Advanced Algebra I
Sep. 19,20, 2003 (Fri. Sat.) 1. Set Theory
We recall some set theory that will be frequently used in the sequel or that is not covered in the basic college course.
1.1. Zorn’s Lemma.
Definition 1.1. A set S is said to be partially ordered if there is a relation ≤ such that
(1) x ≤ x
(2) if x ≤ y and y ≤ x, then x = y.
(3) if x ≤ y and y ≤ z then x ≤ z.
We usually call a partially ordered set to be a POSET.
Definition 1.2. A pair of elements is said to be comparable if either x ≤ y or y ≤ x. A set is said to be totally ordered if every pair is comparable.
We also need the following definition:
Definition 1.3. A maximal element of an poset S is an element m ∈ S such that if m ≤ x then m = x.
Foe a given subset T ⊂ S, an upper bound of T is an element b ∈ S such that x ≤ b for all x ∈ T .
One has
Theorem 1.4 (Zorn’s lemma). Let S be a non-empty poset. If ev- ery non-empty totally ordered subset (usually called a ”chain”) has an upper bound, then there exists a maximal element in S.
Example 1.5. Let R be a 6= 0 commutative ring. One can prove that there exists a maximal ideal by using Zorn’s lemma. The proof goes as following: Let S = {I C R|I 6= R} equipped with the ⊂ as the partial ordering. S 6= ∅ because 0 ∈ S. For a chain {Ij}j∈J, one has a upper bound I = ∪Ij. Then we are done by Zorn’s lemma.
1.2. cardinality. In order to compare the ”zise of sets”, we introduce the cardinality.
Definition 1.6. Two sets A, B are said to have the same cardinality if there is a bijection between them, denoted |A| = |B|. And we said
|A| ≤ |B| if there is a injection from A to B.
It’s easy to see that the cardinality has the properties that |A| ≤ |A|
and if |A| ≤ |B|, |B| ≤ |C|, then |A| ≤ |C|. So It’s likely that the
”cardinality are partially ordered” or even totally ordered.
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Lemma 1.7. Given two set A, B, either |A| ≤ |B| or |B| ≤ |A|.
Sketch. Consider
S = {(C, f )|C ⊂ A, f : C → B is an injection}.
Apply Zorn’s lemma to S, one has an maximal element (D, g), then
one claim that either D = A or im(g) = B. ¤
Theorem 1.8 (Schroeder-Bernstein). If |A| ≤ |B| and |B| ≤ |A|, then
|A| = |B|.
sketch. Let f, g be the injection from A, B to B, A respectively. One needs to construct a bijection by using f and g. Some parts of A use f and some parts not. So we consider the partition
A1 := {a ∈ A|ahas parentless ancestor in A}, A2 := {a ∈ A|ahas parentless ancestor in B},
A3 := {a ∈ A|ahas infinite ancestor}.
And so does B.
Then we claim that f restricted to A1, A3 are bijections to B1, B3. And g restricted to B2, B3 are bijections to A2, A3. So the desired
bijection can be constructed. ¤
We need some more properties of cardinality. If |A| = |{1, .., n}, then we write |A| = n. And if |A| = |N then we write |A| = ℵ0.
Proposition 1.9. If A is infinite, then ℵ0 ≤ |A|.
Proof. By Axiom of Choice. ¤
Definition 1.10.
|A| + |B| := |A q B|,
|A| · |B| := |A × B|.
We have the following properties:
Proposition 1.11. (1) If |A| is infinite and |B| is finite, then |A+
B| = |A|.
(2) If |B| ≤ |A| and |A| is infinite, then |A + B| = |A|.
(3) If |B| ≤ |A| and |A| is infinite, then |A × B| = |A|.
Proof. For (1), take an countable subset A0 in A, one sees that |A0| =
|A0| + |B| by shifting. Then we are done.
