23.
Quotient groups II
23.1.
Proof of the fundamental theorem of homomorphisms (FTH).
We start by recalling the statement of FTH introduced last time.
Theorem
(FTH)
.
Let
G
,
H
be groups and
ϕ
:
G
→
H
a homomorphism.
Then
G/
Ker
ϕ
∼
=
ϕ
(
G
)
.
(
* * *
)
Proof.
Let
K
= Ker
ϕ
and deﬁne the map Φ :
G/K
→
ϕ
(
G
) by
Φ(
gK
) =
ϕ
(
g
) for
g
∈
G.
We claim that Φ is a well deﬁned mapping and that Φ is an isomorphism.
Thus we need to check the following four conditions:
(i) Φ is well deﬁned
(ii) Φ is injective
(iii) Φ is surjective
(iv) Φ is a homomorphism
For (i) we need to prove the implication “
g
1
K
=
g
2
K
⇒
Φ(
g
1
K
) = Φ(
g
2
K
).”
So, assume that
g
1
K
=
g
2
K
for some
g
1
,g
2
∈
G
. Then
g

1
1
g
2
∈
K
by
Theorem 19.2, so
ϕ
(
g

1
1
g
2
) =
e
H
(recall that
K
= Ker
ϕ
). Since
ϕ
(
g

1
1
g
2
) =
ϕ
(
g
1
)

1
ϕ
(
g
2
), we get
ϕ
(
g
1
)

1
ϕ
(
g
2
) =
e
H
. Thus,
ϕ
(
g
1
) =
ϕ
(
g
2
), and so
Φ(
g
1
K
) = Φ(
g
2
K
), as desired.
For (ii) we need to prove that “Φ(
g
1
K
) = Φ(
g
2
K
)
⇒
g
1
K
=
g
2
K
.” This
is done by taking the argument in the proof of (i) and reversing all the
implication arrows.
(iii) First note that by construction Codomain(Φ) =
ϕ
(
G
). Thus, for
surjectivity of Φ we need to show that Range(Φ) = Φ(
G/K
) is equal to
ϕ
(
G
). This is clear since
Φ(
G/K
) =
{
Φ(
gK
) :
g
∈
G
}
=
{
ϕ
(
g
) :
g
∈
G
}
=
ϕ
(
G
)
.
(iv) Finally, for any
g
1
,g
2
∈
G
we have
Φ(
g
1
K
·
g
2
K
) = Φ(
g
1
g
2
K
) =
ϕ
(
g
1
g
2
) =
ϕ
(
g
1
)
ϕ
(
g
2
) = Φ(
g
1
K
)Φ(
g
2
K
)
where the ﬁrst equality holds by the deﬁnition of product in quotient groups.
Thus, Φ is a homomorphism.
So, we constructed an isomorphism Φ :
G/
Ker
ϕ
→
ϕ
(
G
), and thus
G/
Ker
ϕ
is isomorphic to
ϕ
(
G
).
±
1
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23.2.
Applications of FTH.
In most applications one uses a special case
of FTH stated last time as Corollary 22.5:
If
ϕ
:
G
→
H
is a surjective homomorphism, then
G/
Ker
ϕ
∼
=
H
. (***)
Typically this result is being applied as follows. We are given a group
G
,
a normal subgroup
K
and another group
H
(unrelated to
G
), and we are
asked to prove that
G/K
∼
=
H
. By (***)
to prove that
G/K
∼
=
H
it
suﬃces to ﬁnd a surjective homomorphism
ϕ
:
G
→
H
such that
Ker
ϕ
=
K
.
Example 1:
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 Winter '11
 PhanThuongCang
 Group Theory, Normal subgroup, Cyclic group, Coset

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