Answer key No.2 to selected homework problems: Math. 110B
7.6
1.
aK
=
K
means
ae
=
k
∈
K
; so,
a
∈
K
.
Conversely, if
a
∈
K
, then
ak
∈
K
for all
k
∈
K
because
K
is a subgroup.
This shows
aK
⊂
K
.
Pick any
k
∈
K
, then
k
=
ek
= (
aa

1
)
k
=
a
(
a

1
k
)
∈
aK
, because
K
is subgroup. Thus
K
⊂
aK
. This combined the earlier reverse inclusion
shows that
K
=
aK
.
2.
(a) (1) Every left coset
aK
is a subset of
G
; so,
a
aK
⊂
G
. For each
g
∈
G
,
g
=
ge
∈
gK
; so,
G
⊃
a
aK
.
(2) Define
f
:
K
→
aK
by
f
(
k
) =
ak
. To show
f
is injective, assume
that
f
(
x
) =
f
(
y
). Then we need to show that
x
=
y
.
f
(
x
) =
f
(
y
)
implies
ax
=
ay
; then, by leftcancellation, we find
x
=
y
.
x
∈
aK
implies
x
=
ak
for
k
∈
K
; so,
x
=
f
(
k
); therefore,
f
is surjective.
Thus
f
is a bijection. In particular

aK

=

K

if

K

is finite.
(b) We need to show that there are only two possibilities:
aK
=
bK
or
aK
∩
bK
=
∅
. Thus we need to show:
aK
∩
bK
=
∅ ⇒
aK
=
bK
.
Pick
x
=
ak
=
bk
in the intersection.
Then
b

1
a
=
k k

1
∈
K
.
Thus
a

1
bk
∈
K
for all
k
∈
K
; in particular,
bk
=
ebk
= (
aa

1
)
bk
=
a
(
a

1
bk
)
∈
aK
; thus
aK
⊂
bK
. Interchanging the role of
a
and
b
in the above argument, we find the reverse inclusion
aK
⊃
bK
; so,
aK
=
bK
as desired. We are going to prove that
G
is a union of left
cosets
c
1
K
,
c
2
K
,... without overlapping (such a union is written as
G
=
c
1
K
c
2
K
· · ·
c
m
K
(here
m
= [
G
:
K
]) using the disjoint
union symbol “
”. Now start with
K
=
eK
, and write
c
1
=
e
. Pick
c
2
outside
c
i
K
=
K
if any. If there is none, we have
G
=
K
. Otherwise,
c
1
K
=
c
2
K
because
c
2
is not in
c
1
K
; so, we have
c
1
K
∩
c
2
K
=
∅
.
Suppose we have found
c
1
, . . ., c
n
so that
c
i
K
∩
c
j
K
=
∅
for any pairs
(
i, j
) with
i
=
j
. By induction on
n
, we try to find
c
n
+1
with the same
disjointness property. Take
c
=
c
n
+1
outside
c
1
K
∪
c
2
K
∪ · · · ∪
c
n
K
if any.
If there is none, we have
G
=
c
1
K
c
2
K
· · ·
c
n
K
; so,
we are done, getting a disjoint union expression of
G
by left cosets.
Otherwise, we have
c
j
K
=
cK
for all
j
= 1
,
2
, . . ., n
because
c
∈
c
j
K
.
Thus only possibility is to have
cK
∩
c
j
K
=
∅
; so, by induction on
n
,
we can keep going. Since
G
is a finite group, the sequence
c
1
, c
2
, . . .
has to end at some point (exhausting all elements), so at the end, we
find the disjoint union expression. Then

G

=
∑
m
j
=1

c
j
K

. By (a),

c
j
K

=

K

, we find

G

=
m

K

; so, we are done.
(c) By (b), the number of left cosets is

G

/

K

. The number of right
cosets is also

G

/

K

by Lagrange theorem; so, they are equal to each
other.
4.
•
Note that
e
is made of single element
e
. Then
xex

1
=
xx

1
=
e
;
so,
x e x

1
⊂
e
; so, by Theorem 7.34 (3),
e
is normal.
•
Since
G
is a group
xGy
⊂
G
; in particular,
xGx

1
⊂
G
; so, again by
Theorem 7.34 (3), it is normal.
5.
(a) Hint: Compute the product of two upper triangular matrices to show
it is again upper triangular. Also inverse of an upper triangular matrix is
again upper triangular by computation.
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 Winter '08
 HIDA
 Math, Algebra, Abelian group, Subgroup, Cyclic group, Coset, Index of a subgroup

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