**Proof:**

(i) )(ii) If *N*<sup>p</sup> is a partial normal subgroup of *G*<sup>p</sup> then it is easy that for all *x*p∈ *N*<sup>p</sup> and gp∈ *G*<sup>p</sup> we have gp.*x*p.gp �1 ∈ *N*<sup>p</sup> and from just before the theorem gp.*N*p.gp �1 ∈ *N*p.

(iii))(ii) Let gp be an element of *G*p: We need to see gp.*N*p¼ *N* p.gp. Assume that *x*p∈gp.*N*p. Then *x*<sup>p</sup> = gp.*n*<sup>1</sup> is satisfied for n1∈ *N*p. Since*x* p.gp �<sup>1</sup> = gp.*n*p.gp �1 and gp.*N*p.gp �1⊆ *N*<sup>p</sup> we have *x*pgp �1∈ *N* pand so that there exists an *n*2∈ *N*<sup>p</sup> such that xp.gp �<sup>1</sup> = n2. If we product from right with gp then we have the <sup>ð</sup>*x*p.gp �1 Þ.gp = *n*2.gp equality.

Using the associativity property the equality becomes xp.(gp �1 .gp)=n2.gp and using identity element property and converse element property we get *x*<sup>p</sup> = *n*2. gp∈ *N*p.gp. It implies that gp.*N*p⊆ Np.gp. In a similar way gp �1 .*N*p.gp⊆ *N*<sup>p</sup> and we have *N*p.gp⊆ gp.*N*p. Therefore we obtain gpNp = Np.gp.

(ii))(i) Supposing that gp∈ *G*<sup>p</sup> we have to prove gp.*n*p.gp �<sup>1</sup> is contained in *N*<sup>p</sup> for all *n*p∈ *N* p. Since gp.*N*<sup>p</sup> = *N*p.gp, we can say that gp.*n*p∈ *N*p.gp for all *n*p∈ *N*p. Then, by associativity gp.*n*p.gp �<sup>1</sup> is contained in *N*p.gp.gp �<sup>1</sup> and then for all np∈ *N*p, gp.*n*p.gp �1 .

**Lemma 3.1.** The center of a partial group *G*<sup>p</sup> is a partial normal subgroup of *G*p. **Proof:** The center of a partial group is defined as below:

Ζ (Gp)=f*x*p∈ *G*jIf for every gp∈ *G*<sup>p</sup> xp.gp and gp.xp are defined, *x*p.gp¼ gp.*x*pg. Let gp∈ *G*<sup>p</sup> and *x*p∈ *Ζ* (*G*p) then we need to show gp.xp.gp �<sup>1</sup> is contained in Ζ ð*G*pÞ. Since xp∈ Ζ ð*G*p) for every gp∈ *G*<sup>p</sup> if gp.*x*<sup>p</sup> and *x*p.gp is defined then *x*p. gp = gp.*x*p. Using this argument gp.*x*p.gp �1 =*x*p.gp.gp �1 =*x*p∈ *Ζ* ð*G*p) and then we have *Z*(*G*p) is a partial normal subgroup of *G*p.

**Proposition 3.2.** Let *φ*: *G*p! *H*<sup>p</sup> be a partial group homomorphism. Then the kernel of *φ* is partial normal subgroup of *G*p.

**Proof:** Let K = *Ker*(*φ*Þ. We know that *Ker*(*φ*) is a partial subgroup of *G*p. Suppose that *y*∈*K* and gp∈ *G*p. Then using the fact *φ* is a partial group homomorphism

$$\begin{aligned} \boldsymbol{\varrho} \cdot \left( \mathbf{g}\_{\mathbf{p}} \,\boldsymbol{\mathscr{y}}\_{\mathbf{p}} \,\mathbf{g}\_{\mathbf{p}}^{-1} \right) &= \boldsymbol{\varrho} \left( \mathbf{g}\_{\mathbf{p}} \right) . \boldsymbol{\varrho} \left( \boldsymbol{\mathscr{y}}\_{\mathbf{p}} \right) . \boldsymbol{\varrho} \left( \mathbf{g}\_{\mathbf{p}}^{-1} \right) \\ &= \boldsymbol{\varrho} \left( \mathbf{g}\_{\mathbf{p}} \right) . \boldsymbol{\text{e}} \mathbf{H}\_{\mathbf{p}} . \boldsymbol{\varrho} \left( \mathbf{g}\_{\mathbf{p}}^{-1} \right) \\ &= \boldsymbol{\varrho} \left( \mathbf{g}\_{\mathbf{p}} \right) . \boldsymbol{\varrho} \left( \mathbf{g}\_{\mathbf{p}}^{-1} \right) \\ &= \boldsymbol{e} H\_{\mathbf{p}} \text{we have } \boldsymbol{\varrho} \left( \mathbf{g}\_{\mathbf{p}} . \mathbf{y}\_{\mathbf{p}} . \mathbf{g}\_{\mathbf{p}}^{-1} \right) = \boldsymbol{e} H\_{\mathbf{p}} \end{aligned}$$

