GROUP HOMOMORPHISM
L. MARIZZA A. BAILEY
0.1. Homomorphisms.
Definition 1. Let G and H be a groups.
A group homomorphism is a function φ : G → H such that
φ(g1 g2 ) = φ(g1 )φ(g2 ) for any g1 , g2 ∈ G.
Proposition 1. Let φ : G → H be a homomorphism. Then
(a) φ(1G ) = 1H ;
(b) φ(g −1 ) = φ(g)−1 for every g ∈ G.
Proof. We have φ(1G ) = φ(1G · 1G ) = φ(1G )φ(1G ). Multiplying both sides by
φ(1G )−1 in H, we have 1H = φ(1G ).
Let g ∈ G. Then 1H = φ(1G ) = φ(g −1 g) = φ(g)φ(g −1 ). Multiplying both
sides by φ(g)−1 in H yields φ(g)−1 = φ(g −1 ).
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For example,
φ(n) : R → R∗ defined by φ(x) = ex is a group homomorphism from the
group of real numbers under addition, to the group of nonzero real numbers
under multiplication.
By the properties of exponentiation, we get φ(x + y) = ex+y = ex ey = φ(x)φ(y).
Note that φ(0) = 1, so the identity under addition is sent to the identity under
multiplication.
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Also, we see that φ(−x) = e−x = x . So the additive inverse of x is sent to the
e
multiplicative inverse of ex .
Proposition 2 (The restriction of a homomorphism onto a subgroup is still a
homomorphism). Let φ : G → H be a homomorphism and let K ≤ G.
Then φ ¹K : K → H is a homomorphism.
The proof to this is very simple, and should be a mental exercise.
Date: September 29, 2006.
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Proposition 3 (The image of a subgroup is a subgroup). Let φ : G → H be a
homomorphism and let K ≤ G.
Then φ(K) ≤ H.
Proof 1. Recall, to prove φ(K) ≤ H, we need to verify
(IDENT) Since 1G ∈ K, by definition of K ≤ G, then φ(1G ) ∈ φ(K) by definition
of the image of K. By lemma, φ(1G ) = 1H ∈ φ(K). Therefore, the
identity is in φ(K).
(INV) Let h ∈ φ(K). Then h = φ(k) for some k ∈ K. Therefore, by definition of K ≤ G, we get k −1 ∈ K. Hence, φ(k −1 ) ∈ φ(K), and finally,
φ(k −1 ) = h−1 ∈ φ(K) by lemma.
(CLOS) Let h1 , h2 ∈ φ(K). Then there exists k1 , k2 ∈ K such that φ(k1 ) = h1
and φ(k2 ) = h2 . By definition of subgroup k1 k2 ∈ K which implies
φ(k1 k2 ) ∈ φ(K). By definition of group homomorphism, φ( k1 k2 ) =
φ(k1 )φ(k2 ) = h1 h2 .
Proposition 4 (The Preimage of a subgroup is a subgroup). Let φ : G → H be
a homomorphism and let K ≤ H.
Then φ−1 (K) ≤ G.
Proof. Note that φ−1 (K) is closed under multiplication and inverses since K is
and because φ is a homomorphism.
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Proposition 5 (The composition of homomorphisms is a homomorphism). Let
φ : G → H and ψ : H → K be homomorphisms.
Then ψ ◦ φ : G → K is a homomorphism.
Proposition 6 (The order of the image divides the order of the original element).
Let φ : G → H be a homomorphism and let g ∈ G be an element of finite order.
Then ord(φ(g)) | ord(g).
Proof. Let ord(g) = n. Then φ(g)n = φ(g n ) = φ(1G ) = 1H . Thus n is an
multiple of the order of φ(g).
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Proposition 7 (Generators are mapped to generators). Let φ : G → H be a
homomorphism and let X ⊂ G.
Then hφ(X)i = φ(hXi).
