How To Write An Effective Online Job Posting
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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
- expression écrite
- push applicants into a replicable system
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MA441: Algebraic Structures I
Lecture 23
1 December 2003
1Review from Lecture 22:
Definition:
A subgroup H of a group G is called a normal if aH =Ha for all a∈G.
We denote this by HCG.
Theorem 9.1: Normal Subgroup Test
A subgroup H of G is normal in G iff
1xHx H for all x∈G.
2Definition:
For HCG, the set of left (or right) cosets of
H in G is a group called the quotient group
(or factor group) of G by H.
We denote this group by G/H.
Theorem 9.2:
Let HCG. The set G/H is a group under the
operation (aH)(bH)=abH.
3Example 11:
Let G=U(32)={1,3,5,7,...,27,29,31}.
LetH =U (32)={1,17}. (171 (mod 16))16
G/H is abelian of order 16/2=8.
In Chapter 11, we’ll learn about the Funda-
mental Theorem of Abelian Groups. This the-
orem tells us that an abelian group of order 8
must be isomorphic to Z , Z Z , or8 4 2
Z Z Z .2 2 2
Which of these three direct products is isomor-
phic to G/H?
4We will determine the elements of G/H and
their orders.
1H ={1,17}, 3H ={3,19}, 5H ={5,21}
7H ={7,23}, 9H ={9,25}, 11H ={11,27}
13H ={13,29}, 15H ={15,31}.
2We can rule out Z Z Z , because (3H) =2 2 2
9H, so |3H|4.
2 2Now (7H) = (9H) = H, so these are two
distinct elements with order 2. This rules out
Z , which has only one element of order 2.8
Therefore, U(32)/U (32) Z Z , which is16 4 2
isomorphic to U(16).
5Theorem 9.3:
Let G be a group with center Z(G). If G/Z(G)
is cyclic, then G is Abelian.
Proof:
Letg ∈GbesuchthatgZ(G)generatesG/Z(G).
For any a,b∈G, let
i iaZ(G)=(gZ(G)) =g Z(G),
and let
j jbZ(G)=(gZ(G)) =g Z(G).
6i jThen a=g x and b=g y, ∃x,y ∈Z(G).
iSince x,y commute with everything, and g
jcommutes with g , we see that
i j i+j j iab=(g x)(g y)=g xy =(g y)(g x)=ba.
Since the a,b were arbitrary, we have that G is
abelian.
7Theorem 9.4:
For any group G, G/Z(G)Inn(G).
Proof:
Define the map T :G/Z(G)→Inn(G) via
gZ(G)7→ .g
(We’ll use Gallian’s definition of inner auto-
1morphism: (x)=gxg .)g
We need to show that T is a well-defined func-
tion, that is one-to-one, onto, and preserves
the group operation.
8ToshowthatT iswell-defined,supposegZ(G)=
hZ(G). We’ll show that both cosets map to
the same inner automorphism.
1Now gZ(G) = hZ(G) implies h g ∈ Z(G).
1 1Then for all x∈G, h gx=xh g.
Bymultiplyingontheleftby handontheright
1 1 1by g , we have gxg =hxh .
Therefore = .g h
9One-to-one: reverse the argument. = g h
implies gZ(G)=hZ(G).
Onto: by definition of T. For any , the cosetg
gZ(G) is a preimage under T.
Group operation:
T(gZ(G)hZ(G))=T(ghZ(G))= .gh
T(gZ(G))T(hZ(G))= = .g h gh
1 1(Recall: (x)= (hxh )=(gh)x(gh) .)g gh
10