and we get gp.yp.gp �<sup>1</sup> <sup>∈</sup>*K*. So Ker(*φ*<sup>Þ</sup> is a partial normal subgroup of Gp. **Partial normal subgroups**

**Theorem 3.3.** Suppose *G*<sup>p</sup> is a partial group and *H*<sup>p</sup> is a partial subgroup of *G*p.*H* <sup>p</sup> is a partial normal subgroup of *G*<sup>p</sup> if and only if (*aH*p) (*bH*p) = *abH*<sup>p</sup> equality holds for all *a*, *b*∈ *H*p.

#### **Proof:**

():) Suppose Hp is a partial normal subgroup of *G*p. We need to see the equality (*aH*p) (*bH*p) = *abH*<sup>p</sup> holds.

(⊆:) Let *x*∈ (*aH*p) (*bH*p) then ∃*h*,*h* <sup>2</sup>∈ *H* such that *x* = (*ah*1) (*bh*2). By using associativity property, we get *x* ¼ (*h*1*bh*2). From the identity element,

x = abb�<sup>1</sup> h1bh2 is obtained. Considering associativity property we can write *x* ¼ ab (b�<sup>1</sup> h1b)h2. Since *H*<sup>p</sup> is a partial normal subgroup of *G*p, b�<sup>1</sup> h1b ∈ *H*<sup>p</sup> Then x ∈*abH* <sup>p</sup> and this implies that (*aH*p) (*bH*p) ⊆*abH* p.

( ⊇:Þ Conversely, let *y* ∈*abH* <sup>p</sup> then there exists an *h*∈ *H* <sup>p</sup> such that *y* ¼ *abh*. So we can write y = *abh* as follows; *y* ¼ ð Þ *ae* ð Þ *bh* ∈ (*aH*p) (*bH*p). It implies that *abH*<sup>p</sup> ⊆ (*aH*p) (*bH*p). Therefore ð*aH*p) (*bH*p)¼ *abH* <sup>p</sup>

((:) Let us consider (*aH*p) (*bH*p) ¼ *abH* <sup>p</sup> for all *a*, *b*∈ *G*p. If *h*∈ *H*<sup>p</sup> and *g* ∈ *G* <sup>p</sup> then we must see whether or not ghg�1∈ *H:* Using associativity property and considering hypothesis; g*h*g�<sup>1</sup> = (g*h*)(g�<sup>1</sup> *e*) ∈ (gHp)(g�<sup>1</sup> Hp) = gg�<sup>1</sup> *H*<sup>p</sup> = *H*p. This implies that g*h*g�1∈ *H*p. So that *H*<sup>p</sup> is a partial normal subgroup of *G*p.

**Theorem 3.4.** Suppose *G*<sup>p</sup> be a partial group and *H*, *K* are partial subgroups of *G*p. If *K* is a partial normal subgroup of *G*p, then the following cases are satisfied:


#### **Proof:**


we have, h.k.h�<sup>1</sup> .k�<sup>1</sup> <sup>∈</sup> *<sup>H</sup>* <sup>∩</sup>*<sup>K</sup>* <sup>¼</sup> f g*<sup>e</sup>* . Then h.k.h�<sup>1</sup> .k�<sup>1</sup> =*e* and so, *hk* ¼ *kh*; for all *h*∈ *H*, *k*∈ *K*. Thus, we prove that *HK* ¼ *KH*.

**Proposition 3.3.** Let *G*<sup>p</sup> be a partial group and *H*<sup>p</sup> be a partial subgroup of index 2 in *G*p: Then *H*<sup>p</sup> is a partial normal subgroup in *G*p.

