THE UPPER BOUND of COMPOSITION SERIES 1. Introduction. a Composition Series, That Is a Series of Subgroups Each Normal in the Pr

THE UPPER BOUND OF COMPOSITION SERIES

ABHIJIT BHATTACHARJEE

Abstract. In this paper we prove that among all ﬁnite groups of or- α1 α2 αr der n∈ N with n ≥ 2 where n = p1 p2 ...pr , the abelian group with elementary abelian sylow subgroups, has the highest number of composi- j Pr α ! Qr Qαi pi −1 ( i=1 i) tion series and it has i=1 j=1 Qr distinct composition pi−1 i=1 αi! series. We also prove that among all ﬁnite groups of order ≤ n, n ∈ N α with n ≥ 4, the elementary abelian group of order 2 where α = [log2 n] has the highest number of composition series.

1. Introduction. A composition series, that is a series of subgroups each normal in the previous such that corresponding factor groups are simple. Any finite group has a composition series. The famous Jordan-Holder theorem proves that, the composition factors in a composition series are unique up to isomorphism. Sometimes a group of small order has a huge number of distinct composition series. For example an elementary abelian group of order 64 has 615195 distinct composition series. In [6] there is an algo- rithm in GAP to find the distinct composition series of any group of finite order.The aim of this paper is to provide an upper bound of the number of distinct composition series of any group of finite order. The approach of this is combinatorial and the method is elementary. All the groups considered in this paper are of finite order. 2. Some Basic Results On Composition Series. We start with some basic results. As they are known, they are given without proof. Theorem. 2.1 Among all finite groups of order n , the abelian group with elenentary abelian sylow subgroups , has the highest number of distinct composition series. 3. Some New Results On Composition Series. In this section, we study some new results on composition series of finite groups.

Deﬁnition. 3.1 We deﬁne CG as the set of all distinct composition series of the group G.

α1 α2 αr Theorem. 3.2 Let n ≥ 2 be a positive integer where n = p1 p2 ...pr . Then Pr ( i=1 αi)! | CZn |= Qr . i=1 αi!

2000 Mathematics Subject Classiﬁcation. 20D20. Key words and phrases. ﬁnite group, composition series, sylow subgroup,elementary abelian group. 1 2 ABHIJIT BHATTACHARJEE

Proof. We know that for each positive divisor d of n there exists a unique subgroup of Zn of order n. Let S be the set of sequence of primes from the set {p1, ..., pr} of length α1 + α2 + ... + αr where pi occurs αi times. Pr ( i=1 αi)! Then | S |= Qr . i=1 αi! Now CZn is the set of distinct composition series of Zn. Now deﬁne a function f : S → CZn by f β1β2...βα1+α2+...+αr−1 βα1+α2+...+αr = {e} E Zβ1 E Z ... Z = Z , β1β2 E E β1β2...βα1+α2+..+αr n where β1, β2, ..., βα1+α2+...+αr are primes such that βi = pi has exactly αi solutions for 1 ≤ i ≤ α1 + α2 + ... + αr.

So β1β2...βα1+α2+...+αr−1 βα1+α2+...+αr ∈ S. Now {e} Z Z ... Z = Z i.e.{e} E β1 E β1β2 E E β1β2...βα1+α2+...+αr n E Z Z ... Z = Z is a composition series of Z where θ1 E θ2 E E θα1+α2+...+αr n n Qi θi = j=1 βj, 1 ≤ i ≤ α1 + α2 + ... + αr.

Then f : S → CZn is an injective mapping by its construction. Now we will prove that it is surjective also.

Let {e} = G0 E G1 E ... E Gα1+α2+...+αr = Zn be a composition series of Zn. Since for each divisor d of n, there exists a unique subgroup of order d of Zn and any subgroup of Zn is cyclic then |Gi| is a prime number. |Gi−1| |Gi| Deﬁne g : CZ → S by g (CZ ) = q1q2...qα +α +...+α , where qi = , n n 1 2 r |Gi−1| for 1 ≤ i ≤ α1 + α2 + ... + αr. Then each qi is a prime for 1 ≤ i ≤ α + α2 + ... + αr.

