Review Article - European Journal of Applied Engineering and Scientific Research ( 2018) Volume 6, Issue 1

- Corresponding Author:
- M Bello

Mathematics and Computer Science Department

Federal University Kashere, Gombe State, Nigeria

This paper involves an investigation on the maximal subgroups of the groups of order pqr, a theorem has been stated and prove concerning the nature and behavior of the maximal subgroup of such groups. The group algorithm and programming (GAP) has been applied to enhance and validate the results.

Element, Abelian, Permutation groups

The concept of maximal subgroup is playing a vital role in the application of group series more especially in determining the solvability and nil potency of a group. This research has investigated the maximal subgroups of a permutation group of unique order and carries out research on the behavior of the maximal subgroup.

*Definition 1.1*

Let G be a group. A subgroup H of G is said to be a maximal subgroup of G if H ≠ G and there is no subgroup K ? G such that H<K<G. Note that a maximal subgroup of G is not maximal among all subgroups of G, but only among all proper subgroups of G. For this reason, maximal subgroups are sometimes called maximal proper subgroups.

Similarly, a normal subgroup N of G is said to be a maximal normal subgroup of G if N ≠ G and there is no normal subgroup K of G, such that N<K<G.

*Definition 1.2*

A subgroup N of a group G is normal in G if the left and right cossets are the same, that is if gH = Hg ∀g ∈G and a subgroup H of G.

*Definition 1.3*

The factor group of the normal subgroup N in a group G written as G/N is the set of cosets of N in G.

*Definition 1.4*

The factor group of the normal subgroup N in a group G written as G/N is the set of cosets of N in G.

*Definition 1.5 [1]*

A solvable group is a group having a normal series such that each normal factor is abelian.

*Definition 1.6 [2]*

A group G is nilpotent if it has a normal series G=G_{0} ≤ G_{1} ≤ G_{2} ≤…≤ G_{n}=(1) where G_{i}/G_{i+1} ≤ Z(G/G_{i+1}) (centre of G/ G_{i+1}).

*Definition 1.7*

A finite group is simple when its only normal subgroups are the trivial subgroup and the whole group.

*Definition 1.8 [3]*

A group is a p-group if the order is a prime or prime power.

*Definition 1.9*

Let G be a group, and let p be a prime number. A group of order Pk for some k ≥ 1 is called a p-group.

*Theorem 2.1 (Lagrange’s theorem)*

Let G be a finite group and H ≤ G. Then

**Proof**

Let G be a finite group and let H ≤ G such that For any g ? G, define a map Ø_{g}:H → gH by Ø_{g} (h)=gh. This map is clearly surjective.

Furthermore, for any distinct h_{1}≠h_{2} ? H, we have that gh_{1} ≠ gh_{2}. Thus Ø_{g} is a bijection,

It is easy to see that the set of left cossets of H in G form a partition of G. First note that for any g ? G, g ? gH.

Thus, and clearly Therefore,

To show that left cossets are disjoint, suppose that for distinct cossets g_{1}H ≠ g_{2}H ? G/H.

There exists an element x ? g_{1}H ∩ g_{2}H. Then there exist elements h_{1},h_{2} ? H such that, x=g_{1}h_{1}=g_{2}h_{2}.

Therefore, g_{1}=g_{2}h_{2}h_{1}^{-1}, so for any g_{1}h ? g_{1}H,

g_{1}h=(g_{2}h_{2}h_{1}^{-1})h= g_{2}(h_{2}h_{1}^{-1}h)? g_{2}H

Hence, g_{1}H ? g_{2}H. But we have seen that , so g_{1}H = g_{2}H, which is a contradiction. Thus, the k left cosets of H in G are in fact disjoint and, hence, partition G. Since each has cardinality m, it follows that . Therefore divides and .

**Example**

If =14 then the only possible orders for a subgroup are 1, 2, 7 and 14.

*Theorem 2.2 (Cauchy’s theorem for abelian groups)*

Let A be a finite abelian group. If p is a prime number that divides its order, then A must have an element of order p.

*Theorem 2.3 (Sylow’s first theorem)*

Let G be a finite group and p be a prime number such that its power by α is the largest power that will divide Then there exist at least one subgroup of order P^{α}. Such groups are called Sylow p-subgroups.

*Theorem 2.4 (Sylow’s second theorem)*

Let n_{p} be the number of sylow p-subgroups of a finite group G. Then n_{p≡}1 mod.

