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 permutations 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 6 |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
The beginning of the ideas of group theory lay in the solution of equations. Galois
(1811 – 1832)simplified his solution in about 1830 and discovered a great deal about groups
in connection with the solution of equations, besides investigating normal subgroups and
the theory of fields. The theory of groups was elaborated by Lagrange (1789 – 1813), Caley
(1821– 1895) and particularly Cauchy (1789 – 1857) in about (1844 –1846) and by many
other mathematicians.
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 negative 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.
36
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 1 an2 2 ank k |ai ∈ S, ni ∈ Z for 1 ≤ i ≤ k} if S 6= φ
and
< φ > = {e} clearly.
A group G is said to be isomorphic to a group G 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 6= 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)
2 > 1. So H is a normal subgroup of G . Hence group G is
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
Ḡ .
If G contains an odd permutation then by corollary 2.4, 2 6 |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.
Proof: G is isomorphic with a subgroup Ḡ = {tg : g ∈ G} of A(G) where tg (x) = gx for
all x ∈ G under the map g → tg . [Refer proof of Cayleys theorem from any reference book].
By Cauchys theorem, G has an element ‘a ’of order 2. Hence o(ta ) = 2.
Observe that ta : G → G has no fixed element and ta is a product of disjoint cycles,
so each cycle in this product must be a transposition (otherwise o(ta ) 6= 2). Thus ta is the
product of m disjoint transpositions, that is, ta is an odd permutation. By corollary 2.3, G
and hence G has a subgroup of index 2.
CONVERSE OF LAGRANGES THEOREM AND SOLVABLE GROUPS
3
39
SOLVABLE GROUPS
A group G is said to be solvable if there exists a decreasing sequence{Gi }ni=0 of subgroup of G as,
G = G0 ⊇ G1 ⊇ G2 ⊇ · · · ⊇ Gi ⊇ Gi+1 ⊇ . . . Gn = e such that,
(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 }ni=0 with (i) and (ii) is called a solvable series for G . Thus G
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 6= 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 6 |0(G).
Proof: 4 6 | o(G) ⇒ 2 6 | o(G) or 2|0(G) and4 6 | 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 6 | 0(G) or 2| 0(G) but4 6 | 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 6= 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.
Alternately we can prove it by using (3.7) and (3.9). The integers between 84 and 120
which are divisible by 4 are 88 = 23 × 11, 92 = 22 × 23, 96 = 25 × 3, 100 = 22 × 52 , 104 =
23 × 13, 108 = 22 × 33 , 112 = 24 × 7, 116 = 22 × 29. All these are of the forms pm q n where
p, q are primes.
Proposition 3.13 Let G be a finite group for which converse of Lagranges theorem for
finite group is true. Then G is solvable.
Proof:
If 4 6 | o(G) then G is solvable by (3.9). Consider 4| o(G). Then o(G) = 2n q
where integer n ≥ 2 and q is an odd number. Here G has subgroups (by hypothesis)
G1 , G2 , . . . , Gn with o(G1 ) = 2n−1 q, o(G2 ) = 2n−2 q, . . . , o(Gn ) = q and G ⊇ G1 ⊇ G2 ⊇
· · · ⊇ Gn . Now o(Gn ) is odd, so Gn is solvable and hence Gn has a solvable series of
subgroups {Gk }m
k=n as Gn ⊇ Gn+1 ⊇ Gn+2 . . . Gm = {e} with abelian groups Gk /Gk+1 ,
since o(Gk /Gk+1 ) = 2, for n ≤ k ≤ m − 1. Hence, G = G0 ⊇ G1 ⊇ G2 ⊇ · · · ⊇ Gm = {e}
is a solvable series for the group G . Therefore, G is solvable. If G is any group of order
42
Salunke J. N.and Gotmare A.R.
pn where p is a prime number and n ∈ N, then (by sylow theorem) converse of Lagranges
theorem holds for G . Hence by (3.7), the group G is solvable. This proves (3.13).
Proposition 3.14 If H and K are normal solvable subgroups of a group then H K is solvable.
Generalization of this result is : Product of finite number of normal solvable subgroups of
a group is solvable.
Proposition 3.15 If H, K are solvable subgroups of a group and H or K is normal then
HK is solvable.
Proof: Suppose H is a normal solvable subgroup. Then HK|H ≈ K|H ∩ K and K|H ∩ K
is solvable by 3.3 (ii).
HK|H and H are solvable gives HK is solvable by (3.4)
Generalization of the result (3.15) is :
Corollary 3.16 Let H1 , H2 , . . . , Hn be solvable subgroups and also all these are normal
except possible one, then the product H1 H2 . . . , Hn is a solvable subgroup.
Proof:
Suppose H2 , H3 , . . . , Hn are normal solvable subgroups. Then K = H2 H3 . . . Hn
is a normal solvable subgroup by (3.14) and H1 is solvable subgroup. By (3.15), H1 H2 . . . Hn =
H1 K = KH1 is a solvable subgroup.
Note: Converse of the result (3.13) is not true. For example, the group A4 is solvable
(see [1]). Also 6 is a proper divisor of o(A4 ) = 12, but there is no subgroup of A4 of order
6. For if there is a subgroup H of order 6 in A4 , then [A4 : H] = 2 and hence H will be
normal in A4 . Since there are only six elements in H, and there are eight 3-cycles in A4 , H
must contain at least one 3-cycle. But H being normal in A4 , H will then contain all the
3-cycles which is impossible since H has only six elements. For details refer [3]. Thus the
converse of the Lagranges theorem does not hold for the solvable group A4 .
References
[1] Gopalkrishnan N.S., University Algebra, second edition, New Age International (P)
Limited, Publishers, New Delhi, 2001.
[2] Herstein I. N.,
Topics in Algebra, Blaisdell, New York 1965.
[3] Khanna V .K. and Bhambri S. K., A Course in Abstract Algebra, second revised
edition, Vikas publishing house, Pvt, Ltd, New Delhi 2002.