Chapter 2

Question 1

Definition 3.1. Let G be a group and H a subset of G. H is said to be a subgroup of G if…

Answer 1

H is itself a group under the multiplication in G, this is denoted: H ≤ G

Question 2

Theorem 3.4. The following three conditions are equivalent for a non-empty subset H of a group G:

Answer 2

(i) H is a subgroup of G;
(ii) for any two x,y∈H we have xy∈H and x^(−1) ∈H;
(iii) for any two x,y∈H we have x(y^(-1))∈H.
Try proving this! pg 15

Question 3

A permutation σ ∈ Sn is said to be even (respectively, odd) if…

Answer 3

σ can be written as a product of an even (respectively, odd) number of transpositions. The set of all even permutations is denoted by An.

Question 4

An is a subgroup of

Answer 4

SN.
Try proving this (short) pg 16

Question 5

Theorem 3.7. A permutation σ ∈ Sn cannot be

Answer 5

both even and odd.
Long proof pg 16+17

Question 6

The group An is called the alternating group on {1, . . . , n}. Its order is

Answer 6

n!/2

Question 7

Let G be a group, and let Hi, i ∈ I(not the empty set), be a collection of sub- groups of G.

Answer 7

Then the intersection 􏰅of the Hi’s is also a subgroup of G.
Try proving this pg 17

Question 8

Definition 4.2. Let G be a group,and let X be a subset of G.Let Hi, i∈I, be the family of all subgroups of G which contain the set X. Then

Answer 8

The intersection of the Hi is called the subgroup generated by X, and is denoted by ⟨X⟩.

Question 9

⟨X⟩ is the smallest subgroup of G which contains X. The elements of X are called …

Answer 9

generators for ⟨X⟩.
A subgroup may be generated by several of its subsets. If X = {a1,…,an} then we write ⟨a1,…,an⟩ instead of ⟨X⟩.

Question 10

Theorem 4.3. Let G be a group and let X ⊆ G. The subgroup ⟨X⟩ consists of…

Answer 10

all possible products of elements from X and their inverses.

Question 11

In the special case where X contains a single element a, we have ⟨a⟩ = {ai : i ∈ Z}. The order of a is defined to be…
Clearly this is finite if and only if…

Answer 11

The order of this subgroup.

there exists n > 0 such that a^n = e, and is equal to the smallest such n.

Question 12

Corollary 4.4. If all elements from X have finite orders then

Answer 12

⟨X⟩ = {x1x2 …xn : n ≥ 1,x1,…,xn ∈ X}.
I.e. we don’t need the inverses.

In particular this is the case when G itself is finite.

Question 13

Example 4.5. Z = ⟨1⟩; Zn = ⟨1⟩; K4 = ⟨a,b⟩; Q8 = ⟨i,j⟩.

Question 14

EX 4.6

Question 15

EX 4.7

Question 16

Definition 4.11. Let n ≥ 3. The dihedral group of degree n (denoted by Dn)
is the subgroup of Sn generated by…

Answer 16

the permutations α = (1 2 . . . n) and β=(2n)(3n−1)…(in+2−i)
Check if beta has more terms after this?????

Question 17

Theorem 4.12. The following statements are true for the dihedral group Dn:

Answer 17

(i) |α|=n,|β|=2;
(ii) βα = α^(−1)β;
(iii) Dn ={α^I β^j : 0≤i≤n−1, 0≤j≤1}and|Dn|=2n;
(iv) Dn is non-abelian.
proof pg 21

Question 18

Definition5.1. A group is said to be cyclic if…

The significance of cyclic groups is that on one hand one can describe them completely, and that, on the other, every group contains them as subgroups. This is particularly significant for abelian groups, as we will see later in the course.

Answer 18

it can be generated by a single
element.

Question 19

Theorem 5.2. Let G be a cyclic group. If G is infinite then…
While if G has finite order n then…

Answer 19

G ∼= Z
G ∼= Zn
proof pg 22

Question 20

Theorem 5.3. Every subgroup of a cyclic group is

Answer 20

Also cyclic

proff pg 23

Question 21

Let G be a group, let H ≤ G, and let a ∈ G. The right coset of H in G determined by a is the set…
The left coset is…

Answer 21

Ha = {ha : h ∈ H}. The corresponding left coset is aH = {ah : h ∈ H}.

