Inter national Journal of Pure and Applied Mathematics Volume 113 No. 10 2017, 37 – 46 ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version) url: http://www.ijpam.eu Special Issue
ijpam.eu
Counting Subgroups of a Non-Abelian p-group Zpn o Zp
4
Amit Sehgal1 ,Sarita Sehgal2 , P.K. Sharma3 and Manjeet Jakhar4 1 Ch. Bansi Lal University, Bhiwani, Haryana, India amit sehgal
[email protected] 2 Govt. College, Matanhail (Jhajjar), Haryana, India
[email protected] 3 D.A.V. College, Jalandhar City, Punjab, India
[email protected] Department of Mathematics, NIILM University, Kaithal, Haryana, India
[email protected] January 27, 2017 Abstract The main goal of this paper is to count subgroups which are isomorphic to cyclic p-group, internal direct product of two cyclic p-group or semi direct product of two cyclic p-group of the non-Abelian p-group Zpn o Zp , n ≥ 2 where p may be even or odd prime, by using simple-theoretical approach. AMS Subject Classification: Primary 20D60; Secondary 20D15 Key Words and Phrases: subgroup, cyclic subgroups, internal direct product, semi-direct product, number of subgroups
1
Introduction
One of the famous problems in combinatorial group theory of 20th century as well as 21th century is to count subgroups of non-abelian finite p-group. In Burnside [10], if p is a prime greater than two, then there are 15 groups of order p4 up to isomorphism. Five of these are abelian and remaining ten are the non-abelian groups. Out of these
ijpam.eu
37
2017
International Journal of Pure and Applied Mathematics
Special Issue
15 groups, only ten are non-abelian groups and one of non-abelian group is given below: 3
2
Zp3 o Zp = ha, b : ap = bp = e, bab−1 = a1+p i n
Extended form of above said group can be written as Zpn o Zp = ha, b : ap = bp = n−1 e, bab−1 = a1+p i which is a non-abelian p-group of order pn+1 for n ≥ 2 where p may be even or odd prime. In present work, by using simple number-theoretical approach, the authors deduce formula for the number of subgroups isomorphic to cyclic p-group, internal direct product of two cyclic p-group or semi direct product of two cyclic p-group of the p-group Zpn o Zp n ≥ 2 where p may be odd or even prime. One can easily find n−1 n−1 out that hap i and hbaip i where i = 1, 2, 3, ...., p, are cyclic subgroups of order p in group Zpn o Zp except for the group Z22 o Z2 . Firstly we partition the set S (non-trivial cyclic subgroups of Zpn o Zp except for the group Z22 o Z2 ) into (p+1) partitions.
1.1
Notations and assumptions:-
(i) N (H, G) is denote the number of subgroups of G isomorphic to H. (ii) Zpn o Zp , n ≥ 2 except for the Z22 o Z2 is denoted by G. n n−1 (iii) Zpn o Zp is denoted by ha, b : ap = bp = e, bab−1 = a1+p i n−1 n−1 (iv) Rewrite ba = a1+p b as ab = ba(p−1)p +1 (v) s = (p − 1)pn−1 + 1
2
Partition the set S (non-trivial cyclic subgroups of G)
Define two cyclic subgroups H and K in S are equivalent, denoted by H ∼ K, if and only if H ∩ K contains a subgroup of order p (clearly such subgroup is unique and cyclic). Theorem 1. relation on S.
The relation ∼ between elements of the S is an equivalence
Proof. Obvious. Corollary 2. An equivalence relation ∼ on a non-empty set S partitions the set S into the disjoint union of distinct equivalence class.
ijpam.eu
38
2017
International Journal of Pure and Applied Mathematics
Special Issue
Here group G has only p + 1 cyclic subgroups of order p , using above theorem we can partition set S into (p + 1) distinct equivalence class and these partition are as follows: n−1 n−1 (a)hap i = {H ∈ S : H ∼ hap i} and denoted by class-0. n−1 n−1 (b) hbaip i = {H ∈ S : H ∼ hbaip i} and denoted by class-i where (1 ≤ i ≤ p).