For (2), It’s enough to see that |A + A| ≤ |A|. Pick an maximal subset X ⊂ A having the property that |X + X| ≤ |X| (by Zorn’s Lemma). One claim that A − X is finite, and then we are done by (1).
For (3), we leave it as an exercise to the readers. ¤
Advanced Algebra I
Sylow Theorems
We are now ready to prove Sylow theorems. The first theorem re- gards the existence of p-subgroups in a given group. The second theo- rem deals with relation between p-subgroups. In particular, all Sylow p-subgroups are conjugate. The third theorem counts the number of Sylow p-subgroups.
Theorem 0.1 (First Sylow theorem). Let G be a finite group of order pnm (where (n, m) = 1). Then there are subgroups of order pi for all 0 ≤ i ≤ n.
Furthermore, for each subgroup Hi of order pi, there is a subgroup Hi+1 of order pi+1 such that HiC Hi+1 for 0 ≤ i ≤ n − 1.
In particular, there exist a subgroup of order pn, which is maximal possible, called Sylow p-subgroup. We recall the useful lemma which will be used frequently.
Lemma 0.2. Let G be a finite p-group. Then
|S| ≡ |S0| (mod p).
proof of the theorem. We will find subgroup of order pi inductively. By Cauchy’s theorem, there is a subgroup of order p. Suppose that H is a subgroup of order pi. Consider the group action that H acts on S = G/H by translation. One show that xH ∈ S0 if and only if x ∈ NG(H). Thus |S0| = |NG(H)/H|. If i < n, then
|S0| ∼= |S| ∼= 0 (mod p).
By Cauchy’s theorem, the group NG(H)/H contains a subgroup of or- der p. The subgroup is of the form H1/H, hence |H1| = pi+1. Moreover, H C H1.
¤ Example 0.3. If G is a finite p-group of order pn, then one has a series of subgroups {e} = H0 < H1 < ... < Hn = G such that |Hi| = pi and HiC Hi+1, Hi+1/Hi ∼= Zp. In particular, G is solvable.
Definition 0.4. A subgroup P of G is a Sylow p-subgroup if P is a maximal p-subgroup of G.
If G is finite of order pnm then a subgroup P is a Sylow p-subgroup if and only if |P | = pn by the proof of the first theorem.
Theorem 0.5 (Second Sylow theorem). Let G be a finite group of order pnm. If H is a p-subgroup of a G, and P is any Sylow p-subgroup of G, then there exists x ∈ G such that xHx−1 < P .
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Proof. Let S = G/P and H acts on S by translation. Thus by the Lemma, one has |S0| ≡ |S| = m(mod p). Therefore, S0 6= ∅. One has
xP ∈ S0 ⇔ hxP = xP ∀h ∈ H ⇔ x−1hx < P.
¤ An immedaitely but important consequence is that two Sylow p- subgroups are conjugate.
Theorem 0.6 (Third Sylow theorem). Let G be a finite group of order pnm. The number of Sylow p-subgroups divides |G| divides |G| and is of the form kp + 1.
Proof. Let S be the conjugate class of a Sylow p-subgroup P (this is the same as the set of all Sylow p-subgroups). We consider the action that G acts on S by conjugation, then the action is transitive. Hence
|S| | |G|.
Furthermore, we consider the action P × S → S by conjugation.
Then
Q ∈ S0 ⇔ xQx−1 = Q ∀x ∈ P ⇔ P < NG(Q).
Both P, Q are Sylow p-subgroup of NG(Q) and therefore conjugate in NG(Q). However, Q C NG(Q), Q has no conjugate other than itself.
Thus one concludes that P = Q. In particular, S0 = {P }. By the
Lemma, one has |S| = 1 + kp. ¤
Example 0.7. Group of order 200 must have normal Sylow subgroups.
Hence it’s not simple. (let rp := number of Sylow p-subgroups. Then r5 = 1).
Example 0.8 (Classification of groups of order 2p (p 6= 2)). Let G be a group of order 2p. If it’s abelian, then it’s cyclic by fundamental the- orem of abelian groups plus Chinese remainder theorem. Let’s suppose that it’s non-abelian.