Question 22

In the following theorem we list some basic properties of cosets.
Theorem 6.2. Let G be a group, and let H ≤ G.
i)
ii)
iii)
iv)

Answer 22

(i) Every right coset of H has the same number elements as H itself; in other
words |Ha| = |H| for all a ∈ G.
(ii) G is the union of all right cosets of H, i.e. G = 􏰄a∈G Ha.
(iii) Any two right cosets of H are either identical, or else they are disjoint; in other words, for any a, b ∈ G either Ha = Hb or Ha ∩ Hb = ∅.
(iv) For any a,b∈G,Ha=Hb if and only if a(b^(−1)) ∈H. Analogous statements hold for left cosets.

try proving: pg 23+24

Question 23

Remark 6.3. Given a subgroup H of a group G, one can define a relation
≡R (mod H) on G as follows:
x≡R y(modH) ⇐⇒xy−1 ∈H.

This relation turns out to be an equivalence relation, and right cosets of H in G are the equivalence classes of this relation.
Similarly, the equivalence relation ≡L defined by
x≡L y(modH) ⇐⇒x−1y∈H. has left cosets of H in G as its equivalence classes.

Question 24

Theorem 6.4.
Let R = {Ha : a∈G} be the set of all right cosets of H in G, and let L = {aH : a∈G}be the set of all left cosets of H in G .Then…

Answer 24

|R|=|L|.

Question 25

Definition 6.5.
The index of H in G (denoted by [G : H]) is the number of

Answer 25

Right cosets of H in G, and is equal to the number of left cosets of H in G.

Question 26

For any group G , [G:G]=
[G:⟨e⟩]=

Answer 26

1
|G|.

Question 27

Theorem 6.9 (Lagrange)

Answer 27

Let G be a finite group and H ≤ G.
Then |G| = [G : H]|H|. In particular, order of a subgroup divides the order of the group.

Question 28

Corollary 6.10.
The order of an element a ∈ G divides

Answer 28

the order of the group G,
and
a^|G| = e_G.

Try proving pg 25

Question 29

Corollary 6.11.
Every group of prime order p is

Answer 29

isomorphic to Zp
Try proving pg 25

Question 30

Remark 6.12. The following generalisation of Lagrange’s Theorem holds: if K ≤ H ≤ G then [G : K] = [G : H][H : K]. If G is finite this can be proved from Lagrange’s Theorem: [G : H][H : K] = (|G|/|H|)(|H|/|K|) = |G|/|K| = [G : K]. In the infinite case more work has to be done.

Question 31

Definition7.1.
Let G and H be groups. A mapping f : G −→ H is a homomorphism if

Answer 31

f (xy) = f (x)f (y) holds for all x, y ∈ G.

Question 32

Example 7.3. The mapping
f:Z→Zn,x􏰀→x (modn)
is a homomorphism.
prove this

Answer 32

indeed, to prove this, we need to verify that for all x, y ∈ Z we have
x+y (modn)=((x (modn))+(y (modn))) (modn), which is an elementary number-theoretic fact.

Question 33

Definition 7.6. Let f : G −→ H be a homomorphism of groups.
The kernel of f is…
The image of a set X ⊆G under f is…
The inverse image of X ⊆ H under f is …

Answer 33

ker f ={x∈G : f(x)=eH}.
f(X)={f(x) : x∈X}.
f−1(X)={x∈G : f(x)∈X}.

Question 34

Remark 7.8. The notation f−1(X) for the inverse image of a set under f does not implicitly assume that f has an inverse. Indeed, it is not nec- essarily true that f−1(f(X)) = X and f(f−1(Y )) = Y (X ⊆ G, Y ⊆ H). However, we do always have
X ⊆f−1(f(X))(X ⊆G), andf(f−1(Y))⊆Y (Y ⊆H).

