Cyclic groups Algebraic structures
(Monday, May 3, 2010)
Lemma: If G is a group and a e G, where a ¹ e (the identity element of G), the set A of elements an, where n e Z (the set of all integer) forms a group.
G1: If m, G, and am.an e A, then am.an =
am+n e A, as l+m+n e Z (i.e., closure
property)
G2: If l, m, n e Z and al.am.an e A, then,
al. (am . an)=(al . am). an = al+m+n e A as
l+m+n e Z. (i.e., Associative law)
G3: If n e Z and an e A, then a0. an=an. a0
= an and obviously a0 e A
(i.e., identity element)
G4: (an)-1 = a-n e A, where -n e Z such
that an . a-n = a0. (i.e., existence of
an inverse element).
Hence, the set A = {an} n e Z forms a group.
Posted in Posted by waytofeed at 7:32 AM
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