# 4.4: Special Groups

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#### 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

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

 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 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:

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 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

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