So q1q2...qα1+α2+...+αr is a sequence of primes such that qi = pi has exactly αi solutions for 1 ≤ i ≤ α1 + α2 + ... + αr. −1 So, q1q2...qα1+α2+...+αr ∈ S and therefore f = g and hence f is surjective and hence bijective also. Pr Pr ( i=1 αi)! ( i=1 αi)! Therefore | S |=| CZn |= Qr i.e. Zn has Qr distinct compo- i=1 αi! i=1 αi! sition series. Example. 3.3 Let us try to understand the above theorem for Z360. 3 2 1 (3+2+1)! Solution. NOTE : 360 = 2 3 5 . Now 3!2!1! = 60 distinct sequence of primes can be formed of length 6 such that each sequence contains 3 times prime 2,2 times prime 3 and 1 time prime 5. Some of them are 232253,323522,523322. Now we will construct a composition series corresponding to these numbers. (1) The composition series corresponding to 232253 will be {e} E Z2 E Z6 E Z12 E Z24 E Z120 E Z360. (2) The composition series corresponding to 323522 will be {e} E Z3 E Z6 E Z18 E Z90 E Z180 E Z360. THE UPPER BOUND OF COMPOSITION SERIES 3

(3) The composition series corresponding to 523322 will be {e} E Z5 E Z10 E Z30 E Z90 E Z180 E Z360. Theorem. 3.4 Let G be an abelian group of order n where n ≥ 2 be a natural Pr α ! α1 α2 αr Qr ( i=1 i) number such that n = p1 p2 ...pr . Then | CG |= ti Qr where ti i=1 i=1 αi! are the numbers of distinct composition series of the sylow pi- subgroups of G. Proof. Note that in ﬁnite abelian group there exists at least one subgroup of order h for each divisor h of the order of the group n and also each composition factor of an abelian group is group of prime order. The order of composition factors of an ﬁntite abelian group is a prime number and using them we can make a sequence of primes which belongs to S ( which is described in the THEOREM 3.2 ). Therefore the number of distinct composition series of G is a multiple of Pr ( i=1 αi)! Qr . i=1 αi! Now let {e} = G0 E G1 E ... E Gm E Gm+1 E ... E Gα1+α2+...+αr = G be a composition series of G where j be the least positive integer such that s1 s1+1 p1 divide | Gj |and k be the least positive integer such that p1 divide | Gk | when 1 ≤ j < k ≤ α1 + α2 + ... + αr , 1 ≤ s < s + 1 ≤ α1. s1 s1 s1 Therefore | Gj |= p1 q where p1 , q ∈ N with gcd (p1 , q) = 1. | Gk |= s1+1 s1+1 s1+1 p1 u where p1 , u ∈ N with gcd p1 , u = 1. s Let Gj = AjBj where Aj and Bj are two subgroups of G of order p1 and q respectively. Let Gk = CkDk where Ck and Dk are two subgroups of G of s1+1 order p1 and u respectively. We will prove that Aj ⊆ Ck ; if not then, ∃ some x ∈ Aj such that x∈ / Ck. v1 x ∈ Aj ⇒ x = p1 . Now x ∈ Aj =⇒ x ∈ Gj, x ∈ Gj =⇒ x ∈ Gj+1 as Gj is a normal subgroup of Gj+1. x ∈ Gj+1 =⇒ x ∈ Gj+2 as Gj+1 is a normal subgroup of Gj+2. So proceeding in this way we can prove that x ∈ Gk. Now Gk = CkDk where Ck ∩ Dk = {e} . x ∈ Gk =⇒ x ∈ CkDk, i.e. x = cd where c ∈ c1 c1 v1 c1 v1−c1 Ck, d ∈ Dk. Therefore x = p1 d, p1 ∈ Ck. So p1 = p1 d i.e. p1 = d, As α1 v1−c1 v1−c1 v1 gcd (p1 , u) = 1 so P1 ∩ Dk = {e} but p1 ∈ P1 ∩ Dk i.e. p1 = e,p1 = c1 c1 p1 hence x (= p1 ) ∈ Ck , a contradiction. Therefore in the composition series of sylow p1- group, Aj is a subgroup of Ck. So if the sylow p1- group has t1 distinct composition series then we Pr ( i=1 αi)! have t1 Qr such possible composition series due to sylow p1- subgroup i=1 αi! in G. In a similar way for the other sylow p2 - group, ..., sylow pr- group we Pr Qr ( i=1 αi)! have ti Qr distinct composition series for G. i=1 i=1 αi! Theorem. 3.5 In an elementary abelian group of order pn, there are k Qn k Qn p −1 k=1(p −1) k=1 p−1 = (p−1)n distinct composition series. 4 ABHIJIT BHATTACHARJEE

