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 b−1∈Hfor all a,b∈H.

First suppose that His a subgroup of G. Then His closed under multiplication

and taking inverses. Hence if a,b∈H, then b−1∈Hand so ab−1∈H.

Next suppose that ∅ 6=H⊆Gis such that ab−1∈Hfor all a,b∈H. Since

H6=∅, there exists an element a∈Hand hence 1 = aa−1∈H. It follows that if

a∈H, then a−1= 1 a−1∈H. Finally suppose that a,b∈H. Then b−1∈Hand

so ab =a(b−1)−1∈H. 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:A5→Sym(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

s∈S, let Osdenote the corresponding G-orbit.

(a) Prove that if s∈S, 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 s∈Ssuch that

gs =sfor all g∈G.

(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,h∈G, then

g s =h s iff gGs=hGs.

Hence we can deﬁne an injective map ϕ:G/Gs→Osby ϕ(gGs) = g s. To see

that ϕis also surjective, let r∈Osbe arbitrary. Then there exists g∈Gsuch 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.

##### Document information

Uploaded by:
honda-civic

Views: 791

Downloads :
0

University:
Acharya Nagarjuna University

Subject:
Algebra

Upload date:
23/02/2013