Proof. Let X be the collection of subgroups in G which contain X. If K ∈ X,
then φ(K) is a subgroup of H which contains φ(X). On the other hand, if
M ≤ H is a subgroup of H which contains φ(X), then φ−1 (M ) ∈ X. Thus
hφ(X)i = ∩φ(X), and
φ(hXi) = φ(∩X) = ∩φ(X) = hφ(X)i.
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Corollary 1. The homomorphic image of a cyclic group is cyclic.
Definition 2. An injective homomorphism is called a monomorphism. A surjective homomorphism is called an epimorphism. A bijective homomorphism is
called an isomorphism. If there exists an isomorphism between the groups G
and H, we say that G and H are isomorphic, and write G ∼
= H.
Proposition 8. Let G be a group. Then idG : G → G is an isomorphism.
Proposition 9. Let φ : G → H be an isomorphism.
Then φ−1 : H → G is an isomorphism.
Proof. By definition, φ is bijective, so it is invertible. Let h1 , h2 ∈ H. Since φ
is bijective, there exist unique g1 , g2 ∈ G such that φ(g1 ) = h1 and φ(g2 ) = h2 .
Then h1 h2 = φ(g1 )φ(g2 ) = φ(g1 g2 ). Thus φ−1 (h1 h2 ) = g1 g2 = φ−1 (h1 )φ−1 (h2 ).
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Proposition 10. Let G be a collection of groups.
Then isomorphism is an equivalence relation on G.
Proof. The identity map on a group establishes the reflexivity of isomorphism.
The symmetry relation is estabilished by the fact that bijective maps are invertible. The transitivity relation is given by the fact that the composition of
homomorphisms is a homomorphism, and the composition of bijections is a bijection.
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Definition 3. Let φ : G → H be a homomorphism. The kernel of φ is the
subset of G denoted by ker(φ) and defined by
ker(φ) = {g ∈ G | φ(g) = 1H }.
This is easily seen to be a subgroup.
Proposition 11. Let φ : G → H be a homomorphism.
Then φ is injective if and only if ker(φ) = {1G }.
Proof.
(⇒) Suppose the φ is injective. Since the identity of G maps to the identity of
H, no other element of G may map to the identity of H.
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(⇐) Suppose that ker(φ) is trivial. Then
φ(g1 ) = φ(g2 ) ⇔ φ(g1 )φ(g2 )−1 = 1H
⇔ φ(g1 )φ(g2−1 ) = 1H
⇔ φ(g1 g2−1 ) = 1H
⇔ g1 g2−1 = 1G
⇔ g1 = g2 .
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0.2. Cosets.
Definition 4. Let G be a group and H ≤ G. Let g ∈ G.
The left coset at g of H in G is the set
gH = {gh | h ∈ H}.
The right coset at g of H in G is the set
Hg = {hg | h ∈ H}.
Proposition 12. Let G be a group and H ≤ G. Let g, g1 , g2 ∈ G. Then
(a) g ∈ gH;
(b) g ∈ Hg;
(c) gH = H ⇔ g ∈ H;
(d) Hg = H ⇔ g ∈ H;
(e) g1 H = g2 H ⇔ g1−1 g2 ∈ H;
(f ) Hg1 = Hg2 ⇔ g2 g1−1 ∈ H.
Proof. First note that since 1 ∈ G, H is a coset of H in G, specifically, H = 1H.
Also g ∈ gH since g = g · 1. This proves (a).
Thus if gH = H, then g ∈ H.
If g ∈ H, then gH ⊂ H by closure, because H is a group. Since g −1 is also in
H, we have g −1 H ⊂ H, so H ⊂ gH; thus gH = H. This proves (c).
If g1 H = g2 H, then H = g1−1 g2 H, so g1−1 g2 ∈ H. If g1−1 g2 ∈ H, then
g1−1 g2 H = H, so g2 H = g1 H. This proves (e).
The proofs for right cosets are analogous.
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Definition 5. Let G be a group and let H ≤ G. Let g1 , g2 ∈ G.
We say that g1 and g2 are left congruent modulo H if g1−1 g2 ∈ H.
