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Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

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52<br />

ADVANCED ABSTRACT ALGEBRA<br />

.<br />

Hence we finally conclude that our two initial composition series of B are equivalent.<br />

Definition:<br />

A series of sub groups<br />

G = G0 ∆ G1 ∆ G2 ∆ − − − − ∆ G s<br />

= ( 1) of group G is called a subnormal series of G if Gi<br />

+1<br />

∆ Gi<br />

for<br />

each i.<br />

A subnormal series is called a normal series of G if<br />

Solvable groups:<br />

First we define Commutators in a group G. Let a, b<br />

∈ G. The element<br />

Commutator and is denoted by [a, b]. The Commutator [a, b] = 1, only when ab = ba.<br />

is called a<br />

–<br />

[ a, b] 1 = [ b, a],<br />

i.e., the element, inverse to the Commutator is itself a Commutator. But a product of<br />

Commutators need not be a Commutator. Thus, in general, the set of Commutators of a group is not a sub<br />

group. The smallest sub group G 1<br />

of the group G containing all Commutators is called its Commutator sub<br />

group. Note that the commutator sub group G 1<br />

is the set of all possible products of the form [a 1<br />

, b 1<br />

] ---- [a r<br />

,<br />

b r<br />

], where a , b ∈ G , and r is a natural number. From<br />

i<br />

i<br />

which, as a consequence, implies that G 1<br />

∆ G.<br />

Remarks:<br />

1. The commutator sub group G ' of an abelian group is trivial.<br />

2. The Commutator sub group of S n<br />

is A n<br />

, n ≥ 1.<br />

3. The Commutator sub group of GL (n, F) is SL (n, F), F is a field.<br />

4. The Commutator sub group A ' n of A n<br />

is A n<br />

, is A ' n = A n<br />

, because the non-commutative group A n<br />

,<br />

has no non-trivial proper normal subgroups.<br />

Theorem 4:<br />

The Commutator sub group G ' of a group G is the smallest among the normal sub group H of the group G for<br />

which is an abelian group.<br />

b<br />

C<br />

n<br />

G⇔<br />

a<br />

a,<br />

a<br />

Proof:<br />

The Commutator [xH, yH] = [x, y] H<br />

is trivial<br />

is<br />

is abelian<br />

From [a, b] 2 = [a g , b g ], a, b, g ∈ G, we get the second Commutator sub group G", i.e. the Commutator sub

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