2 k−1 k n Proof. Let {e} E p E p E ... E p E p E ... E p be a composition series for the elementary abelian group of order pn. To find the number of distinct composition series, we will find the possible choice from pk−1 to pk at first. k−1 k Let there be Ck−1 choice available for p to p . Now in an elementary abelian group of order pn Qk−1 n i i=0 ((p −1)−(p −1)) k there are Qk−1 k i = A(say) distinct subgroups of order p i=0 (p −p ) each of which is an elementary abelian (By [2], page - 65). In an elementary abelian group of order pn , Qk−2 n i i=0 ((p −1)−(p −1)) also there are Qk−2 k i = B(say) distinct subgroups of order i=0 (p −p ) pk−1 each of which is an elementary abelian(By [2], page-65). Again in an elementary abelian group of order pk Qk−2 k i i=0 ((p −1)−(p −1)) k−1 there are Qk−2 k i = D(say) distinct subgroups of order p i=0 (p −p ) each of which is elementary abelian (By [2],page-65). Now there is an identical algebraic structure of any two subgroups of same order in an elementary abelian group i.e. in elementary abelian group of order pn , any two distinct subgroups of order pk have the same number of subgroup order pk−1. A.B pn+1−k−1 Then Ck−1 = D = p−1 . Qn−1 pn+1−k−1 As a result we have k=1 p−1 distinct possible choice for the composition series of an elementary abelian group of order pn ( as k ≥ 1 ) which n+1−k k Qn k Qn p −1 Qn p −1 k=1(p −1) is same as k=1 p−1 = k=1 p−1 = (p−1)n . Corollary. 3.6 Let G be an abelian group with elementary abelian sylow α1 α2 αr subgroups of order n ≥ 2 where n = p1 p2 ...pr , j Pr α ! Qr Qαi pi −1 ( i=1 i) then | CG |= Qr i=1 j=1 pi−1 i=1 αi! Proof. Which is an obvious result followed from THEOREM 3.4 and THE- OREM 3.5. α1 α2 αr Corollary. 3.7 Let G be a finite group of order n where n = p1 p2 ...pr , j Pr α ! Qr Qαi pi −1 ( i=1 i) then | CG |≤ Qr and the equality is hold if and i=1 j=1 pi−1 i=1 αi! only if G is an abelian group with elementary abelian sylow subgroups. Proof. Which is an obvious result followed from THEOREM 2.1 and COROLLARY 3.6. j Lemma. 3.8 Let X = Qα1+k (2i −1)α !α ! and Y = Qαr ( p −1 )(α +α )! i=α1+1 1 r j=1 p−1 1 r αr ;where p ≥ 5 is a prime and α1 ≥ 0 is an integer ; αr ∈ N, such that t = p THE UPPER BOUND OF COMPOSITION SERIES 5

X and [log2 t] = k . Also αr ≥ 2 whenever p = 3. Then Y is a monotone increasing function of α1. Qα1+k i i=1 (2 −1) Proof. Let f(α1) = j Qα1 (2i−1) Qαr ( p −1 ) (α1+αr)! i=1 j=1 p−1 α1!αr! Qα1+k+1 i i=1 (2 −1) then f(α1 + 1) = j . Qα1+1(2i−1) Qαr ( p −1 ) (α1+αr+1)! i=1 j=1 p−1 (α1+1)!αr! α +k+1 f(α1+1) 2 1 −1 α1+1 Therefore = α +1 . f(α1) 2 1 −1 α1+αr+1 α +k+1 In order to prove f(α + 1) > f(α ) we have to prove that 2 1 −1 > 1 1 2α1+1−1 α1+αr+1 . α1+1 α +1+k 2 1 −1 k α1+1+αr As α +1 > 2 and α + 1 ≥ . 2 1 −1 r α1+1 k So it is sufficient enough to prove 2 > αr + 1 ⇐⇒ k > log2(αr + 1). Now log2 = k + f where 0 < f < 1. So k = log2 −f . Therefore we have to prove that log t−f > log (α +1) ⇐⇒ log ( t ) > 2 2 r 2 αr+1 f. As 0 < f < 1 it is sufficient enough to prove log ( t ) > 1 as 2 αr+1 log x is a monotone increasing function for x > 0.i.e. t > 2 ⇐⇒ pαr > 2α + 2 for all prime p ≥ 5 with α ≥ 1 and p = 3 αr+1 r r with αr ≥ 2 , which can easily be proved using mathematical induction. X Hence Y is a monotone increasing function of α1. Theorem. 3.9 Among all p groups of order ≤ n , where n ≥ 4 is a positive α integer ; the elementary abelian group of order 2 where α = [log2 n] , has the highest number of composition series. Proof. We know that among all finite p− groups of equal order the elementary abelian group has the highest number of composition series. α Let G2 be an elementary abelian group of order 2 where α = [log2 n]. β Let Gp be an elementary abelian p group of order p where β = logp n with p ≥ 3 be a prime and n ≥ 4. Now we know that if logq r = t then logq qr = t + 1 for all prime q ≥ 2 and r, t ∈ N with q ≤ r. Obviously 2 divides p − 1. [logp n] [log2 n] p −1 Therefore we get 2 − 1 > p−1 whenever n ≥ 4...(A). log n i Q[log2 n] i Q[ p ] p −1 | CG2 |= i=1 2 − 1 , | CGp |= i=1 p−1 . log n i Q[log2 n] i Q[ p ] p −1 From ( A ) it implies that i=1 2 − 1 > i=1 p−1 i.e.| CG2 |>|

