Chapter 3- Constructing Groups

Question 1

Theorem 10.1 (Cayley)
Let G be a group, and let SG be the symmetric group on the set G. For each a ∈ G the mapping σ_a : G −→ G defined by xσ_a = xa is a permutation of G. The set H = {σ_a : a ∈ G} is a subgroup of SG and is isomorphic to G.

Answer 1

Therefore, every group is isomorphic to a subgroup of a symmetric group.

proof:show bijective homomorphism, pg 37(The proof is probably more important

Question 2

Learning:
Remark 10.3. Cayley’s theorem implies that a group of order n is isomor- phic to a subgroup of Sn. However, it often happens that G is isomorphic to a subgroup of Sm for smaller than n. For example, Cayley’s theorem implies that the alternating group An is isomorphic to a subgroup of Sn!/2. However, An is also a subgroup of Sn (that is how it is defined).

Question 3

A subgroup of a symmetric group is called

Answer 3

a permutation group.
So the alternating groups an dihedral groups are permutation groups.

Question 4

Theorem 10.4.
Let F be a field, and let n be a positive integer. For a permutation σ∈Sn. define an n×n matrix Aσ =(aij) by
aij = 1 if iσ=j,
aij = 0 otherwise
The mapping f : Sn −→ GL(n,F), f(σ) = Aσ is…
Therfore…

Answer 4

a monomorphism. Therefore, every group of order n is isomorphic to a subgroup of the general linear group GL(n, F).
Proof pg 38+39

Question 5

Definition11.1.
An action of a group G on a set X is a function:
X×G−>X, (x, g) 􏰀→ xg, such that the following hold:

Answer 5

xe=x, x(gh)=(xg)h forallx∈X, g,h∈G. We also say that G acts on X,or that X is a G-set.

Question 6

Example 11.2. The symmetric group SX acts on X via…

Answer 6

(x, σ) 􏰀→ xσ.

Question 7

Any group G acts on itself by
And also via

Answer 7

Right translation, (x, g) 􏰀→ xg,
where x, g ∈ G, and xg is understood as the product in G.
Conjugation (x, g) 􏰀→ g−1xg.

Question 8

Let G be a group and H≤G.Then G acts on the set
G/H = {Ha : a∈G} of all right cosets of H via…

Answer 8

(Ha,g)􏰀→Hag.

Question 9

Theorem 11.6.
Let G be a group that acts on a set X. The relation ∼ on X defined by…
is an equivalence relation.

Answer 9

x ∼ y ⇔ xg = y for some g ∈ G
Simple proof, pg 40

Question 10

Let G be a group that acts on a set X. The relation ∼ on X defined by x ∼ y ⇔ xg = y for some g ∈ G is an equivalence relation.
Definition 11.7.
The equivalence classes of the relation ∼ from the previous theorem are called
Thus the … of x ∈ X is…

Answer 10

Orbits
C_x= ={xg : g∈G}.

Question 11

If there is only 1 orbit, then we say that the action is

Answer 11

transitive

Question 12

Definition 11.8.
The action of SX on X, given by (x, σ) 􏰀→ xσ is…
Likewise, the action of G on itself by right translations

Answer 12

transitive

Question 13

The orbits of the conjugation action of G on itself are precisely

Answer 13

the conjugacy classes of G.

Question 14

Theorem 11.9.
Let G be a group that acts on a set X. Then for any x ∈ X the set
G_x ={g∈G : xg=x}
is…

Answer 14

A subgroup of G

proof: subgroup test pg 40

Question 15

Definition 11.10.
The subgroup Gx
G_x ={g∈G : xg=x}
is called …

Answer 15

the stabiliser of x.

Question 16

Theorem 11.12 (Orbit–Stabiliser Theorem)
Let G be a group that acts on a set X. Then for any x∈X, the size of the orbit of x is…

Answer 16

equal to the number of cosets of its stabiliser:
C_x=[G : Gx]
proof short: make bijection pg 41

It follows that the sizes of orbits of G divide the order of G.

