Selected solutions to Math 4107 midterm 2,
Fall 2009
December 11, 2009
1.
You can look these up.
2.
This is a routine computation, so I won’t bother to do it.
3.
The center of
S
2
is all of
S
2
, since
S
2
is cyclic of order 2, which is abelian.
We claim that the center of
S
n
for
n
≥
3 is just
{
e
}
, the identity: To
see this, suppose
ϕ
is an element of the center. If
ϕ
is not the identity, then
there exists
a,b
∈ {
1
,
2
,...,n
}
,
a
negationslash
=
b
, such that
ϕ
(
a
) =
b.
So, if we let
c
∈ {
1
,
2
,...,n
}
be distinct from
a
and
b
(possible, since
n
≥
3),
we have that
ϕ
◦
(
a c
)
◦
ϕ

1
= (
b ϕ
(
c
))
negationslash
= (
a c
)
.
This contradicts the fact that
ϕ
was in the center; and so, the center contains
only the identity.
4.
If

G

=
pq
,
q > p
, then
G
has Sylow
p
and Sylow
q
subgroups of orders
p
and
q
, respectively.
But how many Sylow subgroups? The number of
Sylow
q
’s is 1
,q
+ 1
,
2
q
+ 1
,...
. But it can’t be 2
q
+ 1 or higher, since if any
pair of Sylow
q
’s shared a nonidenitty element they would have to be the
same subgroup; so, if there were 2
q
+ 1 or more of them, then they would,
collectively, contribute at least
(2
q
+ 1)(
q
−
1)
> pq
=

G

elements of order
q
, impossible. So, there is only one Sylow
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 Fall '08
 Staff
 Math, Algebra, Group Theory, φ, Subgroup, Cyclic group, routine computation

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