MTH2010 GROUPS, RINGS + FIELDS

Question 1

DEFINITION: a group

Answer 1

A group is a pair (G, .) where G is a (non empty) set and . is a composition law (.) : G × G → G,
(g1, g2) |→ g1.g2
which satisfies the following conditions.
1• (.) is associative: (a.b).c = a.(b.c).
2• (.) admits a neutral element eG:
a.eG = eG.a = a, ∀a ∈ G.
3• ∀a ∈ G, ∃a^−1 ∈ G : a.a−1 = a^−1.a = eG (a^−1
is called the inverse of a)

Question 2

DEFINTION: cardinality/order of G

Answer 2

the number of elements of G, denoted |G|

Question 3

DEFINITION: if a group is commutative or abelian

Answer 3

if the composition law (.) further satisfies a.b = b.a, ∀a, b ∈ G

Question 4

DEFINITION: a subgroup (H is a subgroup if H is closed under the composition law (.) )

Answer 4

Let (G, .) be a group. A subgroup of G is a (non
empty) subset H ⊂ G which is closed under the composition law (.) and taking inverses:
• x ∈ H =⇒ x^−1 ∈ H,
• if x, y ∈ H =⇒ x.y ∈ H.
In particular if x ∈ H then x.x−1 = eG ∈ H

Question 5

DEFINITION: order of an element

Answer 5

Let G be a group and a ∈ G. The order of a is,
ord(a) = min {n ≥ 1 : a^n = eG},
if ∃n ≥ 1 with a n = eG.

Question 6

ord(a) = 1 if and only if …

Answer 6

a = eG

Question 7

DEFINITION: cyclic groups

Answer 7

A group G is called cyclic if ∃a ∈ G such that
G = < a > equals the (sub)group generated by a, and a is called a generator of G. [ if a generates G then so does
a^−1 ]

Question 8

If G = < a > is cyclic and ord(a) = n is finite then |G| =

Answer 8

n

Question 9

DEFINITION: product of groups

Answer 9

Let (G, ◦) and (H, *) be groups. We define a new group (G × H, .), called the product group of G and H, as follows,
G × H = {(g, h) : g ∈ G, h ∈ H}
is the set-theoretic product of G and H. The composition law (.) is defined by
(g1, h1).(g2, h2) = (g1 ◦ g2, h1 * h2),
it is a group law on G × H

Question 10

LEMMA: Let (G, ◦) and (H, *) be groups. If G and H are abelian then so is
G × H. If both G and H are finite then so is G × H and |G × H| = |G| |H|

Answer 10

PROOF:
Assume G, H are abelian, and let g1, g2 ∈ G, h1, h2 ∈ H.
Then (g1, h1).(g2, h2) = (g1 ◦ g2, h1h2) = (g2 ◦ g1, h2h1) = (g2, h2).(g1, h1).
If both G and H are finite then the number of elements of G × H is the number of pairs (g, h) with g ∈ G and h ∈ H,
there are |G| possibilities for g and |H| possibilities
for h which gives |G||H| for the number of such pairs

Question 11

DEFINTION: group homomorphism

Answer 11

Let (G, ◦) and (H, *) be groups. A map f : G → H is called a homomorphism if
f(g1 ◦ g2) = f(g1) * f(g2), ∀g1, g2 ∈ G. [ if f is a bijection it is called an isomorphism]

Question 12

PROPERTIES: homomorphism (2) - . Let f : (G, ◦) → (H, *) be a group homomorphism.

Answer 12

1• Let n ∈ Z and g ∈ G then f(g^n) = (f(g))^n.
2• Suppose f is an isomorphism. Then the followings hold.
(i) G is finite ⇐⇒ H is finite.
(ii) G is abelian ⇐⇒ H is abelian.
(iii) ord(g) = ord(f(g)), ∀g ∈ G.

Question 13

DEFINITION: image of homomorphism

Answer 13

Let f : (G, ◦) → (H, *)
be a group homomorphism. We define the image of f to be
Im(f) = {h ∈ H : ∃g ∈ G, h = f(g)}.

Question 14

DEFINITION: kernel of homomorphism

Answer 14

Let f : (G, ◦) → (H, *)
be a group homomorphism. We define the kernel of f to be
Ker(f) = {g ∈ G : f(g) = eH}

Question 15

DEFINITION: subgroup relation

Answer 15

Let G be a group and H ⊂ G a subgroup.
We define the relation R among elements of G by
xRy ⇐⇒ x^−1
y ∈ H ⇐⇒ y = xh, h ∈ H.

