4.4: Special Groups
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- 132499
External Direct Product
External direct product. (works like .
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 1: Consider the following
\((\mathbb{Z}_2 \times \mathbb{Z}_2, +\pmod{2})\).
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Cayley Table \((\mathbb{Z}_2 \times \mathbb{Z}_2, +\pmod{2})\) |
||||
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\(+\pmod{2}\) |
(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 2: 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:
The group \( \mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{mn} \) iff \(gcd(m,n)=1.\)
Internal Direct Product
Let \(G \) be a group with subgroups \( H \) and \( K \), with the following conditions satisfied:
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\(H\le G\) and \( K \le G.\)
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\(G=HK=\{hk|h\in H, k \in K\}\)
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\(H \cap K =\{e\}.\)
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\(hk=kh, \; \forall h \in H, \; k \in K.\)
Then we say that \( G \) is the internal direct product of \( H \) and \( K \).
Example 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 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 \(G=HK \)?
\(H \leG \) and \(K \le G \).
\(H \cap K=\{e\} \).
By inspection, \(hk=kh, \forall k \in K, h \in H \).
Thus \(G=HK \).
Chapter 8: Exercises