**Proof:** Let *H*<sup>p</sup> be a partial subgroup of index 2 in *G*<sup>p</sup> and gp be an element of *G*p. If gp∈ *H*p, then gp*H*p = *H*pgp is satisfied. If gp is not in *H*p, two left cosets must be as *H*<sup>p</sup> and gp*H*p. Since left cosets are disjoint we know gp.*H*<sup>p</sup> = Gp�Hp. Also, the right cosets are disjoint so we can write *H*p.gp = *G*p�*H*p. Thus gp*H*<sup>p</sup> = *H*pgp for all gp∈ Gp. So *H*<sup>p</sup> is normal.

**Example 3.1.** Let *G*<sup>p</sup> be an abelian partial group. Then any subgroup of *G*<sup>p</sup> is a partial normal subgroup of *G*p.

### **Proof:**

If *G*<sup>p</sup> is an abelian partial group and *x*p.*y*p∈ *G* <sup>p</sup> then *x*p.*y*<sup>p</sup> = *y*p.*x*<sup>p</sup> for every *x*p.*y*p∈ *G*p. If gp.*x*p.gp �1∈ *N*<sup>p</sup> then *N*<sup>p</sup> is a partial subgroup of *G*p. Using the hypothesis we get

$$\begin{aligned} \mathbf{g\_p} \mathcal{X}\_\mathbf{p} \cdot \mathbf{g\_p}^{-1} &= \mathbf{g\_p} \cdot \mathbf{g\_p}^{-1} \mathcal{X}\_\mathbf{p} \\ &= \mathbf{e} \mathbf{G\_p} \\ &= \mathcal{x}\_\mathbf{p} \in \mathcal{N}\_\mathbf{p} \end{aligned}$$

Therefore, the partial subgroup *Np* is the partial normal subgroup of *G*p.

**Example 3.2.** Let *H*p and *K*<sup>p</sup> be any partial normal subgroup of *G*p. Then *H*p�*K*<sup>p</sup> is also a partial normal subgroup of *G*p�*G*p.

#### **Proof:**

*H*p�*K*<sup>p</sup> = {*n*p¼(*h*p,*k*p) j*h*p∈ *H*<sup>p</sup> and *k*p∈*K*p}. We have to show that for all gp in *G*<sup>p</sup> and *n*<sup>p</sup> in *H*p�*K*<sup>p</sup> gp.*n*p.gp �<sup>1</sup> is in *<sup>H</sup>*p�*K*p.

$$\begin{aligned} \mathbf{g\_p}.\mathbf{n\_p} &= \mathbf{g\_p}.(h\_\mathbf{p}, k\_\mathbf{p}) \\ &= \left(\mathbf{g\_p}.h\_\mathbf{p}, \mathbf{g\_p}.k\_\mathbf{p}\right) \end{aligned}$$

and

$$\begin{aligned} \mathbf{g\_p} \cdot \mathbf{n\_P} \mathbf{g\_p}^{-1} &= \left( \mathbf{g\_p} \, h\_\mathbf{p}, \mathbf{g\_p} \, k\_\mathbf{p} \right) \cdot \mathbf{g\_p}^{-1} \\ &= \left( \mathbf{g\_p} \, h\_\mathbf{p} \, \mathbf{g\_p}^{-1}, \mathbf{g\_p} \, k\_\mathbf{p} \, \mathbf{g\_p}^{-1} \right) \end{aligned}$$

and since *H*<sup>p</sup> and *K*<sup>p</sup> are partial normal subgroups of *G*p, then gp.*h*p.gp �1∈ *H* p, and gp.*k*p.gp �1∈*K*<sup>p</sup> and so that (gp.*h*p.gp �1 , gp.*k*p.gp �1 )∈ *H*p�*K*p. Then the Cartesian product of two partial normal subgroups is also a partial normal subgroup.

**Theorem 3.5.** Let be a partial group and *N*<sup>p</sup> be a partial normal subgroup of *G*p. The congruence modulo *N*<sup>p</sup> is a congruence relation for the partial group operation "*:*". **Proof.** Let *x*R*N*p*y* denote that x and y are in the same coset, that is;

*x*R*N*p*y*⟺*x:N* p ¼ *y:N*<sup>p</sup>

Let *x*R*N*p*x* and *y*R*N* <sup>p</sup>*y*. To demonstrate that R*N*<sup>p</sup> is a congruence relation for *:*, we need to show, reflexivity, symmetry, and transitivity. These axioms are obvious from the definition of relation.