Question 16

LEMMA: The relation R is an equivalence relation. If x ∈ G then its equivalence class xBAR for the relation R is
xBAR = xH = {xh : h ∈ H}.

Answer 16

PROOF:
R is reflexive: xRx since x^−1x = eG ∈ H.
R is symmetric: assume xRy,
, then (x^−1y)^−1 = y^−1(x^−1)^−1 = y^−1x ∈ H (since H is a subgroup)
which means yRx.
R is transitive: assume xRy, and yRz, then x^−1yy^−1z = x^−1z ∈ H (since H is a subgroup) which means xRz.
The equivalence class of x ∈ G is xBAR = {y ∈ G : xRy} = {y ∈ G : y = xh, h ∈ H} = xH = {xh : h ∈ H}.

Question 17

DEFINITION: left cosets

Answer 17

The equivalence class xH = {xh : h ∈ H} is called a left coset of H; it is a subset of G. 
We have G =
union(x∈G)xBAR = union(x∈G)xH where the union is disjoint

Question 18

the set of all left cosets of H is denoted …

Answer 18

(G/H)_left = {xH : x ∈ G}

Question 19

if G is finite then G/H is a finite set and its cardinality is denoted …

Answer 19

|G : H| = |G/H| which is also called the index of H in G

Question 20

THEOREM: LAGRANGES THEOREM: Let G be a finite group and H a subgroup. Then |G| = |H| |G : H|.

Answer 20

PROOF:
First show that all left cosets have the same cardinality |H| = |xH|.
The map σ : H → xH defined by σ(h) = xh is a bijection. It is surjective. Assume σ(h1) = xh1 = σ(h2) = xh2 then x^−1xh1 = h1 = x^−1xh2 = h2.
Since σ is bijective H and xH have the same cardinality. We know that G is the disjoint union of the left cosets of H and all these cosets have the same cardinality:
equal that of H. Thus |G| equals the number of left cosets times the cardinality of a coset

Question 21

DEFINITION: a normal subgroup

Answer 21

A subgroup H of G is called normal if:

xH = Hx = {h’x : h’ ∈ H} , ∀x ∈ G.

Question 22

every subgroup of an abelian group is …

Answer 22

a normal subgroup

Question 23

DEFINITION: conjugate elements

Answer 23

Two elements g, g’
of a group G are
called conjugate if ∃x ∈ G with g’ = xgx^−1.

Question 24

if G is an abelian group; xhx^-1 = …

Answer 24

h [ and each element equals its own conjugates ]

Question 25

PROPOSITION: Two permutations in Sn are conjugates if and only if they have the same cycle type.

Answer 25

PROOF: Let σ = (a1 . . . an1 ). . .(b1 . . . bnk) ∈ Sn be a permutation of type (n1, . . . , nk). Let τ ∈ Sn. We claim τ ◦ σ ◦ τ^−1 = (τ (a1). . .τ (an1)). . .(τ (b1). . . τ (bnk)) is the decomposition of τ ◦ σ ◦ τ^−1 into the product of disjoint cycles, hence it also has cycle type (n1, . . . , nk). Show that if σ(i) = j then τ ◦ σ ◦ τ^−1(τ (i)) = τ (j). But τ ◦ σ ◦ τ^−1(τ (i)) = τ(σ(τ^−1(τ (i)))) = τ (σ(i)) = τ (j). Conversely let σ = (a1 . . .an1). . .(b1 . . . bnk) and σ' = (a'1. . . a'n1). . .(b'1. . . b'nk) be two permutations with the same cycle type. Define the permutation τ ∈ Sn by τ (ai) = a'i, . . . , τ (bi) = b'i. Then τ ◦ σ ◦ τ^−1 = (τ (a1). . . τ (an1)). . .(τ (b1). . . τ (bnk)) = (a'1. . . a'n1). . .(b'1. . . b'nk) = σ'.

Question 26

DEFINITION: quotient group law

Answer 26

Let G be a group and H a normal subgroup of G. Define a composition law (.) on the set of left cosets G/H by (.) : G/H × G/H → G/H (xH, yH) |→ xH.yH = xyH.

Question 27

PROPOSITION: (G/H, .) is a group, it is called the quotient group of G by H.

