Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-I
33
This is impossible. So nontrivial elements of N must have cyclic decomposition involving cycles of length 2 or
3. Moreover, such elements can not involve just one 3 – cycle – otherwise by squaring we would contain a 3
– cycle in N.
Assume that N contains a permutation π = ( a, b, c) ( a', b', c ') − − − − (with disjoint cycles). Then N contains
b
Θ ( a', c, b') ( a, b, c) ( a', b', c') − − − − ( a', b', c) = ( a', a, b) ( c, c', b')
Hence N contains
which is impossible. Hence each element of N is a product
of an even number of disjoint 2 – cycles.
g
If
by π.
then N contains
for all c unaffected
Hence N contains
It follows that if
then
c
b
l q l qb
( –
, ', ) ', ', c ', –
) 1
H
Θ π 1' ) π' =
≠ π( ' = a π' , ∈= ac b , ,( Nb ) a ),
( ca b b , 1 ,') c π
b) a , π( ∈a c', ,( ) a')
, N( b', ) ca ') − ( ba , −( ', ) a', b− c , = b') ) c−
') )( = − a( , a − , c', b− cb (, −) ') a, b−', () = a') −b
,(')
b −a c a, −a , b ,
1
= 1( 1), 1 ( 12The 2) ( 34 2number ) , 3 H23 of = 2 – 4( cycles 1), 4 ( 12being ) ( 34)
') c at ', )( ba ( least c ') , ', c −) ', b−4.
= ') b' ( −a −, b− , −c
− ) . = ( a, a', c, b, c') − − − − − − − − .
But then N will also contain
π = ( a , b ) ( a , b ) π ( a , b ) ( a , b )
3 2 2 4 2 1 3 2
g
g
= ( a1, a2) ( a3, b1 ) ( b2 , b3 ) ( a4 , b4 ) − − − − and hence π π' = ( a1, a3 , b2 ) ( a2 , b3 , b1 ) which is final
contradiction.
Hence A n
is simple for n ≥ 5.
As promissed earliar, to give an example that converse of Lagranges theorem is false:
Example 7:
The elements of A 4
, the alternating group of degree 4, are
(1),
(12) (34), (13) (24), (14) (23),
(123), (123) 2 ,
(124), (124) 2 ,
(134), (134) 2 ,
(234), (234) 2
Which are 12 in number.
A 4
has 3 cyclic sub-groups of order 2.
h