Question 35

Theorem7.9.
Let G and H be two groups, and let
f : G−→H be a homomorphism.
(i)f(e_G)=
(ii)f(a^(−1)) =
(iii)ker f ≤
(iv)If K ≤ G then f(K) ≤
(v)If L ≤ H then f−1(L) ≤

Answer 35

(i)e_H
(ii)(f(a))^(−1) for all a ∈ G.
(iii)G
(iv)H
(v)G
proof pg 27

Question 36

Name the 5 types of mapping:

Answer 36

monomorphism—injective homomorphism
epimorphism—surjective homomorphism
isomorphism—bijective homomorphism
endomorphism—homomorphism G −→ G
automorphism—isomorphism G −→ G

Question 37

ker d={A∈GL(n,F) : |A|=1},
Definition7.10.
The subgroup ker d from Example7.7 is denoted by…

Answer 37

SL(n,F), and is called the special linear group.

Question 38

Theorem 8.1.
Let G be a group and N ≤ G. The following 4 conditions are equivalent:
(i) every right coset of N in G is also a left coset of N in G;

Answer 38

(ii) Na=aN for all a∈G;
(iii) for every a ∈ G and every n ∈ N we have (a^(−1))na ∈ N (in other words a−1Na ⊆ N for all a ∈ G; N is closed under conjugation);
(iv) a−1Na=N foralla∈G.
proof pg 29

Question 39

Definition 8.2. A subgroup N of a group G satisfying the equivalent conditions of Theorem 8.1(i.e. right coset=left coset, closed under conjugation) is…

Answer 39

said to be normal; this is denoted by N(triangle with line under)G

Question 40

Theorem 8.4. Let G be a group, let N be a normal subgroup of G, and let G/N be the set of all (left) cosets of N in G. Then G/N is…group (called the factor group or the quotient group) under the binary operation
(aN)(bN) = (ab)N.

Answer 40

Group (called the factor group or the quotient group) under the binary operation
(aN)(bN) = (ab)N.
proof pg 29

Question 41

Theorem 8.6.
(i) If f : G −→ H is a homomorphism of groups then ker f…
ii) Conversely, if N is a normal subgorup of G then the map π : G −→ G/N defined by π(a) = aN is…

Answer 41

i)ker f is a normal subgroup of G
ii) An epimorphism and nd ker π = N .

proof pg 30+31

Question 42

Theorem 8.8. The conjugation relation ∼ is

Answer 42

an equivalence relation.
Try proving: pg 37

Question 43

Definition 8.9. The equivalence classes of the relation ∼ are called conjugacy classes of G. In other words, the conjugacy class of an element x ∈ G is

Answer 43

Cx ={y∈G : x∼y}={a−1xa : a∈G}.

Question 44

Example 8.10.
The conjugacy class of the identity e is

Answer 44

{e}. Indeed, if e ∼ x
then, by the definition, a−1ea = x, and so x = e.

Question 45

Theorem 8.11.
Let G be a group and let H be a subgroup of G. Then H is normal if and only if

Answer 45

H is a union of conjugacy classes of G.
Learn short proof of this pg 31

Question 46

Theorem 9.1 (The first isomorphism theorem)

Answer 46

If f : G −→ H is a homomorphism of groups then
G/kerf ∼= imf.
proof pg 32

Question 47

Theorem 9.5 (The second isomorphism theorem)

Answer 47

Let G be a group, let H ≤ G and let N be a normal subgroup of G.
Then H ∩ N is a normal subgroup of H,
N is a normal subgroup of HN = NH ≤ G and
H / ( H ∩ N ) ∼= HN / N .
proof pg 33
Find out and write down what HN actually means.

Question 48

Theorem 9.6 (The third isomorphism theorem)

Answer 48

Let G be a group ,let H,K􏰏 be normal subgroups of G, and let K ≤ H. Then H/K is a normal subgroup of G/K and
(G/K)/(H/K) ∼= G/H.

Question 49

Theorem9.7 (The correspondence theorem)
(I may need to add more clues for this)

Answer 49

Let G be a group and let N be a normal subgroup of 􏰏G. Also let
A = {H : N ≤ H ≤ G}, B = {K : K ≤ G/N}.
The mapping f : A −→ B defined by
f(H) = H/N = {hN : h ∈ H} (N ≤ H ≤ G) is a bijection. This bijection preserves inclusions and normality:
H1 ≤H2 ⇔f(H1)≤f(H2), H a normal subgroup of G ⇔ f(H) 􏰏a normal subgroup of G/N, for all H1,H2,H ∈ A.

proof pg 35