Download 3.2 (III-14) Factor Groups

3.2. (III-14) FACTOR GROUPS
3.2
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(III-14) Factor Groups
Question: If H is a subgroup of G, when can we have the left coset multiplication (aH)(bH) = abH for all a, b ∈ G?
If such equality exists, then {aH | a ∈ G} inherits a group structure.
3.2.1
Factor groups from homomorphisms
Let φ : G → G0 is a group homomorphism and H := ker(φ). Then
φ−1 [{φ(a)}] = aH = Ha
for a ∈ G.
Therefore, aH ↔ φ(a) gives a bijection between cosets of H in G and the
subgroup φ[G] of G0 . We see that the cosets of H corresponds to a group structure.
Denote by G/H the set of all cosets of H in G, read as “G over H” or as
“G modulo H” or as “G mod H”. The above bijection is G/ ker(φ) ↔ φ[G].
Given x ∈ aH and y ∈ bH, we have φ(x) = φ(a) and φ(y) = φ(b). So
ab = φ(x)φ(y) = φ(xy)
Therefore, xy ∈ φ−1 [{ab}] = abH. Thus (aH)(bH) = abH for H = ker(φ).
Thm 3.23. Let φ : G → G0 be a group homomorphism and H := ker(φ).
Then the cosets of H form a factor group G/H, where (aH)(bH) := abH
is well-defined. The map µ : G/H → φ[G], defined by µ(aH) := φ(a), is an
isomorphism.
Ex 3.24 (Ex 14.2, p.137, Z/5Z ' Z5 ). 3 Define φ : Z → Z5 by φ(x) :=
x mod 5. Then φ is a homomorphism, with ker(φ) = 5Z. The set of all
cosets of 5Z is Z/5Z, which consists of the cosets
5Z = {· · · , −5, 0, 5, 10, · · · },
1 + 5Z = {· · · , −4, 1, 6, 11, · · · },
2 + 5Z = {· · · , −3, 2, 7, 12, · · · },
3 + 5Z = {· · · , −2, 3, 8, 13, · · · },
4 + 5Z = {· · · , −1, 4, 9, 14, · · · }.
We see that Z/5Z has a natural group structure by
(a + 5Z) + (b + 5Z) = a + b + 5Z.
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CHAPTER 3. HOMOMORPHISMS AND FACTOR GROUPS
The identity of Z/5Z is the coset 0 + 5Z. The inverse of a + 5Z in Z/5Z is
−a + 5Z.
The homomorphism φ induces an isomorphism µ : Z/5Z → Z5 by
µ(a + 5Z) := a mod 5.
In general, the factor group Z/nZ ' Zn .
Ex 3.25 (Ex 14.3, p.137). The coset additions in the above example are
independent of the representatives chosen.
Def 3.26. 4 When G/H is a group, the group G/H is often called the
factor group of G modulo H. Elements in the same coset of H are said to
be congruence modulo H.
Ex 3.27 (Hw 9, p.142). Find the order of 5 + h4i in the factor group
Z12 /h4i.
Ex 3.28 (Hw 2, p.142). Find the order of the factor group (Z4 ×Z12 )/h(2, 2)i.
3.2.2
Factor group from normal subgroups
In previous section, we define the factor group G/ ker(φ) from a homomorphism φ : G → G0 . Notice that ker(φ) is a normal subgroup of G. Is this
just an incident?
Thm 3.29. Let H be a subgroup of G. Then (aH)(bH) = abH for all
a, b ∈ H if and only if H is normal, that is, gH = Hg for all g ∈ G.
Proof. If gH = Hg for all g ∈ G, then the set multiplication
(aH)(bH) = a(Hb)H = a(bH)H = abH.
Conversely, if (aH)(bH) = abH for all a, b ∈ G, then
Hg ⊆ (eH)(gH) = egH = gH,
and
gHg −1 ⊆ (gH)(g −1 H) = gg −1 H = H.
implies that gH ⊆ Hg. So gH = Hg for g ∈ G.
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3.2. (III-14) FACTOR GROUPS
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Cor 3.30. 5 If H is a normal subgroup of G, then the cosets of H form a
factor group (or called quotient group) G/H.
Proof. Check the associativity, identity, and inverse properties for G/H.
Ex 3.31. Every subgroup of an abelian group is normal.
Ex 3.32 (Hw 23, True or False).
3.2.3
The Fundamental Homomorphism Theorem
We know that the kernel of a homomorphism is normal. Conversely, the
below example indicates that a normal subgroup is the kernel of a homomorphism.
Thm 3.33. If H is a normal subgroup of G, then H is the kernel of the
canonical quotient homomorphism
γ : G → G/H,
defined by
γ(g) := gH.
Thm 3.34 (The Fundamental Homomorphism Theorem). Let φ :
G → G0 be a group homomorphism with kernel H. Then µ : G/H → φ[G],
defined by µ(gH) := φ(g), is an isomorphism. If γ : G → G/H is the
homomorphism given by γ(g) := gH, then φ(g) = µγ(g) for g ∈ G. (See
Figure 14.10)
Ex 3.35 (Ex 14.12, p.140). Classify the group (Z4 × Z2 )/({0} × Z2 ).
3.2.4
Normal Subgroups and Inner Automorphisms
A subgroup H of G is normal if gH = Hg for all g ∈ G. Multiplying g −1
from the right, we have gHg −1 = H. This section shows that gxg −1 defines
a group automorphism.
Caution: gH = Hg is a set equation. It does NOT imply that gh = hg
for all h ∈ H.
Thm 3.36. Three equivalent conditions for a subgroup H of G to be a
normal subgroup of G:
1. ghg −1 ∈ H for all h ∈ H and g ∈ G.
2. gHg −1 = H for all g ∈ G.
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CHAPTER 3. HOMOMORPHISMS AND FACTOR GROUPS
3. gH = Hg for all g ∈ G.
Ex 3.37. Let G be a group and g ∈ G. Define a map ig : G → G such that
ig (x) := gxg −1 . Then ig is an isomorphism of G with itself.
Def 3.38. An isomorphism φ : G → G of G with itself is an automorphism of
G. The isomorphism ig : G → G, where ig (x) := gxg −1 for all x ∈ G, is the
inner automorphism of G by g. The element ig (x) is called the conjugation
of x by g.
Remark: A subgroup H is normal if and only if ig [H] = H for all g ∈ G,
that is, each g ∈ G induces an automorphism of H. Note that ig [H] = H
does NOT imply that ig (h) = h for all h ∈ H.
3.2.5
Homework, III-14, p.142-p.143
1st 1, 6, 10, 12, 40
2nd 8, 21, 24, 31, 32
3rd 26, 27, 29, 35
OPT 30, 33, 34, 37, 39