Key of Hell: the literature of Cave Hill

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1 Key of Hell: the literature of Cave Hill [Diarmuid Kennedy – Cave Hill has appeared in fiction, poetry and plays since at least 1750. Research for this paper has identified around 40 works in which the hill features but it is not the whimsical image of Napoleon's Nose that dominates, but instead something deeper and darker. Belfast is unusual in having rugged nature so close to hand. Cave Hill challenges the city by starkly presenting freedom, space and nature to the streets below.

irish nobility
cave hill
t. r. robinson
j.w. boyd
yeats letters
w.b. yeats
w. b. yeats
mountain
city
world

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Haughton

Shearman

Allison

Singing school

Ballater

Journey

MacNeill

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Publié par	renef
Nombre de lectures	22
Langue	English

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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