# «Abstract In this paper the following results related to groups are discussed. (i) Every ﬁnite group is isomorphic to a group of even permutations. ...»

Bulletin of the Marathwada Mathematical Society

Vol. 10, No. 1, June 2009, Pages 36–42.

## CONVERSE OF LAGRANGE’S THEOREM

## AND SOLVABLE GROUPS

Salunke J. N.∗and A.R. Gotmare †

Abstract

In this paper the following results related to groups are discussed. (i) Every ﬁnite

group is isomorphic to a group of even permutations. (ii) If a ﬁnite group of permuta- tions has an odd permutation then it has a subgroup of index 2. (iii) Group of order twice an odd number is solvable. (iv) If G is a group of order as product of distinct primes, that is, o( G) is a square free integer, then G is solvable. (v) If G is a ﬁnite non solvable group then 4 |o(G). (vi) Non solvable group has smallest order 60. (vii) If G is a ﬁnite group for which converse of Lagranges theorem is true, then G is solvable, but not conversely. Lastly we obtain three simple alternative proofs of ‘any group of prime power order is solvable’.

## 1 INTRODUCTION

Lagranges theorem: Order of a subgroup of a ﬁnite group divides the order of the group.

Corollary: If a group G is ﬁnite then order of each element of G divides O(G) (= order of G ).

Sylow Theorem: If G is a ﬁnite group and p is a prime number then for non neg- ative integer m satisfying pm |o(G), G has a subgroup of order pm.

From Sylow theorem, the converse of Lagranges theorem is true for groups of prime power order.

Throughout this article G stands for a multiplicative group with identity e or (1) = identity permutation and a0 = e for a ∈ G. The smallest subgroup of G containing S ⊂ G, ∗ *Department of Mathematics, North Maharashtra University,Jalgaon-425 001, India.

† G.D.M. Arts, K,R.N. Commerce and M. D. Science College, Jamner, Dist- Jalgaon 424206.

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## CONVERSE OF LAGRANGES THEOREM AND SOLVABLE GROUPS 37

that is, the subgroup generated by S is, S = ∩{H : His a subgroup of G and S ⊂ H} = {an1 an2 ank |ai ∈ S, ni ∈ Z for 1 ≤ i ≤ k} if S = φ 12k and φ = {e} clearly.A group G is said to be isomorphic to a groupG if there exists a group homomorphism φ : G → G which is one-one and onto.Under this map, o(a) = o(φ(a)) for each a ∈ G, and H is a subgroup (normal subgroup) iﬀ φ(H) is a subgroup (normal subgroup) and O(H) = o(φ(H)). Moreover the relation ‘is isomorphic to’(symbolically denoted by ≈ ) is an equivalence relation on a set of groups. If H is a subgroup of G, K is a normal subgroup of G then, HK|K ≈ H|H ∩ K. If G has only two normal subgroups {e } and

**G, then G is called a simple group. Cayley is best known for the great theorem that:**

“Every ﬁnite group is isomorphic to a group of permutations”.This result is also true for inﬁnite groups. In particular ‘Any group G of order n is isomorphic to a subgroup of Sn ’.For a non empty set S, A(S) = {f |f : S → S is a bijection }is a group under composition of functions. If S = {1, 2, 3,..., n} for some n ∈ N then A(S) is denoted by Sn and o(Sn ) = n!. Note that Sn can be considered as a subgroup of Sn+k for any integer k ≥ 0, where f ∈ Sn can be considered as f ∈ Sn+k such that f (n + j) = n + j if j ≥ 1. The notion of the order of an element and a subgroup are ﬁrstly found in Cauchys paper. Cauchys famous theorem states that, “If the order of a ﬁnite group is divisible by a prime p, then the group has a subgroup (and an element) of order p”. Cauchys role in shaping the theory of permutation groups is central. The two-row notation for permutation groups is central.

The two-row notation for permutations was introduced by Cauchy. He has also deﬁned the product of permutations, inverse permutations, transposition and the cyclic notation. He wrote his ﬁrst paper on this subject in 1815 and in 1844, he proved that ‘every permutation is a product of disjoint cycles’. Disjoint cycles are commutative under the product. f is a cyclic permutation on a set S means there exist elements x1, x2,..., xn ∈ S such that f (x1 ) = x2, f (x2 ) = x3,..., f (xn−1 ) = xn, and all the remaining elements remain ﬁxed under f, that is, f(x) = x for all x ∈ S − {x1, x2,... xn }. Then we write f = (x1, x2,... xn ) and call it as n-cycle or a cycle of length n. Order or Length of any n-cycle is n in a group of permutations. 2-cycles are transpositions. Order of any permutation in Sn is the L.C.M.

of the orders of the disjoint cycles in f. From this if G be a ﬁnite group of permutations and f ∈ G is expressed as product of disjoint cycles c1, c2,..., ck then length of each cycle ci is a factor of o(G). In Sn there are exactly half even permutations and exactly half odd permutations. The set An of all even permutations of Sn (n ≥ 2) is a normal subgroup of Sn, since index of An in Sn is 2. For proof of above results refer [2, 3].