*Theorem 2.5 (Sylow’s third theorem)*

Any two Sylow p-subgroups are conjugate.

*Theorem 2.6 (Sylow’s fourth theorem)*

Any p-subgroup B is contained in a Sylow p-subgroup.

**Example**

Let G be a group of order 12 [4-6]. Then either G has a normal Sylow 3-subgroup, or else it’s isomorphic to A_{4}.

Reason: 12=22 × 3. We know n_{3} has to divide 2^{2}=4 and it also has to be congruent to 1 mod 3. So it can be either 1 or 4.

If n_{3}=1, then G has a normal Sylow 3-subgroup.

If n_{3}=4, then we know that the four Sylow 3-subgroups are acted on by G, by conjugation. Let’s call the set S={P_{1},… , P_{4} } . The action of G gives us a homomorphism Ø : G → S 4 .

We will first show that Ø is injective, then we will show that the image of Ø is A4. This will show that G ≡ imØ = A^{4} .

To show Ø is injective, we need to show that ker Ø =1.

We know that for eachi, so we have here that,

Since P_{i} ≤ N_{G} (P_{i}) and is also 3, it means Pi= NG (Pi). So in our case,

The Pi’s happen, in this case, to be distinct groups of prime order (their order is 3). A general and useful fact about distinct groups of the same prime order is that they can only intersect each other trivially. (Take for example two subgroups P_{1} and P_{2} of order p, then the subgroup P_{1}∩P_{2} has o have order 1 or p. If it has order p then P_{1}=P_{2}, so if P_{1} and P_{2} are not the same subgroup P_{1}∩P_{2} has to have order 1, i.e., it’s the trivial subgroup.)

Applying this to our case we get ker Ø =1. Therefore Ø is injective, and G ≡ imØ

Now G has 4 subgroups, P_{1},…,P_{4}, of order 3. Each of these subgroups has two elements of order 3 and the identity element.

The two elements of order three have to be different for each P_{i} (since different P_{i} ‘s had only the identity element in common).

Therefore G contains 8 different elements of order 3.

Since G is isomorphic to imØ, these 8 different elements of order 3 have to map to 8 different elements of order 3 in S_{4}. The only elements of order three in S_{4} are 3-cycles. And 3-cycles are even permutations, so are elements in A_{4}.

So A_{4} ∩ imØ is a subgroup of both A_{4} and imØ with at least 8 elements. But since both A_{4} and imØ have 12 elements, this intersection subgroup has to also divide 12. The only factor of 12 that is greater than or equal to 8 is 12. So A_{4} ∩ imØ is a subgroup of both A4 and imØ of size 12, and since A_{4} and imØonly have 12 elements anyway, it means A_{4} ∩ imØ = A_{4}=imØ.

**Example**

Let G be a group of order 351. Then G has a normal Sylow p-subgroup for some prime p dividing 351.

Reasoning: 351=3^{3} × 13. So a Sylow 3-subgroup would have order 33=27, and a Sylow 13-subgroup would have order 13. Let us start out with what n_{13} can be n_{13} divides 27, and

n_{13}=1mod 13. Only two possibilities: n_{13}=1 or 27.

If n_{13}=1, then the Sylow 13-subgroup is a normal subgroup of G, and we are done.

If n_{13}=27, then we are going to show that there can only be room for one Sylow 3-subgroup, and therefore the Sylow 3-subgroup is normal in G. We’ll use the fact that distinct subgroups of order p for some prime p can only have the identity element in their intersection.

(Suppose P_{1} and P_{1} are subgroups of order p. Then P_{1}∩ P_{2} ≤ P_{1} and P_{1}∩ P_{2} ≤ P_{1}. So must be either 1 or p, and the only way it can be p is if P_{1}∩P_{2} =P_{1} and P_{1}∩P_{2} =P_{2}, making P_{1}=P_{2}).

Therefore if P_{1} and P_{2} are not the same subgroup, their intersection has order 1, so contains only the identity element. Do be warned, though, that this is only true about subgroups of prime order, so this argument wouldn’t work if, say, the Sylow 13-subgroups had order 13^{2}.

Since the Sylow 13-subgroups are subgroups of order 13, they can only intersect each other at the identity element. Also, every element of order 13 forms a subgroup of order 13, which has to be one of the Sylow 13-subgroups. Each Sylow 13 subgroup contains 12 elements of order 13 (every element except for the identity). There are 27 Sylow 13 sub-groups, so there are a total of 27 × 12=324 elements of order 13 in G.

