(
January 14, 2009
)
[16.1]
Let
p
be the smallest prime dividing the order of a finite group
G
. Show that a subgroup
H
of
G
of
index
p
is necessarily
normal
.
Let
G
act on cosets
gH
of
H
by left multiplication.
This gives a homomorphism
f
of
G
to the group of
permutations of [
G
:
H
] =
p
things. The kernel ker
f
certainly lies inside
H
, since
gH
=
H
only for
g
∈
H
.
Thus,
p

[
G
: ker
f
]. On the other hand,

f
(
G
)

= [
G
: ker
f
] =

G

/

ker
f

and

f
(
G
)

divides the order
p
! of the symmetric group on
p
things, by Lagrange. But
p
is the smallest prime
dividing

G

, so
f
(
G
) can only have order 1 or
p
. Since
p
divides the order of
f
(
G
) and

f
(
G
)

divides
p
, we
have equality. That is,
H
is the kernel of
f
. Every kernel is normal, so
H
is normal.
///
[16.2]
Let
T
∈
Hom
k
(
V
) for a finitedimensional
k
vectorspace
V
, with
k
a field.
Let
W
be a
T
stable
subspace. Prove that the minimal polynomial of
T
on
W
is a divisor of the minimal polynomial of
T
on
V
.
Define a natural action of
T
on the quotient
V/W
, and prove that the minimal polynomial of
T
on
V/W
is
a divisor of the minimal polynomial of
T
on
V
.
Let
f
(
x
) be the minimal polynomial of
T
on
V
, and
g
(
x
) the minimal polynomial of
T
on
W
. (We need the
T
stability of
W
for this to make sense at all.) Since
f
(
T
) = 0 on
V
, and since the restriction map
End
k
(
V
)
→
End
k
(
W
)
is a ring homomorphism,
(restriction of)
f
(
t
) =
f
(restriction of
T
)
Thus,
f
(
T
) = 0 on
W
. That is, by definition of
g
(
x
) and the PIDness of
k
[
x
],
f
(
x
) is a multiple of
g
(
x
), as
desired.
Define
T
(
v
+
W
) =
Tv
+
W
. Since
TW
⊂
W
, this is welldefined. Note that we cannot assert, and do not
need, an
equality
TW
=
W
, but only containment. Let
h
(
x
) be the minimal polynomial of
T
(on
V/W
).
Any polynomial
p
(
T
) stabilizes
W
, so gives a welldefined map
p
(
T
) on
V/W
. Further, since the natural
map
End
k
(
V
)
→
End
k
(
V/W
)
is a ring homomorphism, we have
p
(
T
)(
v
+
W
) =
p
(
T
)(
v
) +
W
=
p
(
T
)(
v
+
W
) +
W
=
p
(
T
)(
v
+
W
)
Since
f
(
T
) = 0 on
V
,
f
(
T
) = 0. By definition of minimal polynomial,
h
(
x
)

f
(
x
).
///
[16.3]
Let
T
∈
Hom
k
(
V
) for a finitedimensional
k
vectorspace
V
, with
k
a field.
Suppose that
T
is
diagonalizable
on
V
. Let
W
be a
T
stable subspace of
V
. Show that
T
is diagonalizable on
W
.
Since
T
is diagonalizable, its minimal polynomial
f
(
x
) on
V
factors into linear factors in
k
[
x
] (with zeros
exactly the eigenvalues), and no factor is repeated. By the previous example, the minimal polynomial
g
(
x
)
of
T
on
W
divides
f
(
x
), so (by unique factorization in
k
[
x
]) factors into linear factors without repeats. And
this implies that
T
is diagonalizable when restricted to
W
.
///
[16.4]
Let
T
∈
Hom
k
(
V
) for a finitedimensional
k
vectorspace
V
, with
k
a field.
Suppose that
T
is
diagonalizable
on
V
, with
distinct eigenvalues
. Let
S
∈
Hom
k
(
V
) commute with
T
, in the natural sense that
ST
=
TS
. Show that
S
is diagonalizable on
V
.