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4.4: On Special Groups

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    132499
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    External Direct Product

    Screen Shot 2023-07-06 at 5.49.37 PM.pngDefinition: 

    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:Screen Shot 2023-07-06 at 5.50.49 PM.png

    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 \(G=HK \).

     


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