3
Number of cyclic Subgroups of order pα (α ≥ 1) in each class
Theorem 3. Prove that there exist unique cyclic subgroup of order p from n−1 class-0 can be represent as < ap > and cyclic subgroups of order pα (α ≥ 2) from j pn−α class-0 can be represent as < b a > where j = 1, 2, ..., p from G. Proof. Suppose that H =< bj ak > be a cyclic subgroup of order pα in class-0 n−1 α−1 n−1 from G, then H ∼< ap >. It means o(bj ak ) = pα and (bj ak )p ∈< ap >. Firstly we find possibilities for j when value of j lies between 1 to p. α−1 α−1 (pα−1 −1)j +...+s2j +sj +1] n−1 Here (bj ak )p = bjp ak[s ∈< ap >=⇒ p|jpα−1 . On the basis of j, we have two cases : Case-1: If j = p Therefore H becomes H =< ak > which cyclic subgroup of order pα in class-0 from G. Now we find possibilities for k when value of k lies between 1 to pn . We have n o(ak ) = pα ⇒ (ppn ,k) = pα ⇒ pn−α = (pn , k) ⇒ possibilities for k are pα−1 (p − 1). Each cyclic subgroup of order pα contains pα−1 (p − 1) elements of order pα . Therefore, number of cyclic subgroups in class-0 of order pα is pα−1 (p−1)/pα−1 (p − 1) = 1. n−α This subgroup can be written as < aβp > where (pα , β) = 1, it is easy to check βpn−α pn−α that < a >=< a >. Hence, we get cyclic subgroup of order pα in class-0 n−α from G which is < ap >. Case 2: If (j, p) = 1 If α = 1, then p . jpα−1 , hence this case is not possible. Hence, number of cyclic n−1 subgroups of order p from group in class-0 is one and which is hap i. Now only possibility left which is α > 1. We have (bj ak )p = bjp ak[s
(p−1)j +....+s2j +sj +1]
= ak[s
(p−1)j +....+s2j +sj +1]
Subcase 1:- If p is even prime Here possibility for j is only 1, now we find possibilities for k when value of k lies j between 1 to pn . (bj ak )2 = ak[s +1]
ijpam.eu
39
2017
International Journal of Pure and Applied Mathematics
Special Issue
n−1
Here sj = 2n−1 + 1, then (bj ak )2 = ak[2 +2] . It is given that o(bj ak ) = 2α , hence o (bj ak )2 =
2α (2α ,2)
= 2α−1 .
2α = 2α−1 (1) (2n , k[2n−1 + 2]) It given that n > 2 when p=2, hence 2n , k[2n−1 + 2] = 2n , 2k . n From (1) (2n2,2k) = 2α−1 =⇒ 2n−α = (2n−1 , k). Hence, the possibilities for k when value of j=1 in this case given are 2α−1 . Therefore, possible element of order 2α (α > 1) from group Z2n o Z2 (n > 2) which generates a cyclic subgroup of class-0 of order 2α are 1 × 2α−1 = 2α−1 . Each cyclic subgroup of order 2α contains 2α−1 (2 − 1) elements of order 2α . Thereα−1 fore, number of cyclic subgroups in class-0 of order 2α is 22α−1 = 1. This subn−α group can be written as hbaβ2 i where (2α−1 , β) = 1 , it is easy to check that n−α n−α hbaβ2 i = hba2 i. n−α Hence, we get cyclic subgroup of order 2α in class-0 from Z2n o Z2 which is ha2 i. o ak[2
n−1 +2]
= 2α−1 =⇒
Subcase 2:- If p is odd prime (p−1)j+.....+s2j +sj +1 ] (b a ) = ak[s , j n−1 Here s ≡ [p (p − 1)j + 1](modpn ) P (p−1)j +.....+s2j +sj +1] k[p+pn−1 p−1 j=1 j] . (bj ak )p = ak[s = a P n ). We have k[p + pn−1 (p − 1) p−1 j=1 j] ≡ kp(modp α j k α j k p It is given that o(b a ) = p , henceo((b a ) = (ppα ,p) = pα−1 Hence, j k p
n
o akp = pα−1 =⇒
pn = pα−1 (pn , kp)
(2)
From (2) (pnp kp) = pα−1 =⇒ pn−α = (pn−1 , k) Hence, possibilities for j are p-1 and possibilities for k when value of j is given in this case are pα−1 (p − 1). Therefore, possible element of order pα (α > 1) from group Zpn o Zp n ≥ 2 which generates a cyclic subgroup of class-0 of order pα are (p − 1) × pα−1 (p − 1) = pα−1 (p − 1)2 . Each cyclic subgroup of order pα contains pα−1 (p−1) elements of order pα . Therefore, α−1 2 number of cyclic subgroups in class-0 of order pα is ppα−1(p−1) = p − 1. (p−1) α Hence, number of cyclic subgroups of order p (n ≥ α > 1) from G in class-0 are 1+ (p-1) when 1 < α ≤ n. T heref ore, number of cyclic subgroups of order pα (n ≥ α > 1) f rom G in class − 0 are 1or p according as α = 1 or 1 < α ≤ n.