Answer 27

PROOF: Associativity (xH.yH).zH = (xy)H.zH = (xy)zH = x(yz)H = xH.(yzH) = xH.(yH.zH). The neutral element is eG/H = eGH: xH.eGH = xeGH = xH. The inverse of xH is x^−1H: xH.x^−1H = xx^−1H = eGH.

Question 28

PROPOSITION: The map p is a surjective group homomorphism and Ker(p) = H.

Answer 28

``` PROOF: The fact that p is surjective is clear gH = p(g). It is a homomorphism since p(g1g2) = g1g2H = (g1H).(g2H) = p(g1).p(g2). Show Ker(p) = H. First H ⊂ Ker(p) since if g ∈ H then e^−1G g = g ∈ H and eGRg hence p(g) = gH = eGH. Conversely let g ∈ Ker(p) meaning p(g) = gH = eGH, then eGRg and e^−1G g = g ∈ H. ```

Question 29

THEOREM: THE FIRST ISOMORPHISM THEOREM: The factor group G/ Ker(f) is isomorphic to Im(f).

Answer 29

PROOF: Consider the map π : G/ Ker(f) → Im(f), defined by π(g Ker(f)) = f(g). Show π is a group isomorphism. First check that π is well defined. Assume g Ker(π) = g 0 Ker(π) meaning g'^−1 g = gTILDA ∈ Ker(π). Then f(g) = f(g'gTILDA) = f(g')f(gTILDA) = f(g')eH = f(g') since gTILDA ∈ Ker(f). Further π is a homomorphism: π(g Ker(f).g' Ker(f)) = π(gg' Ker(f)) = f(gg') = f(g)f(g') = π(gKer(f))π(g' Ker(f)). The homomorphism π is surjective: if f(g) ∈ Im(f), g ∈ G, then f(g) = π(g Ker f) by the definition of π. Finally π is injective: assume f(g) = π(g Ker(f)) = π(g' Ker(f)) = f(g'), then f(g')^−1f(g) = f(g'^−1g) = eH and g'^−1g ∈ Ker(f), [equivalently gRg' meaning g Ker f = g' Ker f]

Question 30

COROLLARY: Let G, H be groups with G finite, and f : G→H a group homomorphism. Then Im(f) is finite and |G| / |Ker(f)| = |Im(f)|.

Answer 30

PROOF: The First Isomorphism Theorem states that G/ Ker(f) is isomorphic to Im(f). Since G is finite, G/ Ker(f) is finite, and thus Im(f) is finite being isomorphic to G/ Ker(f). Further this isomorphism implies |G/ Ker(f)| = |Im(f)|. On the other hand |G/ Ker(f)| = |G : Ker(f)| = |G| / | Ker(f)| by Lagrange’s theorem hence the result.

Question 31

DEFINITION: group action

Answer 31

Let (G, *) be a group and A a set. A group action of G on A is a map: (.) : G × A → A (g, a) |→ g.a satisfying: (A1) (g1 * g2).a = g1.(g2.a), ∀g1, g2 ∈ G, ∀a ∈ A (A2) eG.a = a, ∀a ∈ A

Question 32

left multiplication or the left regular action of G: A group (G, *) acts on itself via the action (.) : G × G → G defined by ...

Answer 32

(g, g') |→ g.g' = g * g'

Question 33

action by conjugation: A group (G, *) acts on itself via the action (.) : G × G → G defined by ...

Answer 33

(g, g') |→ g.g' = g * g' * g^−1 | .

Question 34

DEFINITION: Kernel of representation

Answer 34

The kernel of τ : G → SA Ker(τ) = {g ∈ G : τg = idA} = {g ∈ G : g.a = a, ∀a ∈ A}, is called the kernel of the representation τ , it is a normal subgroup of G.

Question 35

If Ker(τ) = {eG} or τ is injective then the action is called ...

Answer 35

faithful

Question 36

DEFINITION: stabiliser

Answer 36

Let G × A → A be an action of a group (G, *) on a set A and a ∈ A. The stabiliser of a, denoted Stab(a), is Stab(a) = {g ∈ G : g.a = a}

Question 37

PROPOSITION: The stabiliser Stab(a) is a subgroup of G

Answer 37

PROOF: First eG ∈ Stab(a) since eG.a = a. Let g ∈ Stab(a). Then a = eG.a = (g^−1 * g).a = g^−1 . (g.a) = g^−1 . a and g^−1 ∈ Stab(a). Further, let g1, g2 ∈ Stab(a) then (g1 * g2).a = g1.(g2.a) = g1.a = a and g1 * g2 ∈ Stab(a).

Question 38

PROPOSITION: The relation R is an equivalence relation.

Answer 38

R is reflexive since a = eG.a hence aRa, ∀a ∈ A. Let a, b ∈ A with aRb, i.e. a = g.b for some g ∈ G. Then g^−1.a = g^−1.(g.b) = (g^−1 * g).b = eG.b = b hence bRa. Finally, let a, b, c ∈ A with aRb and bRc. Thus ∃g, g' ∈ G such that a = g.b and b = g'.c. Then a = g.b = g.(g'.c) = (g* g').c and aRc.

Question 39

DEFINITION: orbit

Answer 39