There are elements a, b of order p, 2 respectively. By Sylow theorem, rp = 1, hence the subgroup < a > is normal. Then one notices that G =< a >< b > for < a > ∩ < b >= {e}. Moreover, bab−1 = ak for some k. One has
a = b2ab−2 = ak2.
It follows that k = 1, −1. If k = 1, then G is abelian. Thus we assume that k = −1. This gives the group Dp :=< a, b|ap = b2 = e, ab = ba−1 >.
Proposition 0.9. If H, K C G and H ∩ K = {e}, HK = G ,then G ∼= H ⊕ K.
Proposition 0.10. Let G be a group of order pq, with p > q distinct primes. If q - p − 1, then G is cyclic. If q | p − 1 then either G is cyclic or there is a unique model of non-abelian group up to isomorphism.
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(Which is a ”semi-direct” of two cyclic groups, or called a metacyclic groups in this case)
Advanced Algebra I
Group Action
We will define the group action and illustrate some previous known theorem from group action point of view.
Definition 0.1. We say a group G acts on a set S, or S is a G-set, if there is function α : G × S → S, usually denoted α(g, x) = gx, compatible with group structure, i.e. satisfying:
(1) let e ∈ G be the idetity, then ex = x for all x ∈ S.
(2) g(hx) = (gh)x for all g, h ∈ G, x ∈ S.
By the definition, it’s clear to see that if y = gx, then x = g−1y.
Because x = ex = (g−1g)x = g−1(gx) = g−1y.
Moreover, one can see that given a group action α : G × S → S is equivalent to have a group homomorphism ˜α : G → A(S), where A(S) denote the group of bijections on S.
Exercise 0.2. Check the equivalence of α and ˜α.
An application is to take a finite group G of order n, and take S = G.
Then the group multiplication gives a group action. Thus we have a group homomorphism
˜
α : G → A(G) ∼= Sn.
One can check that in this case ˜α is an injection. Thus we have the Cayley’s theorem.
We now introduce two important notions:
Definition 0.3. Suppose G acts on S. For x ∈ S, the orbit of x is defined as
Ox:= {gx|g ∈ G}.
And the stabilizer of x is defined as
Gx:= {g ∈ G|gx = x}.
Then one has the following Proposition 0.4.
|G| = |Ox| · |Gx|.
Sketch. Consider Sy := {g ∈ G|gx = y}. Then G is a disjoint union of Sy for all y ∈ Ox. Furthermore, fix a g such that y = gx, then one has Sy = gGx. Thus
|G| = | ∪y∈Ox Sy| = X
y∈Ox
|Gx| = |Ox| · |Gx|.
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By applying this to the situation that H < G is a subgroup and take S = G/H with the action G × G/H → G/H via α(g, xH) = gxH. For H ∈ S, the stabilizer is H, and the orbit is G/H. Thus we have
|G| = |G/H| · |H|, which is the Lagrange’s theorem.
Another way of counting is to consider the decomposition of S into disjoint union of orbits. Note that if Ox = Oy if and only if y ∈ Ox. Thus for convenience, we pick a representative in each orbit and let I be a set of representatives of orbits. We have:
S = ∪x∈IOx. In particular,
|S| =X
x∈I
|Ox|.
This simple minded equation actually give various nice application.
We have the following natural applications.
Example 0.5 (translation). Let G be a group. One can consider the action G × G → G by α(g, x) = gx. Such action is called translation.
More generally, let H < G be a subgroup. Then one has translation H × G → G by (h, x) 7→ hx. Then |S| = P
x∈I|Ox| gives Lagrange theorem again.
Example 0.6 (conjugation). Let G be a group. One can consider the action G × G → G by α(g, x) = gxg−1. Such action is called conjugation. For a x ∈ G, Gx = C(x), the centralizer. And Ox = {x}
if and only if x ∈ Z(G), the center of G. So, for G finite, the equation
|S| = P
x∈I|Ox| now gives
|G| =X
x∈I
|G|/|C(x)|.
