4.4: On Special Groups

External Direct Product

Definition: 

Let \((G, \star)\) and \((G, \diamond)\) be groups.

Then the external direct product is defined by \(G_1 \times G_2=\{(g_1,g_2) | g_1 \in G_1, g_2 \in G_2\}\).  \((g_1, g_2)(h_1,h_2)=(g_1\star h_1, g_2 \diamond h_2)\).

 

Example \(\PageIndex{1}\)

Consider the following

  \((\mathbb{Z}_2 \times \mathbb{Z}_2, +\pmod{2})\).

Cayley Table \((\mathbb{Z}_2 \times \mathbb{Z}_2, +\pmod{2})\)
\(+\pmod{2}\)	(0,0)	(0,1)	(1,0)	(1,1)
(0,0)	(0,0)	(0,1)	(1,0)	(1,1)
(0,1)	(0,1)	(0,0)	(1,1)	(1,0)
(1,0)	(1,0)	(1,1)	(0,0)	(0,1)
(1,1)	(1,1)	(1,0)	(0,1)	(0,0)

 

Notice that \( | \mathbb{Z}_2  \times  \mathbb{Z}_2| =4\), and every element which is not the identity has order \( 2. \) Therefore, the abelian group \( \mathbb{Z}_2  \times  \mathbb{Z}_2 \) is not isomorphic to the cyclic group \( \mathbb{Z}_4. \)

Note

If you have many groups \( G_i, i=1, \cdots n \), then the external direct product can be written as \(\displaystyle\prod_{i=1}^{n} G_{i}\).

Example \(\PageIndex{1}\)

Consider the following:

\(( \mathbb{Z}_2  \times  \mathbb{Z}_3, +(mod 6)). \)

\( \mathbb{Z}_2 \times  \mathbb{Z}_3 = \{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)\}. \)

Notice that \((1,1)^6(mod \, 2)=(0,0).\) Therefore \( \mathbb{Z}_2 \times  \mathbb{Z}_3 = \langle (1,1) \rangle .\) Hence \(  \mathbb{Z}_2  \times  \mathbb{Z}_3 \) is a cyclic group of order \( 6. \) Therefore, \( \mathbb{Z}_2  \times  \mathbb{Z}_3 \cong  \mathbb{Z}_6. \)

 

Theorem \(\PageIndex{1}\)

The group \( \mathbb{Z}_m \times  \mathbb{Z}_n \cong  \mathbb{Z}_{mn} \) iff \(gcd(m,n)=1.\)

 

Internal Direct Product

Definition: 

Let \(G \) be a group with subgroups \( H \) and \( K \), with the following conditions satisfied:

1. \(H\le G\) and \( K \le G.\)

2. \(G=HK=\{hk|h\in H, k \in K\}\)

3. \(H \cap K =\{e\}.\)

4. \(hk=kh, \; \forall h \in H, \; k \in K.\)

Then we say that \( G \) is the internal direct product of \( H \) and \( K \).

Example \(\PageIndex{1}\)

\(H=\{1,3\}\), \(K=\{1,5\}, G = U(8)=\{1,3,5,7\}.\) Is \(G=HK?\)

Yes.

Proof:

Let \(H=\{1,3\}, K=\{1,5\}, G = U(8)=\{1,3,5,7\}.\)

We will show \(H \le G\) and \(K \le G.\)

\(H \subseteq G, e_H=e, ab^{-1} \in H\) thus \(H \le G.\)

Similarly, \(K \le G.\)

\(H \cap K=\{e\}\) by inspection.

We will show \(hk=kh, \; \forall h \in H, k \in K.\)

Consider \(1\cdot 1=1\cdot 1=1 \in G\)

\(1\cdot 5=5\cdot 1=5 \in G \)

\(3\cdot 1=1\cdot 3=3 \in G \)

\(3\cdot 5=5\cdot 3=7 \in G \)

Thus \(G=HK \).◻

Example \(\PageIndex{2}\)

Let \( G=D_6= \langle r,s|r^6=e, s^2=e,r^{-1}sr=s^{-1} \rangle \).

Let \(H=\{e,r^3\} \), \(K=\{e,r^2,r^4,s,r^2s,r^4s\} \).

Is \(G=HK \)?

\(H \le G \) and \(K \le G \).

\(H \cap K=\{e\} \).

By inspection, \(hk=kh, \forall k \in K, h \in H \).

Thus \(G=HK \).