``` Let a ∈ A. The equivalence class of a for the relation R is aBAR = {b ∈ A : ∃g ∈ G, b = g.a} which is denoted orb(a) ```

Question 40

THEOREM: THE ORBIT-STABILISER THEOREM: Assume (G, *) is a group acting on a set A, and G is finite. Then the orbit orb(a) of an element a ∈ A is finite and |orb(a)| = |G| / |Stab(a)| = |G : Stab(a)|.

Answer 40

PROOF: Consider the map f:orb(a) → (G/ Stab(a))left defined by f(g.a) = g. Stab(a). Check the map is well defined; (g^−1_2 * g1).a = g^−1_2.g1.a) = g^−1_2.(g2.a) = (g^−1_2*g2).a = eG.a = a hence g^−1_2*g_1 ∈ Stab(a) which implies g1. Stab(a) = g2. Stab(a). Further f is a bijection. Indeed f is clearly surjective. Suppose g1.a, g2.a ∈ orb(a) and f(g1.a) = f(g2.a), i.e. g1. Stab(a) = g2. Stab(a) which is equivalent to g^−1_2*g1 ∈ Stab(a). Then a = (g^−1_2*g1).a = g^−1_2.(g1.a) ⇒ g2.a = g2.(g^−1_2.(g1.a)) = (g2*g^−1_2).(g1.a) = eG.(g1.a) = g1.a as required. Now since f is a bijection orb(a) and (G/ Stab(a))left; whose cardinality is |G:Stab(a)|, they have the same number of elements.

Question 41

DEFINITION: regular permutation representation

Answer 41

if g ∈ G, ρ(g) is the permutation defined for i, j ∈ {1, . . . , n} by ρ(g)(i) = j if g*gi = gj.

Question 42

THEOREM: Let τ_H : G → S_G/H be the permutation representation associated to the action of G by left multiplication on the left cosets of H. Thus if g ∈ G, τ_H(g) : G/H → G/H is the bijection defined by τH(g)(g'H) = (g*g')H. 1• G acts transitively on G/H. 2• The stabiliser of eGH is the subgroup H. 3• Ker(τH) = intersection(x∈G) xHx^−1, and Ker(τH) is the largest normal subgroup of G contained in H.

Answer 42

PROOF: To see that G acts transitively on G/H let aH, bH ∈ G/H and g = b*a^−1. Then g.(aH) = (b*a^−1).aH = (b*a^−1*a)H = bH this proves 1. 2• The stabiliser of eGH is {g ∈ G : g.(eGH) = gH = H} = {g ∈ G : gH = H} = H, this proves 2. 3• By definition Ker(π_H) = {g ∈ G : g.(xH) = xH, ∀x ∈ G} = {g ∈ G : (g*x)H = xH, ∀x ∈ G} = {g ∈ G : (x^−1*g*x)H = H, ∀x ∈ G} = {g ∈ G : x^−1*g*x ∈ H, ∀x ∈ G} = {g ∈ G : g ∈ xHx^−1, ∀x ∈ G} = intersection(x∈G) xHx^−1. Further Ker(πH) is a normal subgroup of both G and H. Let N be a normal subgroup of G contained in H then N = xNx^−1 ⊂ xHx−1 , ∀x ∈ G, hence N ⊂ intersection(x∈G) xHx^−1 = Ker(πH). This shows Ker(πH) is the largest normal subgroup of G contained in H.