This leaves 351-324=27 elements of G that do not have order 13. Since a Sylow 3-subgroup would have to have exactly 27 elements in it, this means that all these 27 elements form a Sylow 3-subgroup, and it must be the only one (since there are not any extra elements of G to use). So n3=1 and this Sylow 3-subgroup must be normal in G.

*Proposition 2.7 [5]*

A group G is a direct product of subgroups H_{1},H_{2} if and only if

a) G=H_{1}H_{2}

b) H_{1} ∩ H_{2}={e} and

c) Every element of H_{1} commutes with every element of H_{2}

**Proof**

If G is the direct product of H_{1} and H_{2}, then certainly (a) and (c) hold and (b) holds because, for any g ? H_{1}∩H_{2}, the element (g, g-1) maps to e under (H_{1},H_{2})→H_{1}H_{2} and so equals to (e,e).

Conversely, (c) implies that (H_{1},H_{2})→H_{1}H_{2} is a homomorphism and (b) implies that it is injective:

H_{1}H_{2}=e →H_{1}=H_{2}^{-1} ? H_{1} ∩ H_{2}={e}. Finally, (a) implies that it is surjective.

A group G is a direct product of subgroups H_{1},H_{2} if and only if

a) G=H_{1},H_{2}

b) H_{1}∩ H_{2}={e}

c) H_{1} and H_{2} are both normal in G

**Proof**

Certainly, these conditions are implied by those in proposition 1.37 and so it remains to show that they imply that each element H_{1} of H_{1} commutes with each element H_{2} of H_{2}. Two elements H_{1} and H_{2} of G commute if and only if their commutator e but

Which is in h_{2} because h_{2} is normal and is in H_{1} because H_{1} is normal. Therefore (b) implies that [H_{1},h_{2}]=e.

*Proposition 2.9 [9]*

(i) Subgroups H ⊂ G and quotient groups G/K of a solvable group G are solvable.

If normal subgroup N ? G is solvable and if the quotient G/N is solvable then G is solvable.

**Proof**

H(n) ⊂ G(n) for all n. Also G(n) maps onto (G/N)(n) for all n.

If (G/N)(n) ={1} then G(n) ⊂ N. So if N(m)={1} then G_{(N+M)} ⊂ N_{(M)}={1}

*Corollary 2.10 [8]*

All p-groups are solvable.

**Proof**

Let P be a p-group. We will induct on n, where

If = p, then P is cyclic. Therefore, P is abelian and, hence, solvable. Suppose that for all 1 ≤ k < n, groups of order P^{k} are solvable.

Let = p^{n} By Corollary 1, Z. (P) ≠ 1 Thus, P / Z (P) = p^{k} (for some k<n). So, by the inductive hypothesis, P=Z(P) is solvable. Since Z P is abelian, then Z(P) is solvable. Thus, by Proposition 1, P is solvable. Therefore, by induction, all p-groups are solvable.

*Theorem 2.11 [9]*

A finite group is nilpotent if and only if it is a direct product of its Sylow subgroups.

**Proof**

Finite p-groups are nilpotent by theorem 1.4 and hence so are any direct product of p-groups.

Now suppose that G is a finite nilpotent group and let P be any Sylow subgroup of G. Let N=N_{G} (P). Suppose N<_{G}. Then by Corollary 1 N<N_{G} (N).

Let x ? N_{G} (N)-N Hence x^{-1}Px ≤ x^{-1}NX=N and so x^{-1}Px is a Sylow subgroup of N as is P. It follows that y^{-1}x^{-1}Pxy=P for some y ? N and so xy ? N.

But y∈ N and x∉ N a contradiction. It must therefore be that N=G and so P is normal in G. G is therefore the direct product of its Sylow subgroups.

*Theorem 2.12 [3]*

Finite p-groups are nilpotent.

**Proof**

Z_{r+1}(G)/Z_{r}(G)=Z(G/Z_{r}(G)). Since the centre of a non-trivial p-group is non-trivial Z_{r} (G)<Z_{r+1} (G) unless Z_{r} (G)=G .

*Proposition 2.13 [3]*

Every nilpotent group is solvable.

**Proof**

If G is nilpotent group then the upper central series of G.

is a normal series.

Moreover for every i we have,

Z_{i}(G)/Z_{i-1}=Z(G/Z_{i-1}(G))

So all quotients of the upper central series are abelian.

*Theorem 2.14*

A normal subgroup N of a group G is a maximal normal subgroup if and only if the quotient G/N is simple.