CGp |. Therefore among all p groups of order ≤ n , the elementary abelian group α of order 2 has the highest number of composition series where α = [log2 n]. Theorem. 3.10 Let n ≥ 4 be a positive integer. Among all ﬁnite groups of order ≤ n, the elementary abelian group of order 2α has the highest number of composition series where α = [log2 n]. 6 ABHIJIT BHATTACHARJEE

Proof. n ≥ 4 is a natural number and α = [log2 n] and let G be the elemen- α Q[log2 n] i tary abelian group of order 2 then | CG |= i=1 2 − 1 . As we know that among all ﬁnite groups of order n ≥ 2 , the abelian group with elementary abelian sylow subgroups has the highest number of composition series. So in order to prove our claim we have to prove that j Pq β ! Q[log2 n] i Qq Qβi pi −1 ( i=1 i) i=1 2 − 1 ≥ i=1 j=1 Qq pi−1 i=1 βi! Qq βi where each pi is a prime and βi, q ∈ N such that i=1 pi ≤ n and the equality is hold if and only if q = 1, p1 = 2 and β1 = [log2 n]. The inequality will be proved in the following steps. STEP 1 : Let H be an abelian group of order m with elementary abelian α1 α2 αr sylow subgroupssuch that 4 ≤ m ≤ n where m = p1 p2 ...pr also | CH |denotes the number of distnct composition series of H. αr Let pr be a prime such that pr 6= 2 and pr = t, [log2 t] = k. Then k αr 2 < pr . 0 |H|.2k Let H be an abelian group of order αr with elementary abelian sylow pr subgroups. 0 Then | H |<| H |. We will prove that | CH0 |>| CH |. 0 k α1 αr−1 STEP 2 : If | H |= 2 p1 ...pr−1 i.e. 2 is not a divisor of m then we have k > αr i and Qk 2k − 1 > Qαr pr−1 . i=1 i=1 pr−1 Therefore, obviously | CH0 |>| CH |. α1 α2 αr 0 α1+k α2 αr−1 STEP 3 : If | H |= 2 p2 ...pr then | H |= 2 p2 ...pr−1 . In this case we will prove that | CH0 |>| CH |. αr j Pr α ! Qα1 i Qαr pr −1 Qr−1 Qαi pi −1 ( i=1 i) | CH |= 2 − 1 Qr . i=1 i=1 pr−1 i=2 j=1 pi−1 i=1 αi! j Pr−1 α +k ! Qα1+k i Qr−1 Qαi pi −1 ( i=1 i ) | C 0 |= 2 − 1 r−1 . H i=1 i=2 j=1 pi−1 Q i=2 αi!(α1+k)! Qα1+k i (2 −1)(k+s)!α1!αr! |CH0 | i=α1+1 Pr−1 Therefore, = j where s = i=1 αi. |CH | Qαr pr−1 (α1+k)!(αr+s)! j=1 pr−1 So in order to prove our claim we have to prove that Qα1+k 2i − 1 (k + s)!α !α ! > i=α1+1 1 r j Qαr pr−1 (α + k)! (α + s)! ... ( A ) . j=1 pr−1 1 r Pr−1 NOTE : s = i=1 αi ≥ α1, =⇒ s = α1 + a where a ≥ 0 , k > αr =⇒ k = αr + b,b > 0; a, b ∈ Z. Then inequality ( A ) reduces to Qα1+k 2i − 1 (α + α + a + b)!α !α ! > i=α1+1 1 r 1 r j Qαr pr−1 (α + α + a)! (α + α + b)! ... ( B ) j=1 pr−1 1 r 1 r [ Here we are ignoring the case pr = 3 with αr = 1, as m ≥ 4 at least one other case must exist. ] THE UPPER BOUND OF COMPOSITION SERIES 7