Which is the class equation.
The class equation (we mean the general form |S| = P
x∈I|Ox|) is very useful if the group is a finite p-group. By a finite p-group, we mean a group G with |G| = pn for some n > 0. Consider now G is a finite p group acting on S. Let
S0 := {x ∈ S|gx = x, ∀g ∈ G}.
Then the class equation can be written as
|S| = |S0| + X
x∈I,x6∈S0
|Ox|.
One has the following
Lemma 0.7. Let G be a finite p-group. Keep the notation as above, then
|S| ≡ |S0| (mod p).
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Proof. If x 6∈ S0, then 1 6= |Ox| = pk.
¤ By consider the conjugation G × G → G, one sees that
Corollary 0.8. If G is a finite p-group, then G has non-trivial center.
By using the similar technique, one can also prove the important Cauchy’s theorem
Theorem 0.9. Let G be a finite group such that p | |G|. Then there is an element in G of order p.
Proof. Let
S := {(a1, ..., ap)|ai ∈ G,Y
ai = e}.
And consider a group action Cp×S → S by (1, (a1, .., ap)) 7→ (ap, a1, ..., ap−1).
One claims that S0 = {(a, a, ..., a)|a ∈ G}.
By the Lemma, one has |S| ≡ |S0| (mod . It follows that p | |S0|.
In particular, |S0| > 1, hence there is (a, ..., a) ∈ S0 with a 6= e. One
sees that o(a) = p. ¤
Advanced Algebra I
Nilpotent and solvable groups, normal series, Jordan-H¨older theorem
Let G be a group, the center Z(G) is a normal subgroup of G. And we have the canonical projection G → G/Z(G). Let C2(G) be the inverse image of Z(G/Z(G)) in G. By the correspondence theorem, Z(G/Z(G)) is a normal subgroup of G/Z(G) hence C2(G) is a nor- mal subgroup of G. And then let C3(G) to be the inverse image of Z(G/C2(G)). By doing this inductively, one has an ascending chain of normal subgroups
{e} < C1(G) := Z(G) < C2(G) < ...
Definition 0.1. G is nilpotent if Cn(G) = G for some n.
Proposition 0.2. A finite p-group is nilpotent.
Proof. We use the fact that a finite p-group has non-trivial center. Thus one has Ci Ci+1. The group G has finite order thus the ascending chain must terminates, say at Cn. If Cn 6= G, then G/Cnhas non-trivial center. One has Cn Cn+1 which is impossible. Hence Cn= G. ¤ Theorem 0.3. If H, K are nilpotent, so is H × K.
Proof. The key observation is that Z(H × K) = Z(H) × Z(K). Then inductively, one proves that Ci(H × K) = Ci(H) × Ci(K). If Cn(H) = H, Cm(K) = K then Cl(H × K) for l = max(m, n). ¤
Then we are ready to prove the following:
Theorem 0.4. A finite group is nilpotent if and only if it’s a direct product of Sylow p-subgroups.
Proof. By the previous two results, it’s clear that a direct product of Sylow p-subgroups is nilpotent.
Conversely, if G is nilpotent, then we can claim that every Sylow p-subgroup is a normal subgroup of G. Then by checking the decom- position criterion, one has the required decomposition.
Claim. If P is Sylow p-subgroup, then P C G.
To this end, it suffices to prove that NG(P ) = G. We prove the follow- ing claim:
Claim. If H is a proper subgroup of a nilpotent group G, then H is a proper subgroup of NG(H).
By applying this Claim to NG(H), then it says that NG(H) can’t be a proper subgroup of G since NG(NG(H)) = NG(H). Thus it follows
that NG(H) = G. ¤
We have seen that we have a series of subgroup by taking centers.
Another similar construction is to take commutators.