**Theorem 3.6.** If *N*<sup>p</sup> is a partial normal subgroup of a partial group *G*<sup>p</sup> and *G*p¼ *N*<sup>p</sup> is the set of all cosets of *N*<sup>p</sup> in *G*p, then *G*p∕*N*<sup>p</sup> is a partial group under the operation given by ð*aN*pÞð*bN*pÞ ¼ *abN* p.

**Proof.** Let *aN*p, *bN*p,*cN*p∈ *G*p∕*N*p. We must see partial group axioms are satisfied:

(G1) If (*aN*p)(*bN*p)¼ *abN* p. ð*bN*pÞð*cN*pÞ ¼ *bcN* <sup>p</sup> and *a*.( *bcN*p) are defined then

$$\begin{aligned} a.(\mathit{bcN\_p}) &= (aN\_\textup{p}) \left( \left( bN\_\textup{p} \right) \left( cN\_\textup{p} \right) \right) \\ &= \left( \left( aN\_\textup{p} \right) \left( \left( bN\_\textup{p} \right) \right) \left( cN\_\textup{p} \right) \right) \\ &= (ab)cN\_\textup{p} \end{aligned}$$

(G2) For any *aN*<sup>p</sup> in *G*p∕*N*p, *eN*<sup>p</sup> is a candidate for identity element, i.e.,

$$\begin{aligned} \left(aN\_{\mathbb{P}}\right)\left(eN\_{\mathbb{P}}\right) &= aeN\_{\mathbb{P}}\\ &= aN\_{\mathbb{P}} \end{aligned}$$

and

$$\begin{aligned} \left(eN\_{\mathbb{P}}\right)\left(aN\_{\mathbb{P}}\right) &= eaN\_{\mathbb{P}}\\ &= aN\_{\mathbb{P}} \end{aligned}$$

(G3) Since is a partial group *a*∈ *G*<sup>p</sup> has an inverse *a* ∈́ *G*p. For every *aN*<sup>p</sup> in *G*p∕*N*<sup>p</sup> *a* ́*N*<sup>p</sup> is a candidate for the inverse of *aN*p.

$$\begin{aligned} \left(aN\_{\mathbf{P}}\right)\left(\mathbf{a}'N\_{\mathbf{P}}\right) &= (a\mathbf{\acute{a}})N\_{\mathbf{P}} \\ &= eN\_{\mathbf{P}} \\ &= N\_{\mathbf{P}} \end{aligned}$$

This completes the proof.

**Definition 3.2.** Let *G*<sup>p</sup> be a partial group and Np is a partial normal subgroup of *G*p, then the partial group *G*p∕*N*<sup>p</sup> is called the quotient of the partial group or factor group of *G*<sup>p</sup> by *N*p.

**Proposition 3.4.** Let *f* : *G*p⟶*H* <sup>p</sup> be a homomorphism of partial groups, then the kernel of *f* is a partial normal subgroup of *G*p. Conversely, if *N*<sup>p</sup> is a partial normal subgroup of *<sup>G</sup>*p, then the map <sup>Q</sup>: Gp⟶Gp∕*N*<sup>p</sup> given by <sup>Q</sup>(*a*p)<sup>¼</sup> *aN*<sup>p</sup> is an epimorphism with kernel *N*p.

**Proof:** *Kerf* ¼ f*x* <sup>p</sup>∈ *G*<sup>p</sup> j*f x*ð <sup>p</sup>Þ ¼ *eH* <sup>p</sup>g, we need to show that if *x*p∈*Kerf* and *a*p∈ *G*<sup>p</sup> whether or not *a*p*x*p*a*<sup>p</sup> �1∈*Kerf* if and only if *f* (*a*p*x*p*a*<sup>p</sup> �1 Þ ¼ *f a*ð p)*f* (*x*p) *f a*ð <sup>p</sup> �1 )¼ *eH* <sup>p</sup>

So, *a* <sup>p</sup>*x*p*a*<sup>p</sup> �1∈*Kerf*. Therefore, *Kerf* is a partial normal subgroup of *G*p. It is trivial that Q: Gp⟶Gp∕*N*<sup>p</sup> is surjective. *Ker* Q ð Þ¼ {*x*pj <sup>Q</sup>(*x*p)<sup>¼</sup> *<sup>e</sup>*p*N*p}<sup>¼</sup> *<sup>N</sup>*p.

