Basic Algebra 1: Graduate Algebra [Lecture notes]

128 Pages·2014·9.52 MB·English

by  Andrei Jorza

Graduate Algebra, Fall 2014
Lecture 1
Andrei Jorza
2014-08-27
1 Group Theory
1.1 Basic definitions
Let G be a set and · be a binary operation on G. Say that:
1. ·isassociativeifforanyx,y,z ∈Ghave(x·y)·z =x·(y·z). Withinductionyoucanalsoshowthat
for all x ,...,x ∈ G the value of x ·x ···x is independent of the order in which the · operations
1 n 1 2 n
are performed.
2. · has a unit element e if for all x∈G one has x·e=e·x=x. Unit elements, if they exist, are unique:
indeed, if e,e(cid:48) are units then e=e·e(cid:48) =e(cid:48).
3. an element x∈G has an inverse x−1 if x·x−1 =x−1·x=e. If G is associative then inverses, if they
exist, are unique. Suppose a,b are inverses to x. Then a=ae=a(xb)=(ax)b=eb=b.
4. · is commutative or abelian if xy =yx for all x,y ∈G.
We say that G with · is:
1. a semigroup if · is associative.
2. a monoid if G is a semigroup and there exists a unit.
3. a group if G is a monoid and every element has an inverse.
A list of many examples:
1. Z with + and 0 is a group.
2. Z with + and 0 is a monoid.
≥0
3. Q, R, C with + and 0 are groups.
4. for n≥2 an integer Z/nZ={0,1,...,n−1} with addition modulo n and 0 is a group.
5. for n≥2 an integer (Z/nZ)× ={d∈Z/nZ|(d,n)=1} with multiplication modulo n and 1 as unit is
a group.
6. Q/Z=[0,1)∩Q with unit 0 and addition defined as
(cid:40)
x+y x+y <1
x“+(cid:48)(cid:48)y =x+y mod 1={x+y}=
x+y−1 x+y ≥1
is a group (here {x} represents the fractional part).
1
7. If (G,· ,e ) and (H,· ,e ) are two groups then (G×H,·,e ×e ) is a group where · is defined
G G H H G H
component-wise. For example the Klein group is (Z/2Z)×(Z/2Z)={(0,0),(0,1),(1,0),(1,1)}.
8. For n≥2, S is the group of permutations of a fixed set of n elements. Multiplication is composition
n
of permutations and the identity is the identity permutation.
9. The dihedral group D is the group of symmetries of a regular n-gon. Again, multiplication is
2n
composition of symmetries and the identity map is the identity element.
10. If R is Z,Q,R or C and n ≥ 1 then the set M (R) of n×n matrices with entries in R is a group
n×n
with respect to matrix addition.
11. If R is Q,R or C then the set GL(n,R) of n×n matrices with entries in R and non-zero determinant
is a group with respect to matrix multiplication.
1.2 Cyclic groups
The simplest groups are the cyclic ones. An infinite cyclic group is a group G, written multiplicatively,
whose elements are {1,a±1,a±2,...} where a∈G is such that an (cid:54)=1 for any n∈Z. The element a is called
a generator of G (we write G=(cid:104)a(cid:105)) and say that a has infinite order.
AfinitecyclicgroupofordernisagroupG,writtenmultiplicatively,whoseelementsare{1,a,a2,...,an−1}
wherea∈Gsuchthatan =1butad (cid:54)=1forany0<d<n. Again,aissaidtobeageneratorofG(G=(cid:104)a(cid:105))
and we say that a has order ord(a)=n.
If G is any group and a∈G we can still define the order of a as above.
Proposition 1. Suppose a∈G has order n and d≥1 is an integer. Then ord(ad)=n/(d,n).
Proof. In class I only did the case when (d,n) = 1. Suppose m = ord(ad). Then m is the smallest positive
integer such that (ad)m = adm = 1. Certainly (ad)n/(d,n) = (an)d/(d,n) = 1 and so m ≤ n/(d,n) by the
minimality assumption.