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Definition 0.5. Let G be a group. The commutator of G, denoted G0 is the subgroup generated by the subset {aba−1b−1}.
Roughly speaking, the subgroup G0 measures measure the commu- tativity of a group. The smaller G0, the more commutative it is.
Theorem 0.6. G0C G, and G/G0 is ableian. Moreover, if N C G, then G/N is abelian if and only if G0 < N .
Proof. (1) for all g ∈ G, g(aba−1b−1g−1 ∈ G0, hence gG0g < G0. So G0C G.
(2)
aG0bG0 = abG0 = ab(b−1a−1ba)G0 = baG0 = bG0aG0.
(3) Consider π : G → G/N. If G/N is abelian, then π(aba−1b−1 = e, hence G0 < N. Conversely, if G0 < N, we have a surjective homomorphism G/G0 → G/N. G/G0 is abelian, hence so is it homomorphic image G/N.
¤ Definition 0.7. We can define the the commutator inductively, i.e.
G(2) := (G0)0, etc. The G(i) is called the i-th derived subgroup of G.
It’s clear that G > G0 > G(2) > ....
A group is solvable is G(n) = {e} for some n.
Example 0.8. Take G = S4. The commutator is the smallest subgroup that G/G0 is abelian. Since the only non-trivial normal subgroups of S4 are V, A4. It’s clear that G0 = A4 (Or one can prove this by hand).
Similarly, one finds that G(2) = A04 = V , and G(3) = {e}. Hence S4 is solvable.
Another useful description of solvable groups is the groups with solv- able series.
Definition 0.9. A groups G has a subnormal series if there is a series of subgroups of G
G = H0 > H1 > H2 > ... > Hn, such that HiC Hi−1 for all 1 ≤ i ≤ n.
A subnormal series is a solvable series if Hn = {e} and Hi−1/Hi is abelian for all 1 ≤ i ≤ n.
A subnormal series is a normal series if all Hi are normal subgroups of G.
Theorem 0.10. A group is solvable if and only it has a solvable series.
Proof. It’s clear that G > G0 > ...G(n) = {e} is a solvable series. It suffices to prove that a group with a solvable series is solvable. Suppose now that G has a sovable series {e} = Hn < ... < H0 = G. First observe that G0 < H1 since G/H1 is abelian. We claim that G(i) < Hi
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for all i inductively. Which can be prove by the observation that the intersection of the series {e} = Hn < ... < H0 = G with G(i) gives a
solvable series of G(i). ¤
Example 0.11. A finite p-group has a solvable series, hence is solvable.
Moreover, a nilpotent group is solvable.
Proposition 0.12. Let H be a subgroup of a solvable group G, then H is solvable.
Let N be a normal subgroup of G. Then G is solvable if and only if both N and G/N are solvable.
Example 0.13. A5 is not solvable, hence so is Sn for n ≥ 5.
0.1. simplicity of A5. An element in Snis said to be have cycle struc- ture (m1, .., mr) with m1 ≥ m2 ≥ ... ≥ mr , m1 + ... + mr = n if its cycle decomposition is of length m1, ..., mr respectively. For example, (1, 2)(3, 4) ∈ S4 has cycle structure (2, 2) and (1, 2) ∈ S4 has cycle structure (2, 1, 1).
Remark 0.14. There is a one-to-one correspondence between cycle structures of Sn and partition of the integer n.
A key observation is that any two elements are conjugate to each other if and only if they have the same cycle structure. Let’s call the set of all elements of cycle structure (m1, ..., mr) the cycle class of (m1, ...mr). A consequence of this fact is that a subgroup N < Sn is normal if and only if N is union of cycle classes.
Let’s put it another way, given a group G, we can always consider the group action G × G → G by conjugation. The conjugate classes are the orbits. A subgroup H < G is normal if and only if it is union of orbits. If G = Sn, then orbits are cycle classes.
Example 0.15. In S4, V is the union of class (1, 1, 1, 1) and (2, 2).
