$$f(a_1, a_2, ..., a_n)f(b_1, b_2, ..., b_n).$$

Lastly, to prove our claim, we must show that $f$ is 1-1. To see this, suppose that $f(a_1, a_2, ..., a_n) = e$, i.e., $(a_1, a_2, ... a_n)\in Ker(f)$. Then $ee...e =f_1(a_1)f_2(a_2)...f_n(a_n)$, which by the uniqueness of representation, Theorem 9.1.5, implies that $f_i(a_i) = e$, for all $i$, $1 \leq i \leq n$. Since each $f_i$ is 1-1, Theorem 7.1.6 implies that $a_i = e_i$ for all $i$, $1 \leq i \leq n$, where $e_i$ is the identity of $H_i$. Therefore $(a_1, a_2, ..., a_n) = (e_1, e_2, ..., e_n)$ is the identity of $H$, i.e., $Ker(f)$ is trivial. Thus Theorem 7.1.6 implies that $f$ is 1-1, which proves our claim (that $f$ is an isomorphism). $\Box$ We note, in particular, taking $H_i=G_i$, $i = 1, 2, ..., n$, in Theorem 9.1.5 that if $G = G_1 \times G_2 \times ... \times G_n$ (internal), then forming $H = G_1 \times G_2 \times ... \times G_n$ (external) gives a group isomorphic to $G$.

Theorem 9.1.6   Let $G = G_1 \times G_2 \times ... \times G_n$ (external); also let $H_i = \{(e_1, ..., e_{i- 1}, a_i, e_{i+1}, ..., e_n) \ \vert\ a_i \in G_i\}$, for $i = 1, 2, ..., n$. Then the $H_i \lhd G$ and $G = H_1 \times H_2\times ... \times H_n$ (internal), and $H_i \cong G_i$, $i = 1, 2, ..., n$.

Proof: Clearly each $H_i \leq G$ (Why?) and the mapping $G_i \rightarrow H_i$ given by $a_i\longmapsto (e_1, ..., e_{i- 1}, a_i, e_{i+1}, ..., e_n)$ is also clearly an isomorphism of $G_i$ onto $H_i$. Moreover,

$$\begin{array}{c} (b_1,b_2,...,b_n)(e_1, ..., e_{i- 1}, a_i... ....., e_{i- 1}, b_ia_ib_i^{-1}, e_{i+1}, ..., e_n) \end{array}$$

which shows $H_i \lhd G$. Now let $(a_1, a_2, ..., a_n)$ be an arbitrary element of $G$, we can write this in the form

$$(a_1,a_2,...,a_n)=a_1'a_2'...a_n',$$

where $a_i'= (e_1, ..., e_{i- 1}, a_i, e_{i+1}, ..., e_n)\in H_i$. Hence $G = H_1H_2...H_n$. Finally any element of $H_1...H_{i- 1}H_{i+1}...H_n$ is of the form $(a_1, ..., a_{i- 1}, e_i, a_{i+1}, ..., a_n)$, where $a_j \in H_j$. Thus it follows immediately that

$$Hi \cap (H_1...H_{i- 1}H_{i+1}...H_n) = \{e\},$$

$i = 1, ..., n$, where $e = (e_1, ..., e_n)$ is the identity of G. The result now follows from the definition of the internal direct product. $\Box$ From now on, we drop writing in parentheses after an expression $G_1 \times G_2 \times ... \times G_n$ either ``external'' or ``internal''. It should be clear from the context what is meant.