Next, use division with remainder to write md = qn+r where 0 ≤ r < n. This is a phenomenally
powerful tool that we’ll use many times. Then
1=adm =aqn+r =(an)qar =ar
Since a has order n and r <n it follows that r must be 0. Thus dm=qn and we can rewrite this as
d n
m= q
(d,n) (d,n)
Nowd/(d,n)andn/(d,n)arecoprimeandso,byuniquefactorizationintheintegers,itfollowsthatn/(d,n)|
m. As m > 0 this implies that m ≥ n/(d,n) and so we deduce, from the above, that m = n/(d,n) as
desired.
2
Graduate Algebra, Fall 2014
Lecture 2
Andrei Jorza
2014-08-29
1 Group Theory
1.3 Subgroups
Recall that for a group G and a∈G we defined ord(a) to be the smallest positive exponent of a that equals
the identity element, or infinity if no such exponent exists.
Example 1. The order of 2 in the multiplicative group (Z/15Z)× is 4 because 24 ≡ 1 (mod 15) but no
smaller exponent is 1.
Also we wrote (cid:104)a(cid:105)={an|n∈Z}⊂G. If ord(a)=∞ this was the infinite cyclic group and if ord(a)=n
then (cid:104)a(cid:105) is a set of cardinality 1, consisting of {1,a,a2,...,an−1}.
Definition 2. A subgroup H of a group G is a subset of G, closed under multiplication in G, containing
the identity of G and such that every element of H has an inverse in H.
Proposition 3. Let G be a group and H a nonempty subset of G. Then H is a subgroup if and only if for
all a,b∈H, ab−1 ∈H.
Proof. For a ∈ H, aa−1 = e ∈ H. For a ∈ H, ea−1 = a−1 ∈ H. For a,b ∈ H also b−1 ∈ H and so
ab=a(b−1)−1 ∈H so H is a subgroup.
Definition 4. If X ⊂ G is a subset define (cid:104)X(cid:105) as the smallest subgroup of G containing X. For example
(cid:104)a(cid:105) is the smallest subgroup of G containing a.
Example 5. Computing (cid:104)X(cid:105) is rarely easy, and most of the time relies on complicated combinatorics.
1. If m∈Z then (cid:104)m(cid:105)⊂(Z,+) is the set mZ={km|k ∈Z}.
2. If m,n ∈ Z such that (m,n) = 1 then by the Euclidean algorithm one can find p,q ∈ Z such that
pm+qn=1. Let H =(cid:104)m,n(cid:105). Since m,n∈H and H is a subgroup also pm+qn=1∈H. But then
for all k ∈Z also k =k·1∈H and so H =Z.
3. Here is a complicated example based on combinatorics that has applications in complex analysis. The
set SL(2,Z) of 2×2 matrices with determinant 1 and integer entries is a group (show this!). The
(cid:18) (cid:19) (cid:18) (cid:19)
1 1 0 1
subgroup generated by the matrices and is the entire group SL(2,Z).
0 1 −1 0
4. You’ll see some more examples in the second homework.
1
1.4 Symmetric groups and dihedral groups
1.4.1 S
n
Let S be the set of all bijective functions σ : {1,2,...,n} → {1,2,...,n}. Together with composition of
n
functionsasabinaryoperatorS isagroupwithunittheidentityfunction. ElementsofS areoftenwritten
n n
as
(cid:18) (cid:19)
1 2 ... n
σ(1) σ(2) ... σ(n)
Multiplication of matrices can be done easily visually. Here is a self-explanatory example:
(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
1 2 3 4 1 2 3 4 2 1 4 3 1 2 3 4 1 2 3 4
= =
3 4 2 1 2 1 4 3 4 3 1 2 2 1 4 3 4 3 1 2
Note that S is a subgroup of S consisting of all permutations of {1,2,...,n−1}, fixing n.
n−1 n
On the homework you will show that S has cardinality |S |=n!.
n n
1.4.2 D
2n
Let P be a regular n-gon, whose vertices correspond to the n roots of unity of order n in C. Look at all
symmetries of P, i.e., all operations on P that preserve P but move its vertices around. Two: examples: R
is rotation counterclockwise by 2π/n and F is flip with respect to the x-axis.