Question 43

COROLLARY: Let G be finite group of order n and p the smallest prime number dividing n = |G|, then any subgroup of G of index p is normal. [if G has a subgroup of index 2 then this subgroup must be normal]

Answer 43

PROOF: Suppose H ≤ G is a subgroup and |G : H| = p. Let π_H be the permutation representation afforded by multiplication on the set of left cosets of H in G. Let K = Ker(πH), and |H:K| = k. Then |G:K| = |G:H| |H:K| = pk. Since H has p left cosets, G/K is isomorphic to a subgroup of Sp by the First Isomorphism Theorem. By Lagrange’s Theorem, pk = |G/K| divides |Sp| = p!. Thus k divides p!/p = (p − 1)!. But all prime factors of (p − 1)! are less than p and every possible prime divisor of k is greater or equal to p. This forces k = 1, so H = K is a normal subgroup of G (since K is normal).

Question 44

DEFINITION: conjugate

Answer 44

Let G be a group and S(G) the set of all non-empty subsets of G. Then G acts on S(G) by conjugation φ : G × S(G) → S(G) (g, A) |→ gAg^−1 = {gag^−1 : a ∈ A}. The subset gAg^−1 of G is called the conjugate of A by g.

Question 45

DEFINITION: normaliser

Answer 45

If A ∈ S(G), the stabiliser of A under the above action φ by conjugation is called the Normaliser of A in G and denoted N_G(A) = {g ∈ G : gAg^−1 = A}.

Question 46

DEFINITION: centraliser-centre

Answer 46

If A ∈ S(G), the kernel of the action φ_A is called the centraliser of A in G and denoted C_G(A) = {g ∈ G : gag^−1 = a ∀a ∈ A}.

Question 47

DEFINITION: centre (of centraliser)

Answer 47

The centraliser of G in G is called the centre of G and denoted Z(G) = {g ∈ G : gh = hg ∀h ∈ G}.

Question 48

DEFINITION: class equation

Answer 48

Let G be a group and g ∈ G. Consider the subset {g} ⊂ G, then NG({g}) = CG({g}) = {h ∈ G : hgh−1 = g} = {h ∈ G : hg = gh} is the subgroup of elements of G which commute with g.

Question 49

THEOREM: THE CLASS EQUATION:

Answer 49

Let G be a finite group and {orb(g1), . . . , orb(g_r)} the distinct conjugacy classes of G which are NOT contained in Z(G), then |G| = |Z(G)| + sum(r)(i=1) |G : C_G(gi)|.

Question 50

THEOREM: If p is a prime number and P is a finite group of prime power order p^s for some s ≥ 1, then P has a nontrivial centre meaning Z(P) NOT= {e_P}.

Answer 50

``` PROOF: By the class equation |P| = |Z(P)| + sum(r)(i=1) |P : C_P (gi)| where g1, . . . , gr are representative of the distinct non central conjugacy classes, i.e. classes not contained in Z(P). By definition CP (gi) NOT= P since gi NOT∈ Z(P), hence |P : CP (gi)| = |P| / |CP (gi)| > 1, and p divides |P : CP (gi)|. Since p divides |P| it must also divide |Z(P)| = |P| − sum(r)(i=1) |P : CP (gi)|, in particular Z(P) NOT= {e_P}. ```

Question 51

THEOREM: CAUCHY'S THEOREM: Let G be a finite group and p a prime number which divides |G|. Then there exists an element of G of order p, and a subgroup of G of cardinality p.

Answer 51

PROOF:

Question 52

DEFINITION: simple group

Answer 52

A finite group G is called simple if |G| > 1 and the only | normal subgroups of G are {eG} and G itself.

Question 53

DEFINITION: p-group (and p-subgroups)

Answer 53

Let G be a group and p a prime number. A group of order p^r for some r ≥ 1 is called a p-group. Subgroups of G which are p-groups are called p-subgroups.

Question 54

DEFINITION: Sylow p-subgroup

Answer 54

Le G be a group and p a prime number. If G is a group of order p^rm with gcd(p, m) = 1, then a subgroup of order p^r is called a Sylow p-subgroup of G.

Question 55

DEFINITION: set of Sylow p-subgroups of G

Answer 55

Let G be a group and p a prime number. The set of Sylow p-subgroups of G will be denoted by Syl_p(G) and the cardinality of Syl_p(G), which is the number of Sylow p-subgroups, will be denoted n_p(G).