*Theorem 2.15*

Let G be a group of order pq^{r} where p=r and p,q,r distinct primes. The maximal subgroups of G are of order q^{r} and pq^{r-1} which are solvable.

**Proof**

Suppose G is a group of order pqr. By theorem 2.1, the possibility for the order of subgroups of G are p, pq, pq^{2}, …, pq^{r-1}, pq^{r}, q, q2,…, q^{r-1}, q^{r}. It is clear that pq^{r-1} and qr are the maximal subgroups of G since they are not contained in any other proper subgroups of G.

It remains to show that the two maximal subgroups are solvable and nilpotent. q^{r} is solvable by Corollary 2.10.

What left is to show that pq^{r-1} is solvable. Let G_{1} be a group of order pq^{r-1}, the sylow p and q subgroups of G_{1} are normal in the group. Let H_{1} be the Sylow p-subgroup of G_{1} then, . Since q is prime then G_{1}/H_{1} and H_{1} are solvable by corollary 2.10, it follows that G1 is solvable by (ii) of proposition 2.

*Consider the permutation groups C _{1} and D_{1}*

C_{1}={(1),(123),(132)},D_{1}={(1),(46)} acting on the sets S_{1}={123} and Δ_{1}={46}, respectively.

Let

We can easily verify that G_{1} is a group with respect to the operations

The wreath products of C_{1} and D_{1} is given by W_{1} as it appears in the result validation which is given by the (GAP) below:

gap>C1:=Group ((1, 2, 3));

Group ([(1, 2, 3)])

gap>D1:=Group ((4, 6));

Group ([(4, 6)])

gap>W1:=Wreath Product (C1, D1);

Group ([(1, 2, 3), (4, 5, 6), (1, 4) (2, 5) (3, 6)])

gap>Maximal Subgroup Class Reps (W1);

[Group ([ (1, 3, 2) (4, 6, 5) (1, 2, 3) (4, 6, 5)]), Group ([(1, 4) (2, 5) (3, 6), (1, 2, 3) (4, 6, 5)]), Group ([(1, 4) (2, 5) (3, 6), (1, 3, 2) (4, 6, 5)])]

gap>M1:=Group ([(1, 3, 2) (4, 6, 5) (1, 2, 3) (4, 6, 5)]);

Group([(1, 3, 2) (4, 6, 5) (1, 2, 3) (4, 6, 5)])

gap>Is Solvable (M1);

true

gap>Is Nilpotent (M1);

true

gap>Is Normal (W1, M1);

true

gap>Is Simple (M1);

false

gap>Is Transitive (M1);

false

gap>Is Primitive (M1);

false

gap>M2:=Group ([(1, 4) (2, 5) (3, 6) (1, 2, 3) (4, 6, 5)]);

Group ([(1, 4) (2, 5) (3, 6) (1, 2, 3) (4, 6, 5)])

gap>Is Solvable (M2);

true

gap>Is Nilpotent (M2);

false

gap>Is Normal (W1, M2);

true

gap>Is Simple (M2);

false

gap>Is Transitive (M2);

true

gap>Is Primitive (M2);

false

gap>M3:=Group ([(1, 4) (2, 5) (3, 6) (1, 3, 2) (4, 6, 5)]);

Group ([(1, 4) (2, 5) (3, 6) (1, 3, 2) (4, 6, 5)])

gap>Is Solvable (M3);

true

gap>Is Nilpotent (M3);

true

gap>Is Normal (W1, M3);

false

gap>Is Simple (M3);

false

gap>Is Transitive (M3);

true

gap>Is Primitive (M3);

false

gap>quit;

*Consider the permutation groups C _{2}and D_{2}*

C_{2}={(1), (15432), (14253), (13524), (12345)}

D_{2}={(1), (678), (687)} acting on the sets S_{2} ={1, 2, 3, 4, 5} and Δ_{2} ={6, 7, 8}, respectively.