Question 17

Corollary 11.13.
The size of the conjugacy class of any element in a group G…

Answer 17

divides the order of G.
This follows from thm 11.12

Question 18

Theorem 11.14.
If a group G acts on the set X, then the mapping …….. is a homomorphism.
Conversely, if f : G → SX is a homomorphism, then the mapping ………….. is an action

Answer 18

f :G→SX , g􏰀→τg , xτg =xg (x∈X, g∈G)

X×G􏰀→x, (x,g)􏰀→x(f(g)) (x∈X, g∈G)

proof “short” pg 41

Question 19

Definition 12.1.
Let G and H be two groups. The direct product of G and H is the set G × H with the component-wise multiplication:

Answer 19

(g1, h1)(g2, h2) = (g1g2, h1h2).

Question 20

Theorem 12.2.(Basic properties of Direct Products)
Let G and H be any two groups…
i)
ii)
iii)
iv)
v)

Answer 20

(i) G×H is a group.
(ii) |G × H| = |G||H|.
(iii)The set G_BAR = {(g,eH):g∈G} is a normal subgroup of G×H, and is isomorphic to G. Similarly, the set H_BAR = {(eG,h) : h ∈ H} is a normal subgroup of G × H and is isomorphic to H.
(iv) G_BAR∩H_BAR={(eG,eH)}.
(v) G_BARH_BAR=G×H.

Proof pg 42

Question 21

Definition 12.3.
A group G is directly decomposable if

Answer 21

there exist non-trivial groups H and K such that
G ∼= H × K; otherwise, G is directly indecomposable.

Question 22

Learning:
One may ask when a group is directly decomposable. By Theorem 12.2 it follows that if G is directly decomposable then it contains two normal subgroups H and K such that H ∩K = {eG} and HK = G

Answer 22

We need to check what they really mean by HK

Question 23

Theorem 12.4.
Let G be a group, and let H and K be two normal subgroups of G. If the following two conditions hold
i)
ii)
Then G ∼= H × K.

Answer 23

(i) HK=G;
(ii) H∩K={e};

proof pg 43: should learn

Question 24

Theorem 12.5.
Let G be a group, and let Hi, 1 ≤ i ≤ n, be n normal subgroups of G such that the following two conditions are satisfied:
i)
ii)
Then G ∼= H1 × H2 . . . × Hn.

Answer 24

i) G = H1H2 …Hn;
(ii) Hi ∩(H1…Hi−1Hi+1…Hn)={e},foralli. Need to check what this means

No proof given, sim to thm 12.4

Question 25

Theorem12.6.
Let Gi,1≤i≤n,be groups, and let Hi be a normal subgroup of 􏰏Gi,1≤i≤n.
Then H1 ×…×Hn is a normal subgroup of􏰏G1 ×…×Gn and ….

Answer 25

G1 ×…×Gn/H1×…×HN ∼= G1/H1×G2/H2×…×GN/HN

proof, find bijection pg 44

Question 26

13. Finite abelian groups
    In this section we shall use direct products to classify all finite abelian groups up to isomorphism. For abelian groups it is customary to use the additive notation, writing + for the group operation. The following table is a ‘multiplicative–additive dictionary’:
    1) multiplicative—>
    2) ab—>
    3) a−1—>
    4) e—>
    5) ⟨e⟩—>
    6) a^n—>
    7) ab−1—>
    8) aH—>
    9) G/N—>
    10) HK—>
    11) direct product—>
    12) G×H—>

Answer 26

1) additive
2) a+b
3) −a
4) 0
5) 0
6) na
7) a−b
8) a+H
9) G/N
10) H+K
11) direct sum
12) G⊕H

Note that we retain the multiplicative notation G/N for factors, rather than using the alternative G − N .

