Chapter 2 Flashcards
Definition 3.1. Let G be a group and H a subset of G. H is said to be a subgroup of G if…
H is itself a group under the multiplication in G, this is denoted: H ≤ G
Theorem 3.4. The following three conditions are equivalent for a non-empty subset H of a group G:
(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
A permutation σ ∈ Sn is said to be even (respectively, odd) if…
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.
An is a subgroup of
SN.
Try proving this (short) pg 16
Theorem 3.7. A permutation σ ∈ Sn cannot be
both even and odd.
Long proof pg 16+17
The group An is called the alternating group on {1, . . . , n}. Its order is
n!/2
Let G be a group, and let Hi, i ∈ I(not the empty set), be a collection of sub- groups of G.
Then the intersection of the Hi’s is also a subgroup of G.
Try proving this pg 17
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
The intersection of the Hi is called the subgroup generated by X, and is denoted by ⟨X⟩.
⟨X⟩ is the smallest subgroup of G which contains X. The elements of X are called …
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⟩.
Theorem 4.3. Let G be a group and let X ⊆ G. The subgroup ⟨X⟩ consists of…
all possible products of elements from X and their inverses.
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…
The order of this subgroup.
there exists n > 0 such that a^n = e, and is equal to the smallest such n.
Corollary 4.4. If all elements from X have finite orders then
⟨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.
Example 4.5. Z = ⟨1⟩; Zn = ⟨1⟩; K4 = ⟨a,b⟩; Q8 = ⟨i,j⟩.
EX 4.6
EX 4.7
Definition 4.11. Let n ≥ 3. The dihedral group of degree n (denoted by Dn)
is the subgroup of Sn generated by…
the permutations α = (1 2 . . . n) and β=(2n)(3n−1)…(in+2−i)
Check if beta has more terms after this?????
Theorem 4.12. The following statements are true for the dihedral group Dn:
(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
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.
it can be generated by a single
element.
Theorem 5.2. Let G be a cyclic group. If G is infinite then…
While if G has finite order n then…
G ∼= Z
G ∼= Zn
proof pg 22
Theorem 5.3. Every subgroup of a cyclic group is
Also cyclic
proff pg 23
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…
Ha = {ha : h ∈ H}. The corresponding left coset is aH = {ah : h ∈ H}.
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)
(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
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.
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…
|R|=|L|.
Definition 6.5.
The index of H in G (denoted by [G : H]) is the number of
Right cosets of H in G, and is equal to the number of left cosets of H in G.
For any group G , [G:G]=
[G:⟨e⟩]=
1
|G|.
Theorem 6.9 (Lagrange)
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.
Corollary 6.10.
The order of an element a ∈ G divides
the order of the group G,
and
a^|G| = e_G.
Try proving pg 25
Corollary 6.11.
Every group of prime order p is
isomorphic to Zp
Try proving pg 25
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.
Definition7.1.
Let G and H be groups. A mapping f : G −→ H is a homomorphism if
f (xy) = f (x)f (y) holds for all x, y ∈ G.
Example 7.3. The mapping
f:Z→Zn,x→x (modn)
is a homomorphism.
prove this
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.
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 …
ker f ={x∈G : f(x)=eH}.
f(X)={f(x) : x∈X}.
f−1(X)={x∈G : f(x)∈X}.
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).
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) ≤
(i)e_H
(ii)(f(a))^(−1) for all a ∈ G.
(iii)G
(iv)H
(v)G
proof pg 27
Name the 5 types of mapping:
monomorphism—injective homomorphism
epimorphism—surjective homomorphism
isomorphism—bijective homomorphism
endomorphism—homomorphism G −→ G
automorphism—isomorphism G −→ G
ker d={A∈GL(n,F) : |A|=1},
Definition7.10.
The subgroup ker d from Example7.7 is denoted by…
SL(n,F), and is called the special linear group.
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;
(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
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…
said to be normal; this is denoted by N(triangle with line under)G
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.
Group (called the factor group or the quotient group) under the binary operation
(aN)(bN) = (ab)N.
proof pg 29
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…
i)ker f is a normal subgroup of G
ii) An epimorphism and nd ker π = N .
proof pg 30+31
Theorem 8.8. The conjugation relation ∼ is
an equivalence relation.
Try proving: pg 37
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
Cx ={y∈G : x∼y}={a−1xa : a∈G}.
Example 8.10.
The conjugacy class of the identity e is
{e}. Indeed, if e ∼ x
then, by the definition, a−1ea = x, and so x = e.
Theorem 8.11.
Let G be a group and let H be a subgroup of G. Then H is normal if and only if
H is a union of conjugacy classes of G.
Learn short proof of this pg 31
Theorem 9.1 (The first isomorphism theorem)
If f : G −→ H is a homomorphism of groups then
G/kerf ∼= imf.
proof pg 32
Theorem 9.5 (The second isomorphism theorem)
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.
Theorem 9.6 (The third isomorphism theorem)
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.
Theorem9.7 (The correspondence theorem)
(I may need to add more clues for this)
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
