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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UNIT-I<br />
25<br />
1<br />
Note that αβ = <br />
2<br />
2<br />
1<br />
3<br />
1<br />
≠ <br />
3<br />
3<br />
2<br />
2<br />
3<br />
<br />
1<br />
= βα<br />
Hence S<br />
3<br />
, the group of 6 elements, called symmetric group which is non-abelian. This is the smallest finite<br />
non-abelian group, since groups of order 1, 2, 3, 5 are of prime order, hence cyclic and, therefore, they<br />
are abelian. A group of order 4 is of two types upto isomophism, either cyclic or Klein 4-group, given in<br />
example 2.<br />
Cycle Notation<br />
1<br />
2 3 4<br />
Let σ = <br />
3<br />
4 1 6<br />
This can be seen as:<br />
5<br />
5<br />
6<br />
<br />
2<br />
1<br />
2<br />
5<br />
σ<br />
σ<br />
( { )( ( )}) ()( ( ) ( )( }{ ) )( ( )}) )( )}{ ( )}<br />
σS H A 1 =<br />
13 1 e,<br />
2123<br />
246<br />
S,<br />
13 , H<br />
5132<br />
2,<br />
= 2313<br />
e<br />
,,<br />
1123 13 246 2<br />
, H132<br />
3<br />
3=<br />
e,<br />
23<br />
2 1<br />
β [<br />
3<br />
3<br />
S=<br />
3<br />
: A3<br />
] = ,<br />
= αβ 2,<br />
= ,<br />
α β = <br />
1<br />
3 A 23<br />
2<br />
1 3<br />
3<br />
2<br />
2<br />
3<br />
,<br />
1<br />
σ<br />
6<br />
4<br />
In cycle notation σ can be written as<br />
Therefore from example 6:<br />
It has 4 proper subgroups:<br />
and<br />
so A 3<br />
is a subgroup of S 3<br />
of index 2. It can be easily verified that A3 ∆ S3<br />
. Infact, it can<br />
be generalised, that every subgroup of index 2 is a normal subgroup in its parent group.<br />
group.<br />
is called alternating