Symmetries can be composed, in other words, applied sequentially. Thus F2 is applying twice F and so
F2 = 1 where 1 is the identity map. Moreover Rn is rotation by 2π and again this is the identity map so
Rn =1. Also see that RF =FR−1 =FRn−1. The group D is generated by R and F and consists of
2n
D ={1,R,...,Rn−1,F,FR,...,FRn−1}
2n
Using Rn = 1,F2 = 1,RF = FRn−1 it is clear that any combination of rotations and flips can be written
as Rk or FRk and so D has cardinality |D |=2n.
2n 2n
Note that D is a noncommutative group (when n≥3) of order 2n which contains the cyclic group (cid:104)R(cid:105)
2n
of order n.
1.4.3 Cycles in S
n
Definition 6. A cycle (i ,...,i ) is a permutation σ ∈ S such that σ(j) = j for j ∈/ {i ,...,i }, σ(i ) =
1 k n 1 k u
i foru<k andσ(i )=i . Thelengthofacycleis|(i ,...,i )|=k. Acycleoflength2is(ij), onlyflips
u+1 k 1 1 k
i and j and is called a transposition. All cycles of length 1 are equal to the identity element and instead of
(i) we simply write ().
Two cycles c =(i ,...,i ) and c =(j ,...,j ) are said to be disjoint if i (cid:54)=j for all u,v.
1 1 k 2 1 s u v
Proposition 7. 1. If c ,c are disjoint cycles then c c =c c .
1 2 1 2 2 1
2. A cycle c=(i ,...,i ) of length k has order k.
1 k
3. (i ,...,i )=(i i )(i i )···(i i ).
1 k 1 2 2 3 k−1 k
4. Every σ ∈S can be written as a product σ =c ···c where c are disjoint cycles. This expression is
n 1 k i
unique up to permuting the order of the cycles.
5. Every σ ∈S can be written as a product of transpositions, but no uniquely.
n
2
Proof. Most are straightforward, but let me show the fact that permutations are products of disjoint cycles.
Here is an algorithm. Start with a = 1 and construct the cycle c = (a ,σ(a ),σ2(a ),...). Let a be the
1 1 1 1 1 2
smallest number between 1 and n that does not appear in c and let c =(a ,σ(a ),σ2(a ),...). Once you
1 2 2 2 2
havec ,...,c definea asthesmallestnumberbetween1andnnotappearinginc ∪...c andconstruct
1 j j+1 1 j
c =(a ,σ(a ),...). This way you exhaust all the integers between 1 and n.
j+1 j+1 j+1
Lets show that c and c are disjoint for i < j. Suppose σu(a ) = σv(a ). Then σu−v(a ) = a which
i j i j i j
contradicts the choice of a as not appearing in c , which contains all σr(a ) for r ≥0.
j i i
It is now not difficult to show that σ =c c ···c .
1 2 k
3
Graduate Algebra, Fall 2014
Lecture 3
Andrei Jorza
2014-09-01
1 Group Theory
1.3 Subgroups (supplemental)
Example 1. Some more examples of subgroups:
1. Z⊂Q⊂R⊂C are subgroups.
2. GL(n,Q)⊂GL(n,R)⊂GL(n,C) are subgroups.
3. The following are subgroups (for R=Q,R or C):
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
1 b a b a b
{ |b∈R}⊂{ |a∈R×,b∈R}⊂{ |a,c∈R×,b∈R}⊂GL(2,R)
0 1 0 1 0 c
Example 2. Some special subgroups:
1. {e} and G are the non-proper subgroups of G.