Let

We can easily verify that G1 is a group with respect to the operations

The wreath products of C_{2} and D_{2} is given by W_{1} as it appears in the result validation which is given by the (GAP) below:

gap>C2:=Group ((1, 2, 3, 4, 5));

Group ([(1, 2, 3, 4, 5)])

gap>D2:=Group ((6, 7, 8));

Group ([(6, 7, 8)])

gap>W2:=Wreath Product (C2, D2);

Group ([(1, 2, 3, 4, 5) (6, 7, 8, 9, 10) (11, 12, 13, 14, 15) (1, 6, 11) (2, 7, 12) (3, 8, 13) (4, 9, 14) (5, 10, 15)])

gap>Maximal Subgroup Class Reps (W2);

[Group ([(1, 3, 5, 2, 4) (6, 8, 10, 7, 9) (11, 13, 15, 12, 14) (1, 2, 3, 4, 5) (11, 15, 14, 13, 12) (1, 5, 4, 3, 2) (6, 7, 8, 9, 10)])

Group ([ (1, 6, 11) (2, 7, 12) (3, 8, 13) (4, 9, 14) (5, 10, 15) (1, 2, 3, 4, 5) (11, 15, 14, 13, 12) (1, 5, 4, 3, 2) (6, 7, 8, 9, 10)])

Group ([(1, 6, 11) (2, 7, 12) (3, 8, 13) (4, 9, 14) (5, 10, 15) (1, 3, 5, 2, 4) (6, 8, 10, 7, 9) (11, 13, 15, 12, 14)])]

gap>M4:=Group ([(1, 3, 5, 2, 4) (6, 8, 10, 7, 9) (11, 13, 15, 12, 14) (1, 2, 3, 4, 5) (11, 15, 14, 13, 12) (1, 5, 4, 3, 2) (6, 7, 8, 9, 10)]);

Group ([(1, 3, 5, 2, 4) (6, 8, 10, 7, 9) (11, 13, 15, 12, 14) (1, 2, 3, 4, 5) (11, 15, 14, 13, 12) (1, 5, 4, 3, 2) (6, 7, 8, 9, 10)]) gap>Is Solvable (M4);

true

gap>Is Nilpotent (M4);

true

gap>Is Normal (W2, M4);

true

gap>Is Simple (M4);

false

gap>Is Transitive (M4);

false

gap>Is Primitive (M4);

false

gap>M5:=Group ([(1, 6, 11) (2, 7, 12) (3, 8, 13) (4, 9, 14) (5, 10, 15) (1, 2, 3, 4, 5) (11, 15, 14, 13, 12) (1, 5, 4, 3, 2) (6, 7, 8, 9, 10)]);

Group ([(1, 6, 11) (2, 7, 12) (3, 8, 13) (4, 9, 14) (5, 10, 15) (1, 2, 3, 4, 5) (11, 15, 14, 13, 12) (1, 5, 4, 3, 2) (6, 7, 8, 9, 10)])

gap>Is Solvable (M5);

true

gap>Is Nilpotent (M5);

false

gap>Is Normal (W2, M5);

true

gap>Is Simple (M5);

false

gap>Is Transitive (M5);

true

gap>Is Primitive (M5);

false

gap>M6:=Group ([(1, 6, 11) (2, 7, 12) (3, 8, 13) (4, 9, 14) (5, 10, 15) (1, 3, 5, 2, 4) (6, 8, 10, 7, 9) (11, 13, 15, 12, 14)]);

Group ([(1, 6, 11) (2, 7, 12) (3, 8, 13) (4, 9, 14) (5, 10, 15) (1, 3, 5, 2, 4) (6, 8, 10, 7, 9) (11, 13, 15, 12, 14)])

gap>Is Solvable (M6);

true

gap>Is Nilpotent (M6);

false

gap>Is Normal (W2, M6);

true

gap>Is Simple (M6);

false

gap>Is Transitive (M6);

true

gap>Is Primitive (M6);

false

gap>quit;

*Consider the permutation groups C _{3} and D_{3}*

C_{3}={(1), (15432), (14253), (13524), (12345)}

D_{3}={(1), (678), (687)} acting on the sets S_{3} ={1, 2, 3, 4, 5} and Δ_{3} ={6, 7, 8}, respectively.

Let

We can easily verify that G_{1} is a group with respect to the operations

The wreath products of C_{3} and D_{3} is given by W_{3} as it appears in the result validation which is given by the (GAP) below:

gap>C3:=Group ((1, 2, 3));

Group ([(1, 2, 3)])

gap>D3:=Group ((4, 5, 6, 7, 8));

Group ([ (4, 5, 6, 7, 8)])

gap>W3:=Wreath Product (C3, D3);

Group ([(1, 2, 3) (4, 5, 6) (7, 8, 9) (10, 11, 12) (13, 14, 15) (1, 4, 7, 10, 13) (2, 5, 8, 11, 14) (3, 6, 9, 12, 15)])

gap>Maximal Subgroup Class Reps (W3);

