Key of Hell: the literature of Cave Hill
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Le téléchargement nécessite un accès à la bibliothèque YouScribe Tout savoir sur nos offres
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Le téléchargement nécessite un accès à la bibliothèque YouScribe Tout savoir sur nos offres
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MA441: Algebraic Structures I
Lecture 6
22 September 2003
1Review from Lecture 5:
We defined
the center Z(G) of a group G
the centralizer C(a) of an element a∈ G
2We also proved an important theorem about
the structure of cyclic groups.
i jTheorem 4.1: Criterion for a = a
Let G be a group, and let a belong to G. If
a has infinite order, then all distinct powers of
a are distinct group elements. If a has finite
order, say, n, then
2 n 1hai={e,a,a ,...,a }
i jand a = a if and only if n divides i j.
3We have two immediate consequences of this
theorem.
The first corollary states that the order of an
element equals the order of the subgroup gen-
erated by that element.
Corollary 1:
For any group element a,
|a|=|hai|.
Corollary 2:
Let G be a group and let a ∈ G have order n.
kIf a = e, then n divides k.
4Multiplication (composition) of elements in a
cyclic group of order n is accomplished by ad-
dition modulo n.
Infact,Z/nZisaprototypeforallcyclicgroups.
(A cyclic group hai of order n is isomorphic to
Z/nZ, where a plays the role of 1.)
5Theorem 4.2:
Let a be an element of order n in a group and
let k be a positive integer. Then
k gcd(n,k)ha i=ha i
and
nk|a |= .
gcd(n,k)
Proof:
Let d=gcd(n,k) and k = dr.
k d r k dSince a =(a ) , we have ha iha i.
6Using the Euclidean algorithm, we can find s,t
such that d= ns+kt. Then
d ns+kt n s k t k ta = a =(a ) (a ) =(a ) ,
k dso ha iha i and the two sets are equal.
We prove the second part of the theorem by
dshowing that |a |= n/d for any d|n.
76
d n/d n dClearly, (a ) = a = e, so |a | n/d.
Suppose i is a positive integer less than n/d.
d iThen id < n and therefore (a ) = e. So the
dorder of a is n/d.
kNow apply this to a .
k k d d k dSince |a |=|ha i|, |a |=|ha i|, and ha i=ha i,
kwe have that the order of a is n/d, that is,
k|a |= n/gcd(n,k).
8Corollary 1:
i jLet |a|= n. Then ha i=ha i iff
gcd(n,i)=gcd(n,j).
Proof:
By Theorem 4.2, we have that
i gcd(n,i) j gcd(n,j)ha i=ha i and ha i=ha i.
gcd(n,i) gcd(n,j)We need to prove ha i = ha i iff
gcd(n,i)=gcd(n,j).
9Clearly gcd(n,i)=gcd(n,j) implies
gcd(n,i) gcd(n,j)ha i=ha i.
gcd(n,i) gcd(n,j)Suppose that ha i=ha i.
gcd(n,i) gcd(n,j)This means |ha i|=|ha i|, so
gcd(n,i) gcd(n,j)|a |=|a |.
By the second part of Theorem 4.2, on the
gcd(n,i)LHS |a | = n/gcd(n,i) and on the RHS
gcd(n,j)|a |= n/gcd(n,j). Therefore,
n n
= ,
gcd(n,i) gcd(n,j)
so gcd(n,i)=gcd(n,j).
10