We say that g1 and g2 are right congruent modulo H if g1 g2−1 ∈ H.
Proposition 13. Let G be a group and H ≤ G. Then left and right congruence
modulo H is an equivalence relation.
Proof. Let g ∈ G. Then g −1 g = 1 ∈ H, so g is left congruent to itself modulo
H. Thus left congruence is reflexive.
Let g1 , g2 ∈ G. Suppose that g1−1 g2 ∈ H. Thus g2−1 g1 = (g1−1 g2 )−1 ∈ H since
H is closed under inverses. Thus left congruence is symmetric.
Let g1 , g2 , g3 ∈ G. Suppose that g1−1 g2 ∈ H and g2−1 g3 ∈ H. Then g1−1 g3 =
−1
g1 g2 g2−1 g3 ∈ H since H is closed under multipliation. Thus left congruence is
transitive.
The proof for right congruence is analogous.
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Corollary 2. Let G be a group and let H ≤ G. Then the collection of left (right)
cosets of H in G partition G.
Proposition 14. Let G be a group and let H ≤ G. Let g ∈ G. Then the maps
λg : H → gH given by φ(h) = gh
and
ρg : H → Hg given by φ(h) = hg
are bijective.
Proof. Let gh ∈ gH; then h 7→ gh, so λg is surjective. Let gh1 , gh2 ∈ gH; then
h1 = h2 by cancellation, so λg is injective. The proof for ρg is analogous.
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Corollary 3. Let G be a group and let H ≤ G. Let g ∈ G. Then |gH| = |Hg| =
|H|.
Definition 6. Let G be a group and H ≤ G. The collection of left cosets of H
in G is called the left coset space of H in G. The collection of right cosets of H
in G is called the right coset space of H in G.
Proposition 15. Let G be a group and H ≤ G. Then there is a correspondence
between the left and right coset spaces of H in G given by
gH ↔ Hg −1 .
Proof. The map φ : gH 7→ Hg −1 is well-defined and injective:
g1 H = g2 H ⇔ g2−1 g1 H = H
⇔ g2−1 g1 ∈ H
⇔ Hg2−1 g1 = H
⇔ Hg2−1 = Hg1−1 .
Since φ(g −1 H) = Hg, the map is surjective.
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Corollary 4. Let G be a group and H ≤ G. Then the left coset space of H in
G and the right coset space of H in G have the same cardinality.
Definition 7. Let G be a group and let H ≤ G.
The index of H is G is the cardinality of the left coset space of H in G, and
is denoted by [G : H].
Theorem 1 (Lagrange’s Theorem). Let G be a finite group and H ≤ G. Then
|G| = |H|[G : H].
Proof. The cardinality of each left coset is the cardinality of H. There are [G : H]
of these in G. Since these cosets form a partition of G, the result follows.
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Proposition 16. Let G be a finite group and let g ∈ G.
Then ord(g) divides |G|.
Proof. Since hgi ≤ G and ord(g) = |hgi|, the result follows from Lagrange’s
Theorem.
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Proposition 17. Let G be a finite group such that |G| is prime.
Then G is cyclic.
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Proof. Let G be a group of order p, where p is prime. Let p ∈ G. Then ord(g)||G|
so ord(g) = p or ord(g) = 1. Thus if g 6= 1 then G = hgi.
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0.3. Normal Subgroups.
Definition 8. Let G be a group and H ≤ G.
We say that H is a normal subgroup, and write H / G, if gH = Hg for every
g ∈ G.
Proposition 18. Let G be a group and let H ≤ G. The following conditions
are equivalent:
(1) gH = Hg for every g ∈ G;
(2) g −1 Hg = H for every g ∈ G;
(3) g −1 Hg ⊂ H for every g ∈ G.
Proof. That (1) ⇔ (2) and (2) ⇒ (3) are obvious.
Suppose that g −1 Hg ⊂ H for every g ∈ G. Let g ∈ G; then g −1 ∈ G, so
gHg −1 ⊂ H. Thus H ⊂ g −1 Hg.
