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Solutions to HW #7
1. Let
G
be a group and let
a
be an element of
G
with order
n
.
(a) Prove (
bab

1
)
n
=
e
.
•
(
bab

1
)
n
=
bab

1
·
bab

1
·
bab

1
·
...
·
bab

1

{z
}
n times
=
ba
·
(
b

1
b
)
·
a
·
(
b

1
b
)
·
a
·
(
b

1
b
)
·
...
·
(
b

1
b
)
·
ab

1
=
ba
·
e
·
a
·
e
·
a
·
e
·
...
·
e
·
ab

1
=
ba
n
b

1
=
beb

1
=
bb

1
=
e
(b) Prove that the order of
bab

1
is exactly
n
.
•
By (a), we know that the order of
bab

1
is no greater than
n
. So, suppose that the order
of
bab

1
is
m
with
m < n
. Therefore, it must be true that
e
= (
bab

1
)
m
=
ba
m
b

1
.
Consequently,
b

1
eb
=
b

1
ba
m
b

1
b
. Reducing both sides, we get that
e
=
a
m
. But this
implies that the order of
a
is less than
n
, in violation of our hypothesis. We conclude that
the order of
bab

1
must be
n
.
2. Let
G
be a group such that
a
2
=
e
for all
a
∈
G
. Prove that
G
is abelian.
[Hint: To show
ab
=
ba
use the fact that
a
2
=
b
2
= (
ab
)
2
=
e
.]
•
Since (
ab
)
2
=
e
, we have that
abab
=
e
. Multiplying on the left by
ba
, we get
ba
·
abab
=
bae
ba
2
bab
=
ba
bebab
=
ba
b
2
ab
=
ba
eab
=
ba
ab
=
ba
3. Let
G
be a group of size 6 with 5 elements of order 2 and 0 elements of order 3  i.e. case 2a.
(a) Prove that
G
is abelian.
•
Since
G
has 5 elements of order 2, the 6th element must be the identity. Thus,
a
2
=
e
for all
a
∈
G
. Applying the result from problem 2, we see that
G
is abelian.
(b) Let
a
and
b
be two distinct elements in
G
with order 2. Prove that
ab
does not equal
e
,
a
or
b
.
•
We break it into three cases:
–
Suppose
ab
=
e
. Since
aa
=
e
, we get
ab
=
aa
. By canceling (i.e. multiplying by
a

1
on
the left), we get
b
=
a
which is a contradiction.
–
Suppose
ab
=
a
. Cancellation gives
b
=
e
, a contradiction.
–
Suppose
ab
=
b
. Cancellation gives
a
=
e
, a contradiction.
1
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This note was uploaded on 11/03/2010 for the course MATH 541 taught by Professor Pollack during the Fall '09 term at BU.
 Fall '09
 Pollack
 Math, Algebra

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