**Theorem 3.7.** Let *f* : *G* <sup>p</sup>! *H*<sup>p</sup> is partial group homomorphism and *N*<sup>p</sup> is a partial normal subgroup of *G*<sup>p</sup> contained in the kernel of *f*, then there is exactly unique homomorphism *f*: *G*p∕*N*p⟶*H*<sup>p</sup> such that *f aN*ð <sup>p</sup>Þ ¼ *f a*ð Þ for all *a*∈ *G*p. Besides *Imf* ¼ *Imf* and *kerf* ¼ (*kerf*Þ /*N*p.*f* an isomorphism if and only if *f* is an epimorhism and *N*p¼ *ker f*.

**Proof:** *f* : *G* <sup>p</sup>! *f H* p, Gp⟶Gp∕*N*p, *G*p∕*N*<sup>p</sup> ! *f H*<sup>p</sup> diagram is commutative. If *b*∈*aN*p, then *b* ¼ *an* p,*n* <sup>p</sup>∈ *N* and also *f* (*b*p)¼ *f* (*an*p)¼ *f* (*a*)*f* (*n*p)¼ *f a*ð Þ*e* ¼ *f a*ð Þ, since *N*p≤ *ker f* . Therefore, *f* has the same effect on every element of *aN*<sup>p</sup> and the map *f*: *G*p∕*N*p⟶*H*<sup>p</sup> given by *f* (*aN*p)¼ *f a*ð Þ. It is easily seen that *f* is a well-defined function. Now we need to prove whether or not *f* is a homomorphism of partial groups.

$$f(aN\_\mathbf{p}.bN\_\mathbf{p}) = \overline{f}(abN\_\mathbf{p}) = f(ab).$$

*Algebraic Approximations to Partial Group Structures DOI: http://dx.doi.org/10.5772/intechopen.102146*

$$\begin{aligned} &= f(a)f(b) \\ &= \overline{f}(aN\_{\mathbb{P}}) \overline{f}(bN\_{\mathbb{P}}) \end{aligned}$$

So, *<sup>f</sup>* is a partial group homomorphism. *Imf* <sup>¼</sup> *Imf* and *aN*p<sup>∈</sup> *kerf*⇔*f a*ð Þ¼ *<sup>e</sup>* <sup>⟺</sup> *<sup>a</sup>*∈*kerf*, whence

$$\ker \overline{\mathcal{f}} = \{ \mathtt{a} \mathtt{N}\_{\mathtt{P}} | a \in \ker \mathcal{f} \}$$

$$= \ker \mathcal{f} / N\_{\mathtt{P}}$$

*f* is unique since it is completely determined by *f*. Also, *f* is a partial group if and only if *f* is an epimorphism of partial groups *f* is a monomorphism if and only if for *kerf* ¼ ð Þ *ker f* ∕*N*p*ker f* equal to *N*p.

**Example 3.3.** In Example 2.2, it is stated that *G* ¼ f0,�1,�2, … ,�n} is a partial group with known addition operation on . We can easily say that the subset *N* = {0,�1,�, … ,�*n* � 1g of *G* is a partial subgroup of *G*. Let us show whether or not *N* is a partial normal subgroup. If gp �<sup>1</sup> + *n*<sup>p</sup> + gp∈ *N* for all gp∈ *G*<sup>p</sup> then *N* is a partial normal subgroup. gp �1 ¼ gp in this group so that gp + np + gp∈ *N*<sup>p</sup> i.e. *n*p∈ *N* and then *N* is a partial normal subgroup of *G*.

**Theorem 3.8 (First Isomorphism Theorem for Partial Groups).** Let *G*p, *H*<sup>p</sup> be partial groups and *f* : *G*p⟶*H* <sup>p</sup> be partial group epimorphism then *G*p∕*Ker f* is isomorphic to *H*p.

**Proof:** We know that *ker f* is a partial normal subgroup of *G*p. Then *G*p∕*Ker f* is defined. Let show *Ker f* ¼ *K* and g: *G*p∕*K*⟶*H* <sup>p</sup> mapping defined as gð Þ¼ ð Þ *aK* ð Þ *bK* gð Þ¼ *abK f ab* ð Þ for every *aK*, *bK* ∈ *G*p∕*K*.

