Spring 2010
MATH 4320: Solutions to Prelim 2
Instructor: Yuri Berest
Problem 1.
(
a
) This is straightforward: for example, we have
π
1
[(
x
1
, x
2
)
·
(
y
1
, y
2
))] =
π
1
[(
x
1
y
1
, x
2
y
2
)] =
x
1
y
1
=
π
1
[(
x
1
, x
2
)]
π
1
[(
y
1
, y
2
)]
,
and similarly for
π
2
. Both
π
1
and
π
2
are surjective, because given any
x
1
∈
H
1
and any
x
2
∈
H
2
, we have
π
1
[(
x
1
, x
2
)] =
x
1
and
π
2
[(
x
1
, x
2
)] =
x
2
. Finally,
Ker(
π
1
) = (
e
1
, H
2
) and Ker(
π
2
) = (
H
1
, e
2
) .
(
b
) Given any group
G
with two homomorphisms
f
1
:
G
→
H
1
and
f
2
:
G
→
H
2
, we deﬁne
ϕ
:
G
→
H
1
×
H
2
by
ϕ
(
x
) := (
f
1
(
x
)
, f
2
(
x
)) for
x
∈
G
.
Then, we see at once that (
π
1
◦
ϕ
)(
x
) =
π
1
[
ϕ
(
x
)] =
π
1
[(
f
1
(
x
)
, f
2
(
x
))] =
f
1
(
x
)
and (
π
2
◦
ϕ
)(
x
) =
π
2
[
ϕ
(
x
)] =
π
2
[(
f
1
(
x
)
, f
2
(
x
))] =
f
2
(
x
) for all
x
∈
G
. In
other words,
f
1
=
π
1
◦
ϕ
and
f
2
=
π
2
◦
ϕ
as required.
Now, suppose there is another homomorphism, say
ψ
:
G
→
H
1
×
H
2
,
satisfying
f
1
=
π
1
◦
ψ
and
f
2
=
π
2
◦
ψ
. By deﬁnition of
H
1
×
H
2
, the
elements
ψ
(
x
) in the image of
ψ
can be written as
ψ
(
x
) = (
g
1
(
x
)
, g
2
(
x
)) ,
where
g
1
(
x
)
∈
H
1
and
g
2
(
x
)
∈
H
2
. Since
f
1
=
π
1
◦
ψ
and
f
2
=
π
2
◦
ψ
,
we have
f
1
(
x
) = (
π
1
◦
ψ
)(
x
) =
π
1
[
ψ
(
x
)] =
π
1
[(
g
1
(
x
)
, g
2
(
x
))] =
g
1
(
x
) and
f
2
(
x
) = (
π
2
◦
ψ
)(
x
) =
π
2
[
ψ
(
x
)] =
π
2
[(
g
1
(
x
)
, g
2
(
x
))] =
g
2
(
x
) for all
x
∈
G
.
Thus,
g
1
(
x
) =
f
1
(
x
) and
g
2
(
x
) =
f
2
(
x
) , and therefore
ψ
(
x
) =
ϕ
(
x
) for all
x
∈
G
. This proves that
ψ
=
ϕ
, which means the uniqueness of
ϕ
.
Problem 2.
(
a
) For each
h
∈
H
and for all
x
1
and
x
2
in
N
, we have
f
(
h
)(
x
1
x
2
) :=
h
(
x
1
x
2
)
h

1
=
h x
1
(
h

1
h
)
x
2
h

1
= (
h x
1
h

1
) (
h x
2
h

1
)
.
So