Math 4124
Monday, February 14
Solutions to Sample First Test
1. Since

S
3
×
S
5

=
3!5!
=
6! and

S
n

=
n
!, we see that the only possible
n
for which

S
3
×
S
5

can be isomorphic to
S
n
is 6. Now we look at the orders of the elements.
The order of
((
123
)
,
(
12345
))
is the lowest common multiple of 3 and 5, which is 15;
thus
S
3
×
S
5
has an element of order 15. The cycle shapes of elements of
S
6
are (1),
(12), (123), (1234), (12345), (123456), (12)(34), (12)(34)(56), (12)(345), (12)(3456),
(123)(456), and these have orders 1,2,3,4,5,6,2,2,6,4,3 respectively; in particular
S
6
has no elements of order 15. Therefore
S
3
×
S
5
is not isomorphic to
S
6
, and the result
follows.
2.
(i) First
G
=
/0; this is obvious because
II
t
=
I
which tells us that
I
∈
G
.
(ii) Now let
A
,
B
∈
G
. We next show that
AB
∈
G
. The definition of
G
tells us that
AA
t
=
BB
t
=
I
, and we need to establish that
(
AB
)(
AB
)
t
=
I
. But
(
AB
)(
AB
)
t
=
ABB
t
A
t
=
AIA
t
=
AA
t
=
I
, as required.
(iii) Finally we must show that
G
is closed under taking inverses, so let
A
∈
G
. We
want to prove that
A

1
∈
G
, in other words
A

1
(
A

1
)
t
=
I
. But
A
∈
G
implies
AA
t
=
I
, implies
A
t
=
A

1
, implies
A
= (
A

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 Spring '08
 Staff
 Math, Algebra, 1 g, Subgroup, Cyclic group

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