ijpam.eu
40
2017
International Journal of Pure and Applied Mathematics
Special Issue
T his subgroup can be written as hbj aβp n−α n−α that hbj aβp i = hbj ap i.
n−α
i where (pα−1 , β) = 1, it is easy to check
Corollary 3.1. Prove that every cyclic subgroup from class-0 is normal subgroup of G. Proof. Assume H is cyclic subgroup from class-0. n−1
Case 1:-If H is a cyclic subgroup of order p Therefore, H = hap i. Assume x be an arbitrary element of H and y be an arbitrary element of G, then n−1 x = akp and y = bα aβ . So n−1 n−1 n−1 n−1 yxy −1 = bα aβ akp a−β b−α = bα akp b−α = akp (1+p ) ∈ H. Hence H is normal subgroup of G. Case 2:-Since H is a cyclic subgroup of order pα (n ≥ α > 1) n−α Therefore,H = hbj ap iwherej = 1, 2, 3, .., p. Assume x be an arbitrary element of H and y be an arbitrary element of G, then n−α x = (bj ap )k and y = bα aβ . yxy −1 = bα aβ (bj ap
n−α
)k a−β b−α = (bα aβ bj ap
n−α −β
b−α )k = (bj a[p
n−α −β(sj −1)]sp−α
)k (3)
Here sj ≡ [pn−1 (p − 1)j + 1](mod pn )and sp−α ≡ [pn−1 α+ 1](mod pn ) Hence (pn−α − β(sj − 1) sp−α ≡ pn−α − βpn−1 (p − 1)j sp−α (mod pn )
Subcase 1:- If β = p, then pn−α − β(sj − 1) sp−α ≡ pn−α sp−α (mod pn ) p−α n−α p−α n−α From (3) becomes yxy −1 = (bj as p )k ∈ hbj as p i where (sp−α , p) = 1 n−α Thus we have yxy −1 ∈ hbj ap i. Sub Case 2:-If (β, p) = 1, then p−α n−α j (p − β(s − 1) s ≡ pn−α 1 − βpα−1 (p − 1)j sp−α (mod pn ) Here (sp−α , p) = 1 and ((1−βpα−1 (p−1)j), p) = 1, hence (1−βpα−1 (p−1)j)sp−α , p) = 1, Now equation (3) becomes α−1 p−α α−1 p−α yxy −1 = (bj a[1−βp (p−1)j]s )k ∈ hbj a[1−βp (p−1)j]s i where ((1 − βpα−1 (p − 1)j), p) = 1. n−α Thus finally we get yxy −1 ∈ hbj ap i. Theorem 4. Number of cyclic subgroups of order pα (n ≥ α ≥ 1) from G in class-i other than class-0 is one or zero according as α = 1 or α > 1 respectively n−1 and hbaip i is the only subgroup in class-i other than class-0.
