Math 3124
Tuesday, October 25
Second Sample Test Solutions
1. To show that
a
,
b
=
c
,
d
, we need to show that
c
,
d
∈
a
,
b
and
a
,
b
∈
c
,
d
. Now
(2 3 4) = (1 2)(1 2 3 4) and (1 3 4 2) = (2 3 4)(1 2) = (1 2)(1 2 3 4)(1 2). This tells us
that
(
2 3 4
)
,
(
1 3 4 2
)
⊆
(
1 2
)
,
(
1 2 3 4
)
. On the other hand (2 3 4)(1 3 4 2) = (1
4 3 2), so by taking inverses, we see that
(
1 2 3 4
)
∈
(
2 3 4
)
,
(
1 3 4 2
)
. Finally (2 3
4)(1 4 3 2) = (1 2) and the result follows.
2. First note that
bc
=
c
forces
b
to be the identity. Then
ad
cannot be
a
,
d
,
b
, so must be
c
etc.
a
b
c
d
a
b
a
d
c
b
a
b
c
d
c
d
c
a
b
d
c
d
b
a
3. The order of
Z
3
is 3, and the order of
S
3
is 6.
Therefore the order of
Z
3
×
S
3
is
3
*
6
=
18. Let
H
= ([
1
]
,
(
1 2
))
. The order of [1] in
Z
3
is 3 and the order of (1 2)
in
S
3
is 2. Therefore the order of ([1],(1 2)) is 3
*
2
=
6 and we deduce that

H

=
6.
Thus we expect 3 right cosets with 6 elements each. The right cosets are
H
=
([0],(1)), ([1],(1 2)), ([2],(1)), ([0],(1 2)), ([1],(1)), ([2],(1 2))
H
([0],(1 3))
=
([0],(1 3)), ([1],(1 3 2)), ([2],(1 3)), ([0],(1 3 2)), ([1],(1 3)), ([2],(1 3 2))
H
([0],(2 3))
=
([0],(2 3)), ([1],(1 2 3)), ([2],(2 3)), ([0],(1 2 3)), ([1],(2 3)), ([2],(1 2 3))
4. By Lagrange’s theorem,

A

=
1
,
7 or 49. But 49 is ruled out because
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 Fall '08
 PARRY
 Math, Algebra, Subgroup, Cyclic group, 25 SECOND

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