Question 27

Turn this into separate cards?
If G is an abelian group, and X = {a1, . . . , an}, then, by Theorem 4.3, the subgroup generated by X consists of all sums of elements of X and their inverses. However, because of commutativity, we can collect all the summands containing ai together for i = 1, . . . , n, and hence
⟨X⟩={k1a1 +k2a2 +…+knan : k1,…,kn ∈Z}. Similarly, if H and K are subgroups of G we have
⟨H, K⟩ = H + K.
A simple class of finite abelian groups are cyclic groups Zn. The aim of this section is to show how an arbitrary abelian group can be decomposed into a direct sum of cyclic groups. As the first step we analyse the direct product of two cyclic groups.

Question 28

Theorem 13.1.
If m and n are co-prime natural numbers, then

Theorem13.2.
Let n>1 be a natural number ,and let n=p1^α1.p2^α2….pk^αk be it’s decomposition into a product of primes. Then….

Answer 28

Zm ⊕ Zn ∼= Zmn

proof short pg 45

Zn ∼=Z_(p1^α1) ⊕Z_(p2^α2) ⊕…⊕Z_(pk^αk)

Question 29

Lemma 13.3.
Let G = ⟨a1,…,an⟩ be an abelian group.
If b=k1a1 +k2a2 +…+knan,
with gcd(k1,…,kn) = 1 then…

Answer 29

there exist elements b2,…,bn ∈ G such that G = ⟨b,b2,…,bn⟩.
Learn proof for understanding Long pg 46+47

Question 30

Theorem 13.4.
Every finite abelian group G is isomorphic to the direct sum of

Answer 30

cyclic groups, each of which has a prime power order.
proof long pg 47+48

Question 31

We introduce some notation. Let G be an abelian group, and let n ∈ Z.
Then
nG={na : a∈G}.
Lemma 13.5.
i) nG ≤
ii) If n divides m then
iii) nZ_m ∼=
iv) n(G1 ⊕…⊕Gk)=

Answer 31

i) G
ii) mG ≤ nG.
iii) Z_(m/ gcd(m,n)). In particular, if gcd(m, n) = 1 then nZm ∼= Zm, while if m|n then nZm is trivial.
iv) (nG1)⊕…⊕(nGk).
proof pg 47

Question 32

Theorem 13.6. Let G be a finite abelian group. The decomposition of G into a direct sum of cyclic groups of prime power orders is…

Answer 32

unique up to the order of summands.

proof long pg49+50

Question 33

Theorem 13.7. (The fundamental theorem for finite abelian groups)

Answer 33

Every finite abelian group G is isomorphic to a direct sum of cyclic groups, each of which has a prime power order, and any two such decompositions of G have the same numbers of summands of each order.

Question 34

Example13.9. By Theorem13.7, the following is the list of all non-isomorphic abelian groups of order 1800 = 233252:
Z2 ⊕Z2 ⊕Z2 ⊕Z3 ⊕Z3 ⊕Z5 ⊕Z5 Z2 ⊕Z2 ⊕Z2 ⊕Z3 ⊕Z3 ⊕Z25
Z2 ⊕Z2 ⊕Z2 ⊕Z9 ⊕Z5 ⊕Z5
Z2 ⊕Z2 ⊕Z2 ⊕Z9 ⊕Z25
Z2 ⊕Z4 ⊕Z3 ⊕Z3 ⊕Z5 ⊕Z5 Z2 ⊕Z4 ⊕Z3 ⊕Z3 ⊕Z25
Z2 ⊕Z4 ⊕Z9 ⊕Z5 ⊕Z5
Z2 ⊕Z4 ⊕Z9 ⊕Z25
Z8 ⊕Z3 ⊕Z3 ⊕Z5 ⊕Z5 Z8 ⊕Z3 ⊕Z3 ⊕Z25
Z8 ⊕Z9 ⊕Z5 ⊕Z5
Z8 ⊕Z9 ⊕Z25