ijpam.eu
41
2017
International Journal of Pure and Applied Mathematics
Special Issue
Proof. Let H = hbj ak i be a cyclic subgroup of order pα in class-i other than classn−1 α n−1 0 from G, then H ∼ hbaip i. It means o(bj ak ) = pα and (bj ak )p −1 ∈ hbaip i. α−1 α−1 α−1 (p −1)j +···+sj +1] n−1 Here (bj ak )p = bjp ak[s ∈ hbaip i =⇒ p - jpα−1 =⇒ α = 1 and (j, p) = 1. There is no cyclic subgroup of order pα (n ≥ α > 1) from Zpn o Z p (n ≥ 2) except Z22 o Z2 in class-i other than class-0. Now possibilities left out α = 1 and (j, p) = 1 Now we can say that H = hbj ak i with o(bj ak ) = p and (j, p) = 1. We claim that pn−1 | k (p−1)j +···+sj +1] Then (bj ak )p = bjp ak[s = e =⇒ pn | k[s(p−1)j + · · · + sj + 1] Here sj = [(p − 1)pn−1 + 1]j ≡ [j(p − 1)pn−1 + 1](modpn ) (4) (p − 1)jpn−1 ] ≡ kp(mod pn ) Then k[s(p−1)j + · · · + sj + 1] ≡ k[p + p p−1 2 pn | kp =⇒ k = rpn−1 (mod pn ), where 1 ≤ r ≤ p
(5)
By use of (j, p) = 1, we have a unique solution of jx ≡ 1(mod p) say j0 . (bj ak )j0 = bjj0 ak[s
(j0 −1)j +···+sj +1]
= b1 ak[s
(j0 −1)j +···+sj +1]
∈ hbaip
n−1
i
=⇒ k[s(j0 −1)j + · · · + sj + 1] ≡ ipn−1 (mod pn )
By using (4) and (5) in the above equation, we get rpn−1 [j0 + jj0 (
j0 − 1 )(p − 1)pn−1 ] ≡ ipn−1 (mod pn ) 2
=⇒ rpn−1 j0 ≡ ipn−1 (mod pn ) =⇒ rj0 ≡ i(mod p) =⇒ rj0 j ≡ ij(mod p) =⇒ r ≡ ij(mod p) (6) ipn−1 j rpn−1 Therefore we have hba i = hb a i where r ≡ ij(mod p). Hence we conclude that number of cyclic subgroup of order pα (n ≥ α ≥ 1) from G in class-i other than class-0 is one or zero according α = 1 or α > 1 respectively and n−1 hbaip i is the only subgroup in class-i other than class-0.
4
Main Theorem
Theorem 5. Number of subgroups isomorphic to Zpα × Zp is 1 or 0 according as 1 ≤ α < n or α = n and isomorphic to Zpα oZp is is 0 or 1 according as 1 ≤ α < n or α = n from G.
ijpam.eu
42
2017
International Journal of Pure and Applied Mathematics
Special Issue
Proof. As we have discussed in Section 3 that non-trivial cyclic subgroup H of order pα (n ≥ α ≥ 1) from G becomes normal subgroup if and if this belongs to class-0. We know that intersection of two non-trivial cyclic subgroups of G is identity if and only both belongs to different class. Hence, we have to choose a non-trivial cyclic subgroup K from class other than class-0 so that product of two subgroups H and K is subgroup because H is normal subgroup of groupG. Hence, order K is p. Choice’s of H can be classified into three cases: Case-1: If order of H is p n−1 >. Total possibilities for H (belong to class-0) is 1 which is < ap n−1 Total possibilities for K (belong to class-i except class-0) are p which are < baip >. 2 Hence HK contains p elements and HK is subgroup. Now we show that HK = H × K. HK = {xj y k |x = ap n−1
Now yx = baip ap Hence HK becomes
n−1
n−1
, y = baip
= ba(i+1)p
n−1
n−1
n−1
, ba = a1+p
= ap
n−1 (1+pn−1 )
{xj y k |xy = yx, o(x) = p, o(y) = p} =< ap n−1
n−1
b, o(x) = p, o(y) = p}
baip
n−1
= ap
n−1
× < baip
n−1
n−1
baip
n−1
= xy.