A4 is the union of V and the class (3, 1).
The purpose of this subsection is to show that A5 is a simple non- abelian group, hence a non-solvable group.
Theorem 0.16. A5 is a simple non-abelian group.
Proof. One note that in S5, possible cycle structures are (5), (4, 1),(3, 1 , 1),(3, 2), (2, 2, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1) with 24, 30, 20, 20, 15, 10, 1 el- ements in each class. While A5 is the union of classes of (5), (3, 1, 1), (2, 2, 1), (1, 1, 1, 1, 1).
We consider the actions of conjugation α : S5 × A5 → A5 and its restriction β : A5× A5 → A5. For σ ∈ A5, let Oα,σ be the orbit of the α and Oβ,σ be the orbit of the β. And let Gα,σ, Gβ,σ be the stabilizer.
It’s clear that Gα,σ = CS5(σ) and Gβ,σ = CA5(σ) = CS5(σ) ∩ A5. Thus we have either |Gβ,σ| = 12|Gα,σ| or |Gβ,σ| = |Gα,σ|. Hence |Oβ,σ| =
|Oα,σ| or |Oβ,σ| = 12|Oα,σ|.
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case 1. If σ has cycle structure (5), then |Oα,σ| = 24, |Gα,σ| = 5. It follows that |Gβ,σ| = 5 and hence |Oβ,σ| = 12.
case 2. If σ has cycle structure (3, 1, 1), then |Oα,σ| = 20, |Gα,σ| = 6.
However, one notice that there is an element τ ∈ CS5(σ) − CA5(σ) (e.g.
(45)(123) = (123)(45)). Hence |Gβ,σ| 6= |Gα,σ| and must be 12|Gα,σ| = 3.
Therefore |Oβ,σ| = 20.
case 3. If σ has cycle structure (2, 2, 1),then |Oα,σ| = 15, |Gα,σ| = 8.
It follows that |Oβ,σ| = 15.
Combining all this, if H < A5 is a normal subgroup, then |H| = 1 + 12r1+ 20r2+ 15r3, where ri are integers. Moreover |H| | |A5| = 60, which is impossible unless |H| = 1 or 60.
¤ WE turning back to series a little bit more. A subnormal series is called a composition series if every quotient is a simple group.
Definition 0.17. For a subnormal series, {e} = Hn < ... < H0 = G, the factors of the series are the quotient groups Hi−1/Hi and the length is the number of non-trivial factors. A refinement is a series obtained by finite steps of one-step refinement which is {e} = Hn < . < K <
.. < H0 = G.
Definition 0.18. Two series are said to be equivalent if there is a one-to-one correspondence between the non-trivial factors. And the corresponding factors groups are isomorphism.
It’s clear that this defines an equivalent relation on subnormal series.
The main theorems are
Theorem 0.19 (Schreier). Any two subnormal (resp. normal) series of a group G have a subnormal (resp. normal) refinement that are equivalent.
An immediate corollary is the famous Jordan-H¨older theorem.
Theorem 0.20 (Jordan-H¨older). Any two composition series of a group are equivalent.
The main technique is the Zassenhaus Lemma, or sometimes called butterfly Lemma.
Lemma 0.21 (Zassenhaus). Let A∗C A and B∗C B be subgroups of G. Then
(1) A∗(A ∩ B∗) C A∗(A ∩ B).
(2) B∗(A ∩ B) C B∗(A ∩ B).