## 2 FINITE PERMUTATION GROUPS

Proposition 2.1 If H is any subgroup of Sn (n ≥ 2) then either all permutations in H are even or exactly half are even.For a proof see Khanna and Bhambri (2002).

38 Salunke J. N.and Gotmare A.R.

Proposition 2.2 Every ﬁnite group is isomorphic to a group of even permutations.

Proof: Let G be a ﬁnite group. By Cayleys theorem G is isophormic to a subgroup H of a group Sn for somr n ∈ N. By proposition 2.1, either all permutations in H are even or exactly half are even. If H ⊆ An. Then the proposition follows. Let H = An then H has exactly half even permutations and exactly half odd permutations. For a = n + 1, b = n + 2; (a, b) is a transposition in Sn+2 and it commutes with every f ∈ Sn. Deﬁne f : H → Sn+2 by φ(f ) = f = (a, b) if f ∈ H is even. If f ∈ H is odd, then φ is a e;roup homomorphism which is one-one and each element of φ(H) is an even permutation. Hence H is isomorphic to φ(H), a subgroup of Sn+2. Therefore G is isomorphic to φ(H), a group of even permutations.

Corollary 2.3 If a ﬁnite group of permutations of order 2 has an odd permutation then it has a subgroup of index 2 and hence the group is not simple.

Proof: Let G be a ﬁnite group of permutations, then it has an odd permutation and o(G) 2. As G has an odd permutation then by proposition 2.1, G has exactly half permutations as even and exactly half as odd. Then H = {f ∈ G|f is even } is a subgroup of G of index 2 and o(H) = o(G) 1. So H is a normal subgroup of G. Hence group G is 2 not simple.

Corollary 2.4 If a ﬁnite group of permutations has an odd permutation then its order is even.

Proof follows from Corollary 2.3.

Corollary 2.5 Any ﬁnite group of odd order is isomorphic to a group of even permutations.

Proof: Let G be a group of odd order and G be isomorphic with a group of permutations ¯ G.

If G contains an odd permutation then by corollary 2.4, 2 |o(G), a contradiction.

Following result is also given in [1].

Corollary 2.6 If a group G has order 2m, where m is an odd number, then G has a subgroup of index 2 and hence G is not simple.

(i) Gi+1 is normal in Gi and (ii) Gi /Gi+1 is an abelian group for each i, 0 ≤ i ≤ n − 1.

The decreasing sequence {Gi }n with (i) and (ii) is called a solvable series for G. Thus G i=0 is a solvable group means it has a solvable series.

If G is a group then its center Z(G) = {x ∈ G|xy = yx ∀ y ∈ G} is a normal abelian subgroup of G. If o(G) = pn where p is prime, n ∈ N then o(Z(G)) = pm where1 ≤ m ≤ n, that is G has a nontrivial subgroup as centre. G is abelian iﬀ Z(G) = G. Let H and K be non empty subsets of a group G. We deﬁne [H, K] = {xyx−1 y −1 |x ∈ H, y ∈ K}. If H = {x} and K = {y} then we deﬁne [x, y] = [H, K]. Let G(0) = G, G(1) = commutator subgroup of G, is the subgroup generated by [G, G], that is the smallest subgroup containing the set {xyx−1 y −1 |x, y ∈ G}, G(2) = [G(1), G(1) ],..., G(i+1) = [G(i), G(i) ] for i ≥ 0. Note that G(i+1) is a normal subgroup of G(i), G(i) /G(i+1) is an abelian group and G is abelian if, and only if, G(1) = e. We state following results (3.1) to (3.14), some of them without proofs.

For their proofs please refer [1].

** Theorem 3.1 Let G be a ﬁnite group of order pq, p and q being distinct primes and p q.**

Then, (i) G has a unique sylow subgroup.

(ii) Any group of order p2 is abelian.

** Theorem 3.2 Any abelian group is solvable.**

That is, non solvable groups are non abelian groups. Converse of (3.2) is not true, as the example S4 shows.

** Theorem 3.3 (i) Any subgroup of a solvable group is solvable.**

(ii) Any homomorphic image of a solvable group is solvable. In particular, any quotient group of a solvable group is solvable.

** Theorem 3.4 Let G be a group, K a normal subgroup such that both K and G/K are solvable.**

Then G is solvable.

We give two alternative proofs of the following result (3.5) which are comparatively simpler.

Proposition 3.5 Any group of prime power order is solvable.

Proof: Let Q0 = G be a ﬁnite group and O(G) = pn, where p is prime and n is a non negative integer.

