Math Meth
HW1 Solutions
Szilard Farkas
a)
1.
(i)
No. (
a
·
b
)
·
c
=
a
/
(
bc
),
a
·
(
b
·
c
)
=
a
/
(
b
/
c
)
=
ac
/
b
, so the product is not associative.
(ii)
a
=
(24)(153) and
b
=
(12)(354). We will see that if two permutations are of the same type,
which means that the number of cycles of a given length is the same in both permutations, then
they are related to each other by conjugation. Let
c
be for example the permutation (2
,
4
,
1
,
5
,
3)
7→
(1
,
2
,
3
,
5
,
4), that is,
c
=
(1342). Then
b
=
cac

1
.
(iii) Reﬂexive since
a
=
eae

1
, where
e
is the identity element; transitive because
b
=
gag

1
and
c
=
hbh

1
implies that
c
=
hgag

1
h

1
=
(
hg
)
a
(
hg
)

1
; and symmetric since
b
=
gag

1
means that
a
=
g

1
bg
=
(
g

1
)
b
(
g

1
)

1
.
(iv)
H
is a subgroup if and only if
H

1
=
H
and
HH
=
H
. (
gHg

1
)

1
=
gH

1
g

1
=
gHg

1
and
gHg

1
gHg

1
=
gHHg

1
=
gHg

1
, so
gHg

1
is also a subgroup.
(v) An isomorphism preserves the order of any element.
Z
4
has a fourth order element, whereas in
Z
2
×
Z
2
the order of any element is at most two.
2.
I
2
=
J
2
=
K
2
=

1 means that
I

1
=

I
etc. Inverting both sides in
IJ
=
K
, we get
J

1
I

1
=
K

1
,
or
JI
=

K
, so the product is not commutative. The group, which is usually denoted by
Q
, has
eight elements
{
1
,
I
,
J
,
K
,

I
,

J
,

K
}
. The quaternion group
Q
and the dihedral group
D
4
are the
only eightelement nonabelian groups.
Q
is the smallest noncommutative group which is not a
proper semidirect product. (In order for a group to be a proper semidirect product, it has to have
two nontrivial subgroups whose only common element is the identity. But

1 is the element of any
nontrivial subgroup of
Q
.)
3.
(i)(ii) We have to show that two permutations are of the same type
b)
if and only if they are
conjugate to each other. Consider the permutations of a set
H
. Let
a
and
c
be two arbitrary
permutations, and deﬁne the permutation
b
by
b
(
c
(
i
))
B
c
(
a
(
i
)), where
a
(
i
) etc is the element of
H
that the permutation
a
assigns to
i
∈
H
. Since
c
is a bijection on
H
,
b
is indeed a welldeﬁned
bijection on
H
, so it is a permutation. The type of this permutation is the same as the type of
a
,
since
c
just relabels the elements of
H
. The deﬁnition
bc
(
i
)
=
ca
(
i
) of
b
can also be written as
b
=
cac

1
. So far we have shown that conjugation does not change the type.
Now let