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
n
=
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
))
n
= (
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