2. ThecenterZ(G)ofagroupGisdefinedasZ(G)={g ∈G|gx=γ,∀x∈G}. ThenZ(G)isasubgroup.
3. The commutator [a,b]=aba−1b−1 and [G,G]=(cid:104)[a,b]|a,b∈G(cid:105) is a subgroup. Indeed, [a,b]−1 =[b,a]
butaproductofcommutatorsneednotbeacommutator. ForexampleG=S ={1,(12),(13),(23),(123),(132)}
3
hascenterZ(S )=1andcommutator(cid:104)1,(123),(132)(cid:105)={1,(123),(132)=(123)2}whichisasubgroup
3
since (123) has order 3.
4. If X ⊂G then (cid:104)X(cid:105)={(cid:81)a |a or a−1 ∈X}.
i i i
1.5 Homomorphisms
Suppose (G,· ,e ) and (H,· ,e ) are two groups.
G G H H
Definition 3. Amapf :G→H issaidtobeahomomorphismiff(x· y)=f(x)· f(y)forallx,y ∈G.
G H
Proposition 4. If f :G→H is a homomorphism then:
1. f(e )=e .
G H
2. f(x−1)=f(x)−1.
Proof. f(x) = f(e x) = f(e )f(x) for all x ∈ G and so f(e ) = e . Also e = f(e ) = f(xx−1) =
g G G H H G
f(x)f(x−1) and the second property follows.
1
Definition 5. Suppose f :G→H is a homomorphism of groups. The kernel is ker(f)={g ∈G|f(g)=e}
andIm(f)={f(g)|g ∈G}. Thehomomorphismf issaidtobeanisomorphismifitisbijectiveasafunction.
Proposition 6. Let f :G→H be a homomorphism.
1. kerf ⊂G and Imf ⊂H are subgroups.
2. f is injective iff kerf =1 and surjective iff Imf =H.
3. If f is an isomorphism then f−1 : H → G is also a homomorphism, which is then necessarily an
isomorphism.
4. If f is an injective homomorphism then G∼=Imf.
Proof. If f(x)=1 and f(y)=1 then f(xy−1)=f(x)f(y)−1 =1 and so kerf ⊂G is a subgroup. Similarly,
f(x)f(y)−1 =f(xy−1)∈Imf and so Imf ⊂H is a subgroup as well.
Have f(x)=f(y) iff f(x)f(y)−1 =1 iff f(xy−1)=1 iff xy−1 ∈kerf.
Since f(x)f(y) = f(xy) it follows that f−1(f(x)f(y)) = xy = f−1(f(x))f−1(f(y)) and so f−1 is also a
homomorphism.
The last part is by definition.
Definition 7. TwogroupsGandH aresaidtobeisomorphicifthereexistsanisomorphismbetweenthem.
Example 8. 1. The n-roots of unity in C form a group µ wrt multiplication. The map Z/nZ → µ
n n
given by k (cid:55)→exp(2πik/n) is an isomorphism of groups.
2. The map Z→nZ given by f(x)=nx is an isomorphism of infinite cyclic groups.
3. This example I did in lecture 2 but fits better here. Suppose G is a finite group with n elements. For
g ∈ G let σ : G → G given by σ (h) = gh. This is clearly injective and since σ−1 = σ it is also
g g g g−1
bijective. Note that σ ◦σ = σ and so we get a homomorphism σ : G → S from G to the set
g g(cid:48) gg(cid:48) G
S of permutations of G. Since σ = σ if and only if g = g(cid:48) (evaluate at 1) we get an injective
G g g(cid:48)
homomorphism from G into S =S . Thus we realized G∼=Imf ⊂S .
