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Mathematics LibreTexts

4.4: On Special Groups

( \newcommand{\kernel}{\mathrm{null}\,}\)

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 .

 


4.4: On Special Groups is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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