Nonempty Subset - Abstract Algebra - Exam, Exams for Algebra. Acharya Nagarjuna University

Algebra

Description: This is the Exam of Abstract Algebra which includes Weighted Equally, Group Theory, Vector Spaces, Linear Algebra, Unique Factoriztaion, Normal Subgroup etc. Key important points are: Nonempty Subset, Group, Subgroup, Multiplication, Element, Isomorphic, Symmetric Group, Alternating Group, Contradiction, Subgroup
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MATH 451 FIRST MID-TERM
NAME: John Q. Public
Question Marks
1 12
2 25
3 28
4 25
5 10
1
2 MATH 451 FIRST MID-TERM
Question 1. Let Hbe a nonempty subset of the group G. Prove that His a
subgroup of Giff a b1Hfor all a,bH.
First suppose that His a subgroup of G. Then His closed under multiplication
and taking inverses. Hence if a,bH, then b1Hand so ab1H.
Next suppose that ∅ 6=HGis such that ab1Hfor all a,bH. Since
H6=, there exists an element aHand hence 1 = aa1H. It follows that if
aH, then a1= 1 a1H. Finally suppose that a,bH. Then b1Hand
so ab =a(b1)1H. Thus His a subgroup of G.
MATH 451 FIRST MID-TERM 3
Question 2. (a) Prove that the alternating group A5does not have a sub-
group which is isomorphic to the symmetric group S4.
(b) Prove that the alternating group A5does not have a subgroup of order 15.
(a) Suppose that H6A5is a subgroup such that H
=S4. Then |H|=|S4|= 24.
Applying Lagrange’s Theorem, we must have that 24 = |H|divides |A5|= 60, which
is a contradiction.
(b) Suppose that H6A5is a subgroup such that |H|= 15. Consider the
transitive action of A5on the coset space S=A5/H and let
ϕ:A5Sym(S)
be the associated homomorphism. Let N= ker ϕ. Since A5acts transitively on S
and |S|>1, we have that N6=A5. Hence, since A5is simple, it follows that N= 1.
But this means that ϕis an injection of A5into Sym(S), which is impossible since
|A5|= 60 and |Sym(S)|= 24.
4 MATH 451 FIRST MID-TERM
Question 3. Suppose that Gis a finite group and that Sis a G-set. For each
sS, let Osdenote the corresponding G-orbit.
(a) Prove that if sS, then [ G:Gs] = |Os|.
(b) Prove that if Gis a finite p-group and pdoes not divide |S|, then there
exists a fixed point for the action of G; i.e. an element sSsuch that
gs =sfor all gG.
(Hint: Let s1,· · · stbe representatives of the distinct G-orbits and consider
the equation |S|=|Os1|+· · · +|Ost|.)
(a) It is easily checked that if g,hG, then
g s =h s iff gGs=hGs.
Hence we can define an injective map ϕ:G/GsOsby ϕ(gGs) = g s. To see
that ϕis also surjective, let rOsbe arbitrary. Then there exists gGsuch that
g s =rand hence ϕ(gGs) = g s =r.
(b) Let s1,· · · stbe representatives of the distinct G-orbits. Then
|S|=|Os1|+· · · +|Ost|.
Since pdoes not divide |S|, there exists isuch that pdoes not divide |Osi|. Since
|Osi|= [G:Gsi] = |G|/|Gsi|
and Gis a p-group, it follows that |Osi|= 1 and so siis a fixed point for the action
of G.
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