Since gð Þ¼ *a:bK f ab* ð Þ¼ *f a*ð Þ*f b*ð Þ¼gð Þ *aK* gð Þ *bK* then g is a homomorphism. Since *f* is onto there exists *a*∈ *G*<sup>p</sup> such that *f a*ð Þ¼ *h* then *aK* ∈ *G*p∕*K* and gð Þ¼ *aK f a*ð Þ¼ *h* and g is onto. For one-to-one conditions let *aK*, *bK* ∈ *G*p∕*K* and

$$\mathbf{g}\ (aK) = \mathbf{g}(bK)\ \mathbf{iff}f(a) = f(b)$$

$$f(b)^{-1}f(a) = \mathbf{f}(b)^{-1}f(b)\ \mathbf{iff}f\ (b^{-1}a) = eH\_{\mathbf{p}}$$

$$\mathbf{b}^{-1}\mathbf{a} \in K \text{ or } ke \mathbf{f}\ \mathbf{iff}aK = bK$$

Then, g is an isomorphism and *G*p∕*Ker f* ffi *H*p.

**Lemma 3.2.** Let *G*<sup>p</sup> be a partial group and *H*<sup>p</sup> be a partial subgroup of *G*<sup>p</sup> and *N*<sup>p</sup> be a partial normal subgroup of *G*p. If *H*<sup>p</sup> is a partial normal subgroup of *G*p; then *H*p*N*<sup>p</sup> is a partial normal subgroup of *G*p.

**Proof:** Trivial.

**Theorem 3.9 (Second Isomorphism Theorem for Partial Groups).** Let *G*<sup>p</sup> be a partial group. *H*<sup>p</sup> be a partial subgroup of *G*<sup>p</sup> and *N*<sup>p</sup> be a partial normal subgroup of *G*p, then the following isomorphism holds:

$$H\_{\mathbf{P}}N\_{\mathbf{P}}/N\_{\mathbf{P}} \cong H\_{\mathbf{P}}\left(H\_{\mathbf{P}} \cap N\_{\mathbf{P}}\right),$$

**Proof:** It can be easily seen using First Isomorphism Theorem. So, the proof is left to the reader.

**Theorem 3.10 (Third Isomorphism Theorem for Partial Groups).** Let *G*<sup>p</sup> be a partial group and *H*p,*N*<sup>p</sup> partial normal subgroups of *G*<sup>p</sup> with *H*p≤ *N*<sup>p</sup> Then *H*<sup>p</sup> is also a partial normal subgroup of *N*p.*N*∕*H*<sup>p</sup> is a partial normal subgroup of *G*p∕*H*<sup>p</sup> and also *G*p∕*N*<sup>p</sup> is isomorphic to ð*G*p∕*H*p)∕(*N*p∕*H*p).

**Proof:** Let *define f* : *G*p∕*H*<sup>p</sup> ⟶*G*p∕*N*p, *f aH*ð <sup>p</sup>Þ ¼ *aN* <sup>p</sup> First let show *f* is welldefined: If *aH*<sup>p</sup> = *bH*<sup>p</sup> then we need to prove whether or not *f aH*ð <sup>p</sup>Þ ¼ *f bH*ð <sup>p</sup>Þ. Since *aH*<sup>p</sup> = *bH*<sup>p</sup> then *ab*�1∈ *H*<sup>p</sup> and *H*p≤ *N*p,*ab* �1∈ *N*p. Since *ab* �1∈ *N*<sup>p</sup> then *aN*<sup>p</sup> = *bN*<sup>p</sup> and so that *f aH*ð <sup>p</sup>Þ ¼ *f bH*ð <sup>p</sup>Þ, i.e., *f* is well defined. *f* is homomorphism. And using the First Isomorphism Theorem we can conclude the result.

## **4. Conclusion**

There are some papers such as solvable partial groups, topological structures of partial group, Transitivity Theorem-Thompson Theorem of partial groups, k-partial groups, primitive pairs of partial groups so on related with the partial group in the literature. It is known that every group is partial but the converse is not true. That is why some structures are different from each other for the usual group and partial group.

In this chapter, some structures of partial groups (Clifford Semigroup) are sought to demonstrate algebraically. At the beginning of the chapter (preliminaries section), several fundamental results of partial groups with some numerical examples are given from the literature. For example, if A and B be two usual groups such that the intersection of them is equal to {1 = e}, then the union of subgroups of A and subgroups of B is a partial group. Partial normal groups and partial quotient groups have introduced an analog of the group theory. By using them, a number of isomorphism theorems are proved for partial groups with several other ideas. All results are obtained using closely group theory as algebraic approximations. Readers also may consider/investigate other structures/properties of the partial groups different from the group as algebraically.

### **Conflict of interest**

The author declare no conflict of interest.
