LAGRANGE THEOREM S gorse two

LAGRANGE THEOREM S

Theorem 2 -11 ( Lagrange ) .
The order and index of any subgroup of a finite group
divides the order of the group .
Corollary
If (G, * ) is a group of order n , then the order of any . element a? G is a factor of n , and an = e .
Proof .
Let a have order k.
So ((a) , ?) generated by a must also be of order k .
By Lagrange, s Theorem, k is a divisor of n
n= kr r ? Z+ .
Hence an = ark = (ak)r = er = e
Example.
In example 1 o(H) = 3 , o(G) = 12 and 3|12.
Theorem 2-12.
If (G ,*) is a flnite group of composite order then ( G , ?)
has nontrivial subgroups .
Proof .
If G is not cyclic .
let e? a ?G , then a generates a nontrivial cyclic subgroup
((a) ,? ) . 31
If G is cyclic group of composite order ,
G = (a) such that the order of a is nm (n,m? 1).
Then (an)m = e, while (an)k? e for 0 ? k ? m.
Hence ((an) ,?) is a nontrivial cyclic subgroup of order m .
Corollary .
Every group of prime order is cyclic .
Proof .
Let (G, ?) be group of prime order , and ((a) , ?) be cyclic
subgroup generated by any a?G such that a ?e.
order of (( a) ,?) must divide the order of (G ,?).
(a) Contains more than one element ,
Then order of (a) = order of G .
Hence G = (a) .
Theorem 2 – 13 .
Every cyclic group is commutative .
Proof .
Let ((a) , *) be cyclic group .
If x , y ? (a) then x = am and y = ar where m , r ? Z+.
x * y = am * ar
= am+ r = ar+m
= ar * am = y * x
Then any cyclic group is commutative.
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Theorem 2 – 14 .
Any noncommutative group has at lest six elements .
Proof .
By theorem 2 – 12 a group of prime order is cyclic ,
Then a group of prime order is commutative ?
Hence any group having order 2,3 or 5 will be commutative .
If (G , ?) is a group of order 4,
then by Lagrange, s theorem each element of G distinet from
the identity has order 2 or 4 .
If one of them has order 4 ,
then G is cyclic of order 4 ,
hence G is commutative .
If no one has order 4 , then a? e of order 2.
Then G is commutative .