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Chapter 2

MAT247H1 Chapter Notes - Chapter 2: Non-Abelian Group, Invariant Subspace, Binary Operation

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Fiona T Rahman

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Mat 247 - Definitions and results on group theory
Definition: Let Gbe nonempty set together with a binary operation (usually called mul-
tiplication) that assigns to each pair of elements g1,g2Gan element in G, denoted
by g1g2or g1·g2. We say that Gis a group under this operation if the following three
properties are satisfied:
Associativity: (g1g2)g3=g1(g2g3) for all g1,g2, and g3G.
Existence of identity element: There exists an element e(called an identity) in Gsuch
that g·e=e·g=gfor all gG.
Existence of inverses: Let ebe an identity element in G. For each element gG,
there is an element g1G(called an inverse of g) such that g·g1=g1·g=e.
Examples: (details omitted)
(1) If Fis a field and nis a postive integer, let GLn(F) = {AMn×n(F)|det(A)6= 0 }.
Then GLn(F) is a group under the operation of matrix multiplication.
(2) Let Vbe a finite-dimensional vector space over a field F. Let G={T∈ L(V)|Tis invertible }.
Then GL(V) is a group under the operation of composition of linear transformations.
(3) Let nbe a positive integer. The set Unof (complex) unitary n×nmatrices is a group
under the operation of matrix multiplication.
(4) The set Zof integers is a group under the operation of addition of integers. (Note:
e= 0; the inverse of mZis m.)
(5) The set Z\{0}of nonzero integers is not a group under the operation of multiplication
of integers. The operation is associative and 1 is an identity, but the only nonzero
integers that have inverses in Z\{0}are 1 and 1.
Definition: If Gis a group, we say that Gis abelian (or commutative) if g1g2=g2g1for
all g1and g2G. If Gis not abelian, we say that Gis nonabelian (or noncommutative).
Definition: The order of a group Gis the number of elements in G. If the order of Gis
finite, we say that Gis a finite group. Otherwise, we say that Gis an infinite group.
If Gis an abelian group, the group operation may be written with a plus sign: g1+g2
instead of g1g2.
Examples. If Fis a finite field, then GLn(F) is a finite group. If Fis an infinite field, then
GLn(F) is an infinite group. If n2, then GLn(F) is a nonabelian group. The notation
F×is often used for the group GL1(F) of nonzero elements in F(with the operation of
multiplication in F). The group F×is abelian.
Lemma. If Gis a group, there is a unique identity element in G. If gG, there is a
unique inverse g1of gin G.
Proof. If eand e0are identity elements in G, we have e·e0=e0·e=e, using that e0is
an identity element, and we also have e·e0=e0·e=e0, since eis an identity element.
Therefore e·e0=e=e0. The second part is left as an exercise.
Definition. If His a (nonempty) subset of a group Gand His itself a group under the
operation on G, we say that His a subgroup of G.
The subset {e}of a group Gis a subgroup of G. Clearly, Gis a subgroup of G. The
proof of the following lemma was discussed in class.
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Lemma. (Subgroup test) A nonempty subset Hof a group Gis a subgroup of Gif and
only if h1h1
2Hfor all h1,h2H.
(1) Let G=GLn(F), n2, and let H={AG|Ajk = 0 whenever j > k }. Then H
is a subgroup of G. (Details omitted.)
(2) Let Vbe a vector space of dimension n2, let G=GL(V), and let H={T
G|nullity(TIV)>0}. Let β={x1, . . . , xn}be an ordered basis for V. There exists
a unique T∈ L(V) such that T(x1) = x1and T(xj) = xj, 2 jn. Check that T
is invertible, nullity(TIV) = 1, Tis invertible, and nullity(TIV) = n1 (left
as an exercise). This implies that T,TH. But (T(T))(xj) = T2(xj) = xj
for 1 jn, so T(T) = IV, and nullity(IVIV) = nullity(2·IV) = 0.
That is, T(T)/H. This implies that His not a subgroup of G.
(3) Let G=GLn(F), n2. Let Dnbe the set of diagonal matrices in G. Then Dnis a
subgroup of G, and Dnis abelian. (This example shows that there can be nontrivial
abelian subgroups of nonabelian groups.)
Defintion. A subgroup Hof a group Gis said to be normal in Gif ghg1Hfor all
gGand hH.
Examples: (details omitted)
(1) Let G=GL2(F) and let H={AG|A21 = 0 }. Then His a subgroup of Gbut
His not normal in G. (Note that h=1 1
0 1 H, Let g=0 1
1 0 . Show that
gGand ghg1/H.)
(2) Let G=GLn(F),n2, and let H=SLn(F) = {AG|det(A) = 1 }. Then His a
normal subgroup of G. (This is easily proved using properties of determinants.)
(3) If Gis an abelian group, then any subgroup Hof Gis normal in Gbecause ghg1=
h(g·g1) = h·e=hfor all hHand gG.
Definition. If Gand G0are groups, a map ϕ:GG0is a homomorphism if ϕ(g1g2) =
ϕ(g1)ϕ(g2) for all g1and g2G.
Examples. (details omitted)
(1) Then det : GLn(F)F×=GL1(F) is a homomorphism.
(2) If Gis a nonabelian group, the map ϕ:GGdefined by ϕ(g) = g2is not a
homomorphism. (Here, g2=g·g,gG.)
Notation. If Gis a group, gG, and nZ, define g0=e,gn=g·gn1,n1, and
gn= (g1)n,n≤ −1.
Lemma. Let Gand G0be groups and let ϕ:GG0be a homomorphism.
(1) Let eand e0be identity elements in Gand G0, respectively. Then ϕ(e) = e0.
(2) If gGand nZ, then ϕ(gn) = (ϕ(g))n.
Definition: Let Gand G0be groups and let ϕ:GG0be a homomorphism.
(1) The kernel of ϕis defined to be {gG|ϕ(g) = e0}. Here, e0is the identity element
in G0.
(2) The image of ϕis defined to be ϕ(G) = {ϕ(g)|gG}.
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