MATH 451 FIRST MID-TERM
NAME: John Q. Public
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
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