>∼ = Zp × Zp .
n−1
n−1
We can easily check that < ap > × < baip >=< ap > × < bap > when i 6= 1. n−1 n−1 Therefore, < ap > × < bap > is unique subgroup in which order of H is p and this is isomorphic to Zp × Zp . Case-2: If order of H is pα (n > α > 1) n−α Total possibilities for H (belong to class-0) is p which is < bj ap > where j = 1, 2, ..., p. Total possibilities for K (belong to class-i except class-0) are p which are n−1 < baip >. Hence HK contains pα+1 elements and HK is subgroup. Now we show that HK = H × K. HK = {xj y k |x = bj ap
n−α
, y = baip
n−1
,ba = a1+p
n−1
b,
o(x) = pα , o(y) = p}.
n−1
Here ab = ba(p−1)p +1 and now assume s = (p − 1)pn−1 + 1. Hence sj = [(p − 1)pn−1 + 1]j ≡ [j(p − 1)pn−1 + 1](modpn ). Now yx = baip
n−1
xy = bj ap
ijpam.eu
bj ap
n−α
n−α
= b1+j ap
n−1 (isj +1)
n−1
= bj+1 ap
n−1 (s+i)
baip
= b1+j ap
= bj+1 ap
43
n−1 (ij(p−1)pn−1 +i+1)
n−1 ((p−1)pn−1 +1+i)
= b1+j ap
= b1+j ap
n−1 (i+1)
n−1 (i+1)
.
2017
International Journal of Pure and Applied Mathematics
Special Issue
Hence HK becomes {xj y k |xy = yx, o(x) = pα , o(y) = p} =< bj ap n−α
n−α
> × < baip
n−1
n−1
>∼ = Zpα × Zp .
n−α
n−1
We can easily check that < bj ap > × < baip >=< bap > × < bap > when i 6= 1∀ j. n−α n−1 Therefore, < bap × < bap > is unique subgroup in which order of H is pα (n > α > 1) and this is isomorphic to Zpα × Zp . Case 3: If order of H is pn Total possibilities for H (belong to class-0) is p which is < bj a1 > where j = 1, 2, ..., p. n−1 Total possibilities for K (belong to class-i except class-0) are p which are < baip >. (n+1) n+1 Hence HK contains p elements and HK is subgroup. But Group itself has p elements; hence HK must be group itself. Therefore, < a > o < b > is unique subgroup in which order of H is pn and this is isomorphic to Zpn o Zp . Theorem 6. Show that for Zpn o Zp except Z4 o Z2 where p is any prime as N (H, Zpn o Zp ) is if H ∼ = Z0 or Zpk × Zp or Zpn o Zp where k = 1, 2, ..., n − 1 1 ∼ p + 1 if H = Zp p if H ∼ = Zpk wherek = 2, 3, ..., n. Proof. By use of theorem 3, 4 and 5, we get required result.
References [1] Amit, Sehgal and Yogesh Kumar, On the number of subgroups of finite abelian Zm ⊗ Zn , International Journal of Algebra, 7 no. 19 (2013), 915-923. [2] Grigore Calugareanu, The total number of subgroups of a finite abelian group, Scientiae Mathematicae Japonicae 60 no. 1 (2004), 157-167. [3] J. A. Gallian, Contemporary Abstract Algebra, Narosa, (1999). [4] Laszlo Toth, On the number of cyclic subgroups of a finite Abelian group, Bull. Math. Soc. Sci. Math. Roumanie, 55(103) (2012), 423–428. [5] Marius Tarnauceanu, An arithmetic method of counting the subgroups of a finite abelian group, Bull. Math. Soc. Sci. Math. Roumanie Tome 53(101) No. 4, (2010), 373–386.
ijpam.eu
44
2017
International Journal of Pure and Applied Mathematics
Special Issue
[6] R. Schmidt, Subgroup Lattices of Groups, de Gruyter Expositions in Mathematics 14, de Gruyter, Berlin, (1994). [7] Amit Sehgal , Sarita Sehgal and P.K. Sharma, The number of subgroups of a finite abelian p-group of rank two, Journal for Algebra and Number Theory Academia, 5(1) (2015), 23–31. [8] J. Petrillo, Counting subgroups in a direct product of finite cyclic groups, College Math J., 42 (2011), 215–222. [9] M. Hampejs, and Laszlo Toth, On the subgroups of finite Abelian groups of rank three, Annales Univ. Sci. Budapest., Sect. Comp., 39 (2013), 111–124. [10] William Burnside, Theory of groups of finite order, Cambridge University Press, first edition, (1897). Reprinted (2010) through Nabu Press.
ijpam.eu
45
2017
46