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Proposition 19. Let G be a group. Then G / G and 1 / G.
Proposition 20. Let G be a group and let N be a collection of normal subgroups
of G. Then ∩N is a normal subgroup of G.
Proof. Let N = {Hα / G | α ∈ A}, where A is some indexing set. Then for any
g ∈ G,
g −1 (∩α∈A Hα )g = ∩α∈A g −1 Hα g = ∩α∈A Hα
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Proposition 21. Let G be an abelian group and let H ≤ G. Then H / G.
Proof. For H / G and g ∈ G, h ∈ H we have gh = hg. Thus gH = Hg.
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Proposition 22. Let φ : G → H be a homomorphism and let K ∈ ker(φ). Then
K / G.
Proof. We have φ(g −1 Kg) = φ(g)−1 φ(K)φ(g) = φ(g)1 · 1H φ(g) = 1H . Thus
g −1 Kg ⊂ K.
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Definition 9. Let G be a group and let X, Y ⊂ G. Set
XY = {xy ∈ G | x ∈ X and y ∈ Y }
and
X −1 = {x−1 ∈ G | x ∈ X}.
Proposition 23. Let G be a group and let H, K ≤ G. Then HK ≤ G if and
only if HK = KH.
Proof. If M ≤ G, then M −1 = M . Thus if HK ≤ G, then HK = (HK)−1 =
K −1 H −1 = KH.
Suppose HK = KH. Let h1 , h2 ∈ H and k1 , k2 ∈ K so that h1 k1 and h2 k2
are arbitrary members of HK. Since HK = KH, there exists k3 ∈ K such that
k1 h2 = h2 k3 . Then h1 k1 h2 k2 = h1 h2 k3 k2 ∈ HK.
Let h ∈ H and k ∈ K so that hk is an arbitrary member of HK. Then
(hk)−1 = k −1 h−1 ∈ KH = HK. Thus HK ≤ G.
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Proposition 24. Let G be a group, H ≤ G, and K / G. Then HK = KH, and
HK ≤ G.
Proof. We have hK = Kh for every h ∈ H, so HK = KH. Thus HK ≤ G by
the previous proposition.
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1. Groups of Order Four
Proposition 25. Let G be a finite group and let g ∈ G. Then ord(g) divides
|G|.
Proof. The order of g is equal to the order of the cyclic subgroup generated by
g. By LaGrange’s Theorem, this order divides the order of the group G.
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Proposition 26. Let G be a group such that g 2 = 1 for every g ∈ G. Then G
is abelian.
Proof. For g ∈ G, g 2 = 1 ⇒ g = g −1 . For g, h ∈ G, gh ∈ G so (gh)2 = 1. That
is, 1 = ghgh = g −1 h−1 gh. Thus hg = gh.
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Proposition 27. Every group of order four is abelian.
Proof. Let G be a group of order four. If G has an element of order 4, then
this element generates G, so G is cyclic, and therefore is abelian in this case.
Otherwise, every element of G has order 2 or 1, so g 2 = 1 for every g ∈ G; thus
G is abelian in this case.
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Proposition 28. Let G be a group of order four.
Then either G ∼
= Z4 or G ∼
= Z2 × Z2 .
Proof. Suppose that G has an element g of order 4. Then G = hgi = {1, g, g 2 , g 3 }.
Let φ : G → Z4 be given by g n 7→ n. So φ(g m+n ) = m + n = m + n = φ(g m ) +
φ(g n ), so φ is a homomorphism. It is easily seen that φ is a homomorphism.
Clearly φ is bijective, so φ is an isomorphism.
Suppose that G does not have an element of order 4. Since G has exactly one
element of order 1, that being the identity, and since the order of an element
divides the order of the group, the other element of G have order 2.
Let g, h ∈ G be element of order 2. Then G = {1, g, h, gh}. Let φ : G →
Z2 × Z2 be given by 1 7→ (0, 0), g 7→ (1, 0), h 7→ (0, 1), and gh 7→ (1, 1). Direct
computation shows that this is an isomorphism.
