Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-II
51
If r = 1, then G is simple and so is the only composition series of G. So let r > 1 and assume by
induction that the result holds for any group having some composition series of length less than r.
If G 1
= H 1
, then G 1
has two composition series of respective length r – 1 and s – 1. Therefore by induction we
see that r = s and two composition series of G 1
and equivalent. Hence G has two composition series which
are equivalent.
Therefore, we suppose G
≠ H .
1 1
But is simple, so G1 ≰ H1
, hence
and so
because G H is simple.
1
Let K = G 1
∩ H 1
∆ G. Now
G G1 H1
H1
H1
= ≅ = and
G1
G1
G1 ∩ H1
K
G G1 H1
G1
G1
= ≅ =
H1
H1
G1 ∩ H1
K
Θ K ∆ G and G has a composition series,
HG∆( 1 ) ∆ GH, H G, ∴ G1 H1
∆ G.
1 H<
= 1 1 , K has a composition series, say
∴Θ
GK = KG 0
> ∆GK 1 1
> ∆K 1∆ > K− − ∆ − − > − − K t ∆ = K( s
1 –
). = ( )
G1
NG''∆
0 1 1 2
1
G 1
now have two composition series
G 1
∆G 2
∆G 3
∆ − − − − ∆G r
= ( 1) and
G 1 ∆K ∆K 1 ∆K 2 ∆ − − − − ∆K t = ( 1).
These are of lengths r – 1 and t + 1, respectively. By induction, we get t = r – 2 and that the series are
equivalent. Similarly, H 1
has two composition series:
H 1 ∆H 2 ∆ − − − − ∆H s = ( 1) and
H1 ∆K ∆K1 ∆K2 ∆ − − − − Kr
– 2
= ( 1) ( Θ t = r – 2)
These have respective length s – 1 and r – 1, so by induction we see r = s and the series are equivalent.
We now conclude that the composition series
and G = H0 ∆H1 ∆K ∆K1 ∆ − − − − ∆ K s – 2
= ( 1)
are equivalent, because we have proved above
G G
1
H
≅
1
K and