Introduction to Abstract Algebra
Homework 7
October 25, 2010
Problem 1
Given a group
G
with

G
:
Z
(
G
)

=
n
, we wish to show that all conjugacy classes have at
most
n
elements.
We know that for some
x
∈
G
,
Z
(
G
)
≤
C
G
(
x
)
≤
G,
and in a previous homework we have shown that if
K
≤
H
≤
G
, then

G
:
K

=

K
:
H

G
:
H

.
Thus, we ﬁnd that

G
:
Z
(
G
)

=

Z
(
G
) :
C
G
(
x
)

G
:
C
G
(
x
)

.
By proposition 6 (p. 123), the number of elements in the conjugacy class of
x
is the index
of
C
G
(
x
) in
G
, which we have shown above to divide the index of the center of
G
in
G
. The
center has index
n
, so the number of elements in every conjugacy class divides
n
, and is thus
at most
n
.
Problem 2
Let
G
1
,G
2
,...,G
n
be groups. We show that
Z
(
G
1
× ··· ×
G
n
) =
Z
(
G
1
)
× ··· ×
Z
(
G
n
)
.
Let (
g
1
,...,g
n
)
∈
Z
(
G
1
×···×
G
n
) and let (
x
1
,...,x
n
) be an arbitrary element in
G
1
×···×
G
n
.
Then by deﬁnition
(
x
1
,...,x
n
)(
g
1
,...,g
n
) = (
g
1
,...,g
n
)(
x
1
,...,x
n
)
⇔
(
x
1
g
1
,...,x
n
g
n
) = (
g
1
x
1
,...,g
n
x
n
)
.
Note that the above requirement on the element in position
i
is met by all elements
g
i
∈
Z
(
G
i
),
for all
x
i
∈
G
i
. Thus, the right hand side above is just the cross product
Z
(
G
1
)
× ··· ×
Z
(
G
n
).
Next we show, that
G
1
× ··· ×
G
n
is abelian if and only if
G
i
is abelian for
i
= 1
,...,n
.
We ﬁrst show the if direction. Assume that
G
1
× ··· ×
G
n
is abelian. Let (
g
1
,...,g
n
) and
(
x
1
,...,x
n
) be elements of
G
1
× ··· ×
G
n
. Then, by assumption,
(
x
1
,...,x
n
)(
g
1
,...,g
n
) = (
g
1
,...,g
n
)(
x
1
,...,x
n
)
⇔
(
x
1
g
1
,...,x
n
g
n
) = (
g
1
x
1
,...,g
n
x
n
)
.
From this, we see that for any
i
in the range, with
x,g
∈
G
i
, that
xg
=
gx
, so all
G
i
are
abelian, for
i
= 1
,...,n
.
Conversely, assume that
G
i
is abelian for
i
= 1
,...,n
. Then for
g,x
∈
G
i
:
xg
=
gx
. In
the following, let
x
i
,g
i
∈
G
i
, we then ﬁnd
(
x
1
g
1
,...,x
n
g
n
) = (
g
1
x
1
,...,g
n
x
n
)
⇔
(
x
1
,...,x
n
)(
g
1
,...,g
n
) = (
g
1
,...,g
n
)(
x
1
,...,x
n
)
.
From this we see, that if we consider arbitrary elements
x,g
∈
G
1
× ··· ×
G
n
, then
xg
=
gx
,
so
G
1
× ··· ×
G
n
is abelian.
1/
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View Full DocumentIntroduction to Abstract Algebra
Homework 7
October 25, 2010
Problem 3
Let
G
1
,G
2
,...,G
n
be groups, and let
G
=
G
1
× ··· ×
G
n
. Let
I
and
J
be proper, nonempty
subsets of
{
1
,
2
,...,n
}
such that
I
∩
J
=
∅
. Deﬁne
G
I
to be the set of elements of
G
that
have the identity of
G
j
in positions
j
for all
j
∈
J
.
(a) We prove that
G
I
is isomorphic to the direct products of the groups
G
i
for
i
∈
I
. For
convenience we denote this group
×
i
∈
I
G
i
. Deﬁne
φ
:
G
I
→
×
i
∈
I
G
i
,
s.t.
φ
maps an element in
G
I
to the element in
×
i
∈
I
G
i
where all elements on positions
in
J
are removed. Consider two arbitrary
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 Summer '11
 gelfand
 Chemistry, elements, Ring, gJ

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