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Definition 10. Let K4 = {1, g, h, gh} such that g 2 = 1 and h2 = 1. These
relations completely determine the a group structure on K4 , called the Klein
Four group.
2. Homomorphisms
Proposition 29. Let φ : G → H be a group homomorphism. Let g ∈ G be an
element of finite order. Then ord(φ(g)) divides ord(g).
Proof. Note that 1H = φ(1G ) = φ(g ord(g) ) = φ(g)ord(g) . Thus ord(φ(g))|ord(g).
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Problem 1. Find all injective homomorphisms K4 → D5 .
Proof. Solution The image of an injective homomorphism is a subgroup of the
range which is isomorphic to the domain. The order of a subgroup of D5 divides
10, since |D5 | = 10. Thus D5 has no subgroup of order 4, an there cannot be an
injective homomorphism K4 → D5 .
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Problem 2. Find all surjective homomorphisms Z5 × Z2 → D5 .
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Proof. Solution Since |Z5 × Z2 | = |D5 |, any surjective homomorphism would be
bijective, and hence an isomorphism. But these groups cannot be isomorphic,
since Z5 × Z2 is abelian and D5 isn’t.
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Problem 3. Find all surjective homomorphisms Z10 × Z2 → Z8 .
Proof. Solution It follows from the isomorphism theorem that the order of the
image of a group homomorphism divides the order of the domain. Thus we
cannot have a group of order 8 as a homomorphic image of a group of order
20.
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Problem 4. Find all surjective homomorphisms Z10 × Z2 → D5 .
Proof. Solution The homomorphic image of an abelian group is abelian, so again
there are none.
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Problem 5. Find all homomorphisms K4 to S3 .
Proof. Solution Let σ = (123) and τ = (12). We know that
S3 = {², σ, σ 2 , τ, τ σ, τ σ 2 }.
Let φ : K4 → S3 be a homomorphism. Then ord(φ(g))|2, so φ(g) ∈
{², τ, τ σ, τ σ 2 }; similarly
φ(h) = ²; otherwise, φ(gh) = φ(g)φ(h) is an element of order 3. On the other
hand, we get a homomorphism in both of these cases, as one can check. Similarly,
we may send g to any of 3 reflections, and h to either the same reflection or to
²; the same comments apply to h. Thus we get ten homomorphisms:
•
•
•
•
•
•
•
•
•
•
g
g
g
g
g
g
g
g
g
g
7→ ², h 7→ ²;
7→ ², h 7→ τ ;
7→ ², h 7→ τ σ;
7→ ², h 7→ τ σ 2 ;
7→ τ, h 7→ ²;
7→ τ, h 7→ τ ;
7→ τ σ, h 7→ ²;
7→ τ σ, h 7→ τ σ;
7→ τ σ 2 , h 7→ ²;
7→ τ σ 2 , h 7→ τ σ 2 .
Note that the kernel of each of these homomorphisms contains one of g, h, or
gh, or all three.
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3. Automorphisms
Problem 6. Find an automorphism which is not an inner automorphism.
Proof. Let G = hgi be a cyclic group of order 5. Then G is abelian. The map
φ : G → G given by g n 7→ g 2n is a homomorphism, since φ(g m g n ) = φ(g m+n ) =
g 2(m+n) = g 2m g 2n . It is injective since if g n ∈ ker(φ), then g 2n = 1, so 5|2n since
ord(g) = 5. Since 5 is prime, we have 5|n. Thus g n = 1 in G. Since g 7→ g 2 , φ is
not the identity homomorphism.
Now let h ∈ G. Then innh (g n ) = hg n h−1 = g n , since G is abelian. Thus innh
is the identity homomorphism, and φ is not an inner automorphism.
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Definition 11. Let G be a group and let φ ∈ Aut(G). The fixed set of φ is
Fix(φ) = {g ∈ G | φ(g) = g}.
If H ≤ Aut(G), the fixed set of H is
Fix(H) = {g ∈ G | φ(g) = g for all φ ∈ H}
= ∩φ∈H Fix(φ).
