Algebra , Assignment 2 Solutions
Due, Friday, Sept 24
1. In
S
3
(reminder, this is our standard notation for the permutations
on
{
1
,
2
,
3
}
) show that there are four elements
x
satisfying
x
2
=
e
and
three elements
x
satisfying
x
3
=
e
.
Solution: The flips
parenleftbigg
123
132
parenrightbigg
,
parenleftbigg
123
321
parenrightbigg
,
parenleftbigg
123
213
parenrightbigg
have
x
2
=
e
and
parenleftbigg
123
231
parenrightbigg
and
parenleftbigg
123
312
parenrightbigg
have
y
3
=
e
and
e
satisfies both
x
2
=
e
and
y
3
=
e
.
2. In
Z
*
13
let
H
=
{
1
,
5
,
12
,
8
}
. List the right cosets
Ha
.
Solution:
H
1 =
{
1
,
5
,
12
,
8
}
,
H
2 =
{
2
,
10
,
11
,
3
}
,
H
4 =
{
4
,
7
,
9
,
6
}
and
that is all of them. (As the group
Z
*
13
is Abelian right and left cosets
are the same.)
3. Let
G
be the symmetries of the square. (See the solutions to assign
ment 1 for a table.) Let
H
=
{
I, V
}
. List the right cosets
Ha
and the
left cosets
aH
. Do the same with
H
=
{
I, R, S, T
}
.
Solution:
H
=
{
I, V
}
: The left cosets are
IV, RD, SH, TA
(just go
down the
I
and
V
columns of the table) and the right cosets are
IV, RA, SH, TD
((just go down the
I
and
V
rows of the table). (These
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 Fall '08
 JOSEPHS
 Permutations, Coset, Conjugacy class, right cosets, cosets Ha, right cosets Ha

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