n G n
4. Consider the map f : Sn → GL(n,Q) ∼= AutQ−vs(Qn) taking the permutation σ ∈ Sn to the n×n
matrix with 0-s everywhere except at (i,σ(i)) for all i where there is a 1. For example
 
(cid:18) (cid:19) 1
1 2 3
f( )=1 
3 1 2
1
What is f(σ)f(τ) for σ,τ ∈S ? Let e ,...,e be the standard basis of Qn. Then f(σ) is the matrix
n 1 n
wrt this basis of the linear map T : Qn → Qn taking (cid:80)x e to (cid:80)x e . Thus f(σ)f(τ) is the
σ i i i σ(i)
(cid:80) (cid:80) (cid:80)
matrix of T ◦T which takes x e to T ( x e ) = x e and so T ◦T = T and thus
σ τ i i σ i τ(i) i σ(τ(i)) σ τ στ
f(σ)f(τ)=f(στ) which shows that f is a homomorphism. It’s also clearly injective.
Thus we realized S as a subgroup of GL(n,Q). This is the first instance of realizing a group as a
n
subgroupofamatrixgroupusinga“faithfullinearrepresentation”,whichisaverypowerfultoolabout
which we’ll learn in representation theory.
1.6 The alternating group A
n
Proposition 9. There is a homomorphism ε:S →{−1,1} such that ε((i ,...,i ))=(−1)k−1.
n 1 k
2
Proof. Let f : S → GL(n,Q) as above and take ε(σ) = detf(σ). Then ε is a homomorphism S → Q×.
n n
We’d like to check that ε(σ)=±1 for every permutation σ.
What is detf(σ)? It is the linear map ∧nf(σ) : ∧nQn → ∧nQn. Explicitly, it is ∧nf(σ)e ∧...∧e =
1 n
(f(σ)e )∧...∧(f(σ)e )=e ∧...∧e =±e ∧...∧e a number which is 1 if σ has an even number
1 n σ(1) σ(n) 1 n
of inversions and −1 otherwise.
There is something more general to be said. Suppose that f(σ) has integer entries. Then the above
explanation implies that detf(σ) ∈ Z. Note that f(σ)−1 = f(σ−1) and so I = f(σ)f(σ−1) and so
n
1=ε(σ)ε(σ−1) and each of the two factors is an integer. Thus again we get that ε(σ)=±1 indirectly this
time.
Finallyε((ij))=−1andtheconclusionfollowsfromwritingthecycleasaproductoftranspositions.
Definition 10. Let A ⊂S be the subgroup A =kerε of the sign homomorphism ε.
n n n
3
Graduate Algebra, Fall 2014
Lecture 4
Andrei Jorza
2014-09-03
1 Group Theory
1.7 Group quotients
For g ∈ G and H ⊂ G write gH = {gh|h ∈ H} and Hg = {hg|h ∈ H}. If H and K are subsets then
HK ={hk|h∈H,k ∈K}.
Lemma 1. Let H ⊂ G be a subgroup. If g,h ∈ G then gH and hH are either disjoint, if g−1h ∈/ H, or
coincide if g−1h∈H. The same is true for Hg and Hh, depending on whether gh−1 ∈H or not.
Proof. If gx = hy for some x,y ∈ H then g−1h = xy−1 so if g−1h ∈/ H then the two sets are disjoint. If
g−1h=u∈H then hH =guH =gH as u∈H.
Definition 2. Define G/H as the set of cosets gH. Similarly write H\G for the set of cosets Hg.
Proposition 3. Let H ⊂G be a subgroup.
1. The map f(x) = x−1 gives a bijection G/H → H\G. The common cardinality |G/H| = |H\G| is
denoted [G:H].
2. If G is a finite group then [G:H]=|G|/|H|.
Proof. For the first part: (gH)−1 =H−1g−1 =Hg−1.
For the second part: the finite set G is partitioned into finitely many sets of cardinality |H| and |G/H|
is the number of these sets.
Corollary 4 (Lagrange). Let G be a finite group.
