Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak
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