NOTE : (α1+αr+a+b)!α1!αr! is a monotone increasing function of a for (α1+αr+a)!(α1+αr+b)! fixed α1, αr, b. So it is sufficient enough to prove the inequality for a = 0 j i.e. Qα1+k 2i − 1 α !α ! > Qαr pr−1 (α + α )! ... ( C ) . i=α1+1 1 r j=1 pr−1 1 r j Let X = Qαr+k (2i − 1)α !α ! and Y = Qαr ( pr−1 )(α + α )! . i=α1+1 1 r j=1 pr−1 1 r X Then according to LEMMA 3.8 Y is a monotone increasing function of α1 whenever pr and αr are fixed. So it is sufficiently enough to prove the inequality ( C ) for α1 = 0 j i.e.Qk 2i − 1 > Qαr pr−1 which is following form THEOREM 3.9. i=1 j=1 pr−1 As a result we prove that | CH0 |>| CH |. Now we will repeat the same process in the group H0 and get a group H00 such that | CH00 |>| CH0 |. Since | H |is finite then | H |has only finitely many prime divisors . So after a finite number of steps we get an elementary abelian group of η order 2 where η ≤ [log2 n]. α 1 STEP 4 : Let | H |= 2 1 3 for any α1 ∈ N . Qα1 i Then | CH |= i=1 2 − 1 (α1 + 1). Then we can find an elementary abelian group H0 such that| H0 |= 2α1+1 Qα1+1 i and | CH0 |= i=1 2 − 1 α1+1 Obviously | CH0 |>| CH | as 2 > α1 + 2 for each α1 ∈ N. Among all 2 groups of order ≤ n ,G has the highest number of composition series which proves the theorem. 4. ACKNOWLEDGEMENT. I would like to thank Dr. Nisith Chandra Das , Dr. Shobhan Mondal , Mr. Subhajit Jana for their help which made it possible.

5. APPENDIX. Lemma. 5.1Let G be a ﬁnite solvable group and M be a maximal normal subgroup of G , then G/M is a cyclic group of prime order and hence abelian. Proof. Let G be a ﬁnite solvable group and M be a maximal normal subgroup of G. As G is solvable, both M and G/M are solvable. Now G/M is simple as well as solvable group hence it is abelian also ( a non abelian simple group cannot be solvable ). Hence a simple as well as abelian group must be a cyclic group of prime order. CORRESPONDENCE THEOREM. Theorem. 5.2 Given G.N, denote the the family of subgroups of G that contain N by sub(G; N) and the family of subgroups of G/N by sub(G/N). Then there is a bijection φ : sub(G; N) −→ sub(G/N),H 7→ H/N which preserves all subgroup lattic and normality relationships. 8 ABHIJIT BHATTACHARJEE

Proof. It is a very well known theorem and can be found any standared book on Group Theory. Theorem. 5.3 Among all finite solvable groups of order n , the abelian group with elementary abelian sylow subgroups has the highest number of composition series. Proof. Let K be the abelian group of order n with elementary abelian sylow subgroups. It is sufficient enough to prove, among all finite solvable groups of order n , K has the highest number of maximal normal subgroup. Among all finite p−groups, the elementary abelian group has the highest number of maximal normal subgroups.(By [2], page-65). Now any finite abelian group ( also for nilpotent ) is the direct product of its sylow subgroups and any maximal normal subgroup has prime index. As a result, among all finite abelian groups ( also for nilpotent ) of ordrer n , K has the highest number of maximal normal subgroups. Now let G be a solvable ( not necessarily abelian ) group of order n. Let G has m distinct maximal normal subgroups namely H1,H2, ..., Hm. 0 0 Let C be the commutator subgroup of G. Then C E Hi for 1 ≤ i ≤ m. ( As G/Hi is abelian from the lemma ) . m 0 Let H = ∩i=1Hi . Then H is a normal subgroup of G such that C E H . Therefore G/H is abelian . Now by Correspondence Theorem , there exists a bijection φ : sub(G; H) −→ sub(G/H), which preserves all subgroup lattic and normality relationship. Therefore G/H is an abelian group with exactly m maximal normal subgroups. But it is already proved that among all finite abelian groups of order n, K has the highest number of maximal normal subgroup, which proves the theorem.

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a student of Msc , Dept of Mathematics, INSTITUTE OF SCIENCE ; BA- NARAS HINDU UNIVERSITY, VARANASI ; INDIA-221005 Email address: [email protected]