1. If H is a subgroup of G then |H|||G|.
2. If a∈G has order ord(a)=m then m||G|.
3. If a∈G then a|G| =1.
Proof. The first part follows from the proposition. The second from the fact that |(cid:104)a(cid:105)|=ord(a). The third
part follows from the second part.
Example 5. 1. Foranintegerndenoteϕ(n)thecardinalityof(Z/nZ)×,thenumberofk between1and
n coprime to n. Then for a∈(Z/nZ)× one has aϕ(n) ≡1 (mod n). Indeed, ord(a)|ϕ(n).
2. In particular, if p(cid:45)a then ap−1 ≡1 (mod p).
3. If |G| = p is a prime number then G is cyclic. Indeed, if a ∈ G is nontrivial then ord(a) | p and so
ord(a)=p and so G=(cid:104)a(cid:105).
Proposition 6. From pset 2 if K ⊂H ⊂G are subgroups then [G:K]=[G:H][H :K] (with no finiteness
assumption on G).
1
1.8 Direct products
We already saw that if G and H are groups then G×H is a group, called the exterior direct product.
Definition 7. Suppose G is a group and K,H ⊂G are subgroups. The interior direct product of H and K
is the set HK ={hk|h∈H,k ∈K}⊂G.
Lemma 8. The interior direct product HK is a subgroup of G if and only if HK = KH as sets. Then
HK =(cid:104)H ∪K(cid:105).
Proof. If HK =KH then any ratio of HK is still in HK and so HK is a subgroup. Reciprocally, if HK is
a subgroup then KH =(eK)(He)=(HK)(HK)=HK.
Proposition 9. 1. [H :H ∩K]≤[G:K]
2. If [G:K] is finite then G=HK iff the above ineqality is in fact equality: [H :H ∩K]=[G:K].
Proof. SinceH∩K ⊂K,gH∩K ⊂gK andsowegetamapH/H∩K →G/K bysendingthecosetgH∩K
to the unique coset gK containing it. Suppose that for g,h∈H, gH ∩K and hH ∩K are both sent to the
same coset gK =hK. Then g−1h∈K but also in H and so in H∩K which means that gH∩K =hH∩K.
Thus this map in injective proving the first part.
For the second part we need surjectivity since G/K has finite cardinality. Suppose gK = hK for some
h ∈ H. Then immediately g ∈ hKK−1 = hK ⊂ HK and so surjectivity can only happen if G = HK.
Suppose that G = HK. Then every g ∈ G can be written g = hk in which case gK = hkK = hK and
surjectivity follows.
1.9 Normal subgroups
Definition 10. A subgroup H ⊂ G is said to be normal if gHg−1 = H for every g ∈ G in which case one
writes H (cid:67)G.
Proposition 11. If H (cid:67)G then G/H and H\G become groups.
Proof. Indeed, gHhH = ghh−1HhH = ghHH = ghH and (gH)−1 = H−1g−1 = Hg−1 = g−1gHg−1 =
g−1H.
Lemma 12. Suppose f :G→H is a homomorphism. Them kerf(cid:67)G. Moreover, if H(cid:67)G then G→G/H
sending g to gH is a group homomorphism with kernel H.
Proof. If f(g) = 1 and h ∈ G then f(hgh−1) = f(h)f(g)f(h−1) = f(h)f(h−1) = 1 and so hgh−1 ∈ kerf.
The second part is straightforward.
Example 13. General examples.
1. Every subgroup of an abelian group is normal.
2. If H,K(cid:67)G then H ∩K(cid:67)G.
3. If N (cid:67)G and N is a subgroup of a subgroup H of G then N (cid:67)H.
4. If N (cid:67)G and H is a subgroup of G then N ∩H (cid:67)H.
5. If G is a group then Z(G)(cid:67)G.
Example 14. Specific examples.
1. The alternating group A =kerε is a normal subgroup of S as ε is a homomorphism.
n n
2. For R=Q,R or C, SL(n,R)(cid:67)GL(n,R).
2