Problem 7. Let G be a group and let φ ∈ Aut(G). Show that Fix(φ) ≤ G.
Proof. Solution Let g, h ∈ Fix(φ). Then φ(gh) = φ(g)φ(h) = gh, so gh ∈ Fix(φ).
Also φ(g −1 ) = φ(g)−1 = g −1 , so g −1 ∈ Fix(φ). Thus Fix(φ) ≤ G.
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Problem 8. Let G be a group and assume inng ∈ Aut(G). Find Fix(inng ).
Proof. Solution Let x ∈ Fix(inng ). Then gxg −1 = x, so gx = xg. Thus x ∈
CG (g). On the other hand, if gx = xg, we have inng (x) = x. Thus Fix(inng ) =
CG (g), the set of all elements which commute with g.
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Problem 9. Let G be a group and assume Inn(G) ≤ Aut(G). Find Fix(Inn(G)).
Proof. Solution We have
Fix(Inn(G)) = ∩g∈G Fix(inng )
= ∩g∈G CG (g)
= Z(G),
the set of all elements of G which commute with every element of G.
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Definition 12. Let G be a group and let g ∈ G. The stabilizer of g in Aut(G)
is
Stb(g) = {φ ∈ Aut(G) | φ(g) = g.
Clearly φ ∈ Stb(g) ⇔ g ∈ Fix(φ).
Problem 10. Let G be a group and let g ∈ G. Show that Stb(g) ≤ Aut(G).
Proof. Solution Let φ, ψ ∈ Stb(g). Then (ψ ◦ φ) = ψ(φ(g)) = ψ(g) = g, so
ψ ◦ φ ∈ Stb(g). Also if φ(g) = g, then φ−1 (φ(g)) = φ(g), so φ−1 (g) = g and
φ−1 ∈ Stb(g). Thus Stb(g) ≤ Aut(G).
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4. Lifting
Problem 11. Let G be a finite group and let H / G. Let p be a prime integer.
Suppose that G/H has an element of order p. Show that G has an element of
order p.
Proof. Solution Let G = G/H and for g ∈ G, denote the coset gH by g.
Let g ∈ G be an element of order p. Then g p = g p = 1, which means that
p
g ∈ H. Since G is finite, the element g p has finite order, say ord(g p ) = k. Let
h = g k . Then hp = (g k )p = (g p )k = 1, so ord(h)|p; thus ord(h) = 1 or p. If
ord(h) = 1, this means that h = 1, in which case ord(g) = p. Otherwise h has
order p.
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Problem 12. Let G be a finite group and let g ∈ G. Suppose that CG (g) / G.
Show that every element of G has a power which commutes with g.
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Proof. Solution Since CG (g) is normal, G/CG (g) is a group. Let G = G/CG (g)
and for h ∈ G, let h = hCG (g).
Let h ∈ G. Since G is finite, so is G. Thus h has finite order, say ord(h) = n.
n
Thus hn = h = 1, which means that hn ∈ CG (g). Thus hn commutes with
g.
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Problem 13. Show that Q/Z is an infinite abelian group in which every element
has finite order.
Proof. Solution Since Q is abelian and Z ≤ Q, we have Z / Q. Thus Q/Z is
a group. A factor group of an abelian group is always abelian, so Q/Z is an
abelian group.
Note that two rational numbers represent the same coset if and only if their
difference is an integer. Then Q/Z is infinite, since there are an infinite number
of rational numbers such that none of their differences is an integer.
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For example, let m and n be positive integers. Then n1 − m
= m−n
mn , which is
not an integer unless mn divides m − n. Since |m − n| ≤ mn, this is never the
case. Thus the rational numbers of the form n1 represent an infinite number of
distinct cosets of Q/Z.
m
m
Now let m
n + Z be a coset. Then n n = m ∈ Z, so the order of n divides m,
and in particular, it is finite.
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Department of Mathematics, Arkansas School of Mathematics, Sciences and the
Arts
E-mail address: [email protected]