Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

32
ADVANCED ABSTRACT ALGEBRA
For proving this we shall need a simple fact about 3 – cycles in A n.
Lemma 3:
A n
is generated by cycles of length 3 (3 – cycles) if
Proof.
Every even permutation is the product of an even number if 2 – cycles. Since (a, b) (a, c) = (a, b, c) and
(a, b) (c, d) = (a, b, c) (a, d, c), an even permutation is also a product of 3 – cycles. Further, 3 – cycles are
even and thus belong to A n
.
(Here we have taken product from left to right).
Proof of Theorem:
Suppose it is false and there exists a proper nontrivial normal subgroup N.
Assume that a 3 – cycle
F π =
H G a b c I K J ,
a'
b' c'
If (a', b', c') is another 3 – cycle and
such that
π
– 1
( a, b, c)
π
a' b''
c
a b c
a b c
b c a
a b c
= F H G I K JF H G I K JF H G I K J =
a' b' c'
Θ π ∈ S n
, so π may be odd, hence we replace it by even permutation
∈n
(a ∃=
π
π
where e, f differ from a', b', c' without disturbing the conjugacy relation (here we use the fact n ≥ 5.
Hence (a', b', c') and N = A n
by above lemma 3. Therefore, N can not contain a 3 – cycle.
Assume now that N contains a permutation where disjoint cyclic decomposition involves a cycle of length
at least 4, say
Then N also contains
– 1
b1 2 3g b1 2 3g
1
π = a , a , a π a , a , a
Hence N contains π
=
F
HG
a a a a
2 3 4 5
a a a a
1 2 3 4
π
–1 1
− − − −IKJ− − − − −
− − − −
F
HG
a a a
2 3 1
a a a
3 1 4
= ( , , ): a a a
2 4 1
Note that other cycles cancel here.
− − −
− − −
IKJ− − − −