[Group ([ (1, 4, 7, 10, 13) (2, 5, 8, 11, 14) (3, 6, 9, 12, 15) (1, 2, 3) (13, 15, 14) (1, 2, 3) (10, 12, 11) (1, 2, 3) (7, 9, 8) (1, 3, 2) (4, 5, 6) ])

Group ([(1, 3, 2) (4, 6, 5) (7, 9, 8) (10, 12, 11) (13, 15, 14) (1, 2, 3) (13, 15, 14) (1, 2, 3) (10, 12, 11) (1, 2, 3) (7, 9, 8) (1, 3, 2) (4, 5, 6)])

Group ([(1, 4, 7, 10, 13) (2, 5, 8, 11, 14) (3, 6, 9, 12, 15) (1, 3, 2) (4, 6, 5) (7, 9, 8) (10, 12, 11) (13, 15, 14)])]

gap>M7:=Group ([(1, 4, 7, 10, 13) (2, 5, 8, 11, 14) (3, 6, 9, 12, 15) (1, 2, 3) (13, 15, 14) (1, 2, 3) (10, 12, 11) (1, 2, 3) (7, 9, 8) (1, 3, 2) (4, 5, 6)]);

Group ([(1, 4, 7, 10, 13) (2, 5, 8, 11, 14) (3, 6, 9, 12, 15) (1, 2, 3) (13, 15, 14) (1, 2, 3) (10, 12, 11) (1, 2, 3) (7, 9, 8) (1, 3, 2) (4, 5, 6)])

gap>Is Solvable (M7);

true

gap>Is Nilpotent (M7);

false

gap>Is Normal (W3, M7);

true

gap>Is Simple (M7);

false

gap>Is Transitive (M7);

true

gap>Is Primitive (M7);

false

gap>M8:=Group ([(1, 3, 2) (4, 6, 5) (7, 9, 8) (10, 12, 11) (13, 15, 14) (1, 2, 3) (13, 15, 14) (1, 2, 3) (10, 12, 11) (1, 2, 3) (7, 9, 8) (1, 3, 2) (4, 5, 6)]);

Group ([(1, 3, 2) (4, 6, 5) (7, 9, 8) (10, 12, 11) (13, 15, 14) (1, 2, 3) (13, 15, 14) (1, 2, 3) (10, 12, 11) (1, 2, 3) (7, 9, 8) (1, 3, 2) (4, 5, 6)])

gap>Is Solvable (M8);

true

gap>Is Nilpotent (M8);

true

gap>Is Normal (W3, M8);

true

gap>Is Simple (M8);

false

gap>Is Transitive (M8);

false

gap>Is Primitive (M8);

false

gap>Maximal Subgroup Class Reps (W3);

[Group ([ (1, 4, 7, 10, 13) (2, 5, 8, 11, 14) (3, 6, 9, 12, 15) (1, 2, 3) (13, 15, 14) (1, 2, 3) (10, 12, 11) (1, 2, 3) (7, 9, 8) (1, 3, 2) (4, 5, 6)])

Group ([(1, 3, 2) (4, 6, 5) (7, 9, 8) (10, 12, 11) (13, 15, 14) (1, 2, 3) (13, 15, 14) (1, 2, 3) (10, 12, 11) (1, 2, 3) (7, 9, 8) (1, 3, 2) (4, 5, 6)])

Group ([(1, 4, 7, 10, 13) (2, 5, 8, 11, 14) (3, 6, 9, 12, 15) (1, 3, 2) (4, 6, 5) (7, 9, 8) (10, 12, 11) (13, 15, 14)])]

gap>M9:=Group ([(1, 4, 7, 10, 13) (2, 5, 8, 11, 14) (3, 6, 9, 12, 15) (1, 3, 2) (4, 6, 5) (7, 9, 8) (10, 12, 11) (13, 15, 14)]);

Group ([(1, 4, 7, 10, 13) (2, 5, 8, 11, 14) (3, 6, 9, 12, 15) (1, 3, 2) (4, 6, 5) (7, 9, 8) (10, 12, 11) (13, 15, 14)])

gap>Is Solvable (M9);

true

gap>Is Nilpotent (M9);

true

gap>Is Normal (W3, M9);

false

gap>Is Simple (M9);

false

gap>Is Transitive (M9);

true

gap>Is Primitive (M9);

false

gap>

My acknowledgements go the Allah that gives me the knowledge and the opportunity to write this paper. I also want to acknowledge Federal University Kashere for funding this research.

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