40 Salunke J. N.and Gotmare A.R.

Method I : We prove the result by induction on n. For n = 0, 1, the group G is cyclic (hence abelian) and thus solvable. For n = 2, G is an abelian (by 3.1(ii)) and hence solvable. Assume that any group of order pk is solvable for 0 ≤ k ≤ n. Let n ≥ 3. If G is abelian then it is solvable. Now consider G as a non abelian group. Then its center Z(G) is a non trivial proper abelian normal subgroup and hence solvable. Then Q1 = G/Z(G) is a group of order pm, where 1 ≤ m n. By induction hypothesis Q1 is solvable. Hence by (3.4), G is solvable.

Method II: By (3.3) and (3.4), G/Z(G) is solvable if and only if, G is solvable. If G is abelian then it is solvable (by 3.2). Suppose G is not abelian. If Q1 = Z(G) is abelian (of order less than o(G) then it is solvable and hence G is solvable. If Q1 is not abelian then consider the quotient group Q2 = Q1 /Z(Q1 ).Q2 is solvable iﬀ Q1 is solvable iﬀ G is solvable. Similarly Q3 = Q2 /Z(Q2 ) is solvable iﬀ G is solvable. In general the group G of prime power order is solvable iﬀ Qk is solvable for any k ≥ 1. Note that for some k ∈ N, k ≤ n, Qk is a trivial group and hence solvable. Hence G is a solvable group.

** Theorem 3.6 Every group of odd order is solvable.**

** Theorem 3.6 is due to W.**

Feit and J.G. Thompson, whose proof given by them is of pages 255, appeared in Paciﬁc Jr. Math. 13, pages 775-1029, and this issue contains only this result with proof. By theorem (3.6) we have obviously (i) Any ﬁnite non abelian simple group is of even order.

(ii) Any ﬁnite non solvable group is of even order.

Now we consider a group G of order pm q n where p, q are primes and m, n are positive integers. If p = q, then G is solvable by (3.5). If both p, q are odd primes then by (3.6), G is solvable. If 2 ∈ {p, q}, p = q then G has a normal subgroup of order pm or q n and hence G is solvable by (3.4).

**Now we state:**

** Theorem 3.7 (Burnsides (p, q) Theorem) Any group of order pm q n (where p, q are primes and m, n ∈ N ) is solvable.**

Proposition 3.8 Let m be an odd number. Then any group of order 2m is solvable.

Proof: Let G be a group of order 2m. Then there is a subgroup H of the group G of index 2 (by 2.6) which is normal and H is solvable by (3.6) as o(H) = m, an odd number.

Quotient group G/H is solvable, since its order is 2, that is, G/H is abelian. By (3.4), G is solvable.

Proposition 3.9 If a ﬁnite group G is not solvable then 4 |0(G).

Proof: 4 | o(G) ⇒ 2 | o(G) or 2|0(G) and4 | o(G) ⇒ G is solvable by (3.6) and (3.8).

Corollary 3.10 Let G be a ﬁnite group whose order is product of distinct (non repeated) primes. Then G is solvable.

Proof: Let G be a group of order p1 p2... pn where p1, p2,..., pn are distinct n prime

## CONVERSE OF LAGRANGES THEOREM AND SOLVABLE GROUPS 41

numbers. Then 2 | 0(G) or 2| 0(G) but4 | o(G). By (3.6) and (3.9), G is solvable.**By (3.7) or by (3.1), (ii),(3.2) and (3.9) we have:**

Any group of order pq, P and q being primes, is solvable.

Corollary 3.11 If G is a group of order pqr where p, q, r are prime numbers, then G is solvable.

Proof follows by (3.10) and (3.7).

Corollary 3.12 Non solvable group with smallest order is 60.

Proof: Alternating groups A5 has order 60 and A5 is non solvable because it is non-abelian simple group.(see [1] Consider any positive integer n with n 60, that is, n 22 × 3 × 5. Then n can not be written as a product of four distinct primes. Thus n has at most three distinct prime factors. Thus if 1 ≤ n 60, n = 2 × 3 × 5 then n is of type pr q s where p, q are primes and r, s are non negative integers. Thus by (3.7) and (3.11) any group with order 60 is solvable.

Note: Obviously by (3.12), all proper subgroups of A5 are solvable. In general, all proper subgroups of any group of order n, 1 n ≤ 119, are solvable(since any proper subgroups of non-trivial group of order ‘119 has order less than 60.). A5 is the smallest non solvable group for which converse of Lagranges theorem is not true.

**Example: Any group G of order n, 60 n 84 or 84 n 120 is solvable.**

Solution: Any integer between 60 and 84 is either odd (61, 63,..., 83) or twice an odd number (62 = 2 × 31, 66, 74, 78, 82) or of the form pm (64 = 26 ) or of the form pm q n (68 = 22 × 17, 72 = 23 × 32, 76 = 22 × 19, 80 = 24 × 5). Therefore by theorems 3.5 to 3.8, any subgroup of order n, 60 n 84 is solvable. Similarly we can show that any group of order n, 84 n 120, is solvable.