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MATH 150B WINTER 2010
HOMEWORK 1 SOLUTIONS
ADAM SORKIN
Many thanks to Jeﬀ Ferreira for assistance on this homework, and, I’m sure, many more.
5.8.2
Let
S
be a group on which
G
operates, and let
H
=
{
g
∈
G

gs
=
s
for all
s
∈
S
}
. Prove
that
H
is a normal subgroup of
G
.
Proof.
This is a quick computation. Let
h
∈
H,g
∈
G,s
∈
S
. Then
ghg
−
1
(
s
) =
g
(
h
(
g
−
1
(
s
))
=
g
(
g
−
1
(
s
))
by deﬁnition of
h
=
s.
Hence if
h
∈
H
, then for all
g
∈
G
we have
ghg
−
1
∈
H
, and so
H
is normal.
±
5.8.3
Let
G
be the dihedral group of symmetries of a square. Is the action of
G
on the vertices
a faithful action? on the diagonals?
Proof.
For the action of
G
on the vertices, we do have a faithful action. This is because only the
identity element in
D
4
ﬁxes all the vertices simultaneously. For the action of
G
on the diagonals, we
do not have a faithful action. This is because, for example, rotation by
π
ﬁxes both diagonals.
±
5.8.5
A group
G
operates faithfully on a set
S
of ﬁve elements, and there are two orbits, one of
order 3 and one of order 2. What are the possibilities for
G
?
Proof.
Recall that a
G
action on
S
is equivalent to a homomorphism
ρ
:
G
→
Perm(
S
). Because
the action splits
S
into 2 orbits,
O
3
and
O
2
of size 3 and 2 respectively, the homomorphism splits
into a product
ρ
= (
ρ
1
,ρ
2
):
G
→
Perm(
O
3
)
×
Perm(
O
2
) (cf 5.8.4). That is,
ρ
is a morphism to
S
3
×
S
2
. Furthermore, this mapping is injective, as the action is faithful. Hence
G
is (isomorphic
to) a subgroup of
S
3
×
S
2
, subject to the condition that
G
has some elements of orders 2 and 3.
One quickly sees there are only two subgroups of
S
3
×
S
2
satisfying this condition, namely
S
3
×
S
2
itself, and the subgroup
⟨
((123)
,
(12))
⟩
isomorphic to the cyclic group of order six.
±
5.Misc.2bc
(b) Prove that the map
φ
:
G
→
Aut(
G
) deﬁned by
g
7→
(conjugation by
g
) is a homomorphism,
and determine its kernel.
Proof.
Veriﬁcation that
φ
is a morphism requires checking the maps pointwise. Letting
φ
(
g
) =
φ
g
:
G
→
G
, one has that
φ
gh
(
k
) =
gh
(
k
)(
gh
)
−
1
=
g
(
hkh
−
1
)
g
−
1
=
g
(
φ
h
(
k
))
g
−
1
=
φ
g
(
φ
h
(
k
)). Hence
φ
gh
=
φ
g
φ
h
. The kernel of
φ
is
{
g
∈
G

ghg
−
1
=
h
for all
h
∈
G
}
, which
is precisely
Z
(
G
), the center of
G
.
±
(c) The automorphisms which are conjugation by a group element are called
inner automor
phisms
. Prove that the set of inner automorphisms, the image of
φ
, is a normal subgroup
of Aut(
G
).
1
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ADAM SORKIN
Proof.
Consider some inner automorphism
φ
g
and an arbitrary automorphism
ψ
. By direct
computation pointwise, one has
ψ
◦
φ
g
◦
ψ
−
1
=
φ
ψ
(
g
)
, whence
φ
(
G
) is a normal subgroup
of Aut(
G
).
±
5.Misc.7
Let
G
be a ﬁnite group acting on a ﬁnite set
S
. For each element
g
∈
G
, let
S
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 Winter '10
 VAZIRANI

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