(3) A∗(A ∩ B)/A∗(A ∩ B∗) ∼= B∗(A ∩ B)/B∗(A∗∩ B).
proof of the lemma. It’s clear that A∩B∗ = (A∩B)∩B∗CA∩B. And similarly, A∗∩ B C A ∩ B. Let D = (A ∩ B∗)(A∗∩ B) C A ∩ B. One can have a well-defined homomorphism f : A∗(A ∩ B) → A ∩ B/D with kernel A∗(A ∩ B∗). And similarly for the other homomorphism. ¤
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proof of Schreier’s theorem. Let {e} = Gn+1 < ... < G0 = G and {e} = Hm+1 < ... < H0 = G be two subnormal series. Let G(i, j) =:=
Gi+1(Gi ∩ Hj) (resp. H(i, j) := Hj+1(Gi ∩ Hj)). Then one has a refinement
G = G(0, 0) > G(0, 1) > ... > G(0, m) > G(1, 0) > ... > G(n, m), G = H(0, 0) > H(1, 0) > ... > H(n, 0) > H(0, 1) > ... > H(n, m).
By applying Zaseenhaus Lemma to Gi+1, Gi, Hj+1, Hj, one has G(i, j)/G(i, j + 1) ∼= H(i, j)/H(i + 1, j).
¤
Advanced Algebra I
free groups and presentation of groups
For any given non-empty set X, we would like to construct a free group F (X) associate to X which is universal w.r.t all functions from X to groups. That is:
Theorem 0.1. Let X be a non-empty set. There is a group F (X) and a map ı : X → F (X) satisfying the following ”unerversal property”:
For any function f : X → G to a group G, there is a unique group homomorphism ¯f : F → G such that ¯f ◦ ı = f .
One way to prove it is to construct the free group out of X. The idea is to construct a group out of X without any constrain. To do this, we start with X, and then we need the inverse, and identity. And moreover, we need element of the form x1x2... etc. After we success- fully constructed the group, it’s natural that the group F (X) have the universal property.
Proof. Let X be a non-empty set. Let
X0 := (X × {1, −1}) ∪ {1}.
By abuse the notation, we identify X with X × {1} and X−1 with X × {−1}. And we have the inverse maps µ : X → X−1 and ν : X−1 → X. We denote the image of inverse maps at a ∈ X0 − {1} by a−1.
A word on X is a sequence (a1, a2, ...) such that ai ∈ X0 and there is n0 such that ai = 1 for all i > n0. (We call it a word of length n0.)
Given two words w1 := (a1, a2, ..., an, 1, ...), w2 := (b1, b2, ..., bm, 1...) of length n, m respectively, one can compose them to obtain a word
w1 ◦ w2 := (a1, ..., an, b1, ..., bm, 1, ...).
A word w (of length n) is said to be reduced if (1) ai 6= 1 for i ≤ n.
(2) ai+1 6= a−1i for all i < n.
Let W be the set of all words on X and let F (X) be the set of reduced words. It’s tedious but elementary to show that there is a well-defined map from W → F (X). Moreover, the composition on W induces a well-defined composition on F (X).
One can check that F (X) together with composition of words is a group. Moreover, one has ı : X → F (X) given by x 7→ (x, 1, ...).
For simplicity, given a reduced word (a1, .., an, 1, ...), we denote it by a1a2...an.
Lastly, we prove the universal property. Let f : X → G be a func- tion from X to a group G. One can produce a map ¯f : F (X) → G by f (a¯ 1...an) := f (a1)...f (an). One checks that this is a group homomor- phism and also ¯f ◦ ı = f .
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It remains to prove the uniqueness of the homomorphism ¯f . Suppose one has a group homomorphism g : F (X) → G such that g ◦ ı = f . Then g(x) = f (x) = ¯f (x) for all x ∈ X. One checks that g(w) = ¯f (w) by induction on length of reduced word w ∈ F (X).
¤ Proposition 0.2. Let G be a group, then there is a free group F map onto G. In particular, any group G can be realized as F/N for some free group F and normal subgroup N C G.
Proof. Given a group G. We can take F = F (G). The identity map iG : G → G induces a group homomorphism ¯f : F (G) → G by the universal property. It’s clear that this homomorphism is surjective.
Let N be ker( ¯f . It follows by the first homomorphism theorem that
G ∼= F/N. ¤
In general, given a non-empty set X and a set (possibly empty) of reduced words R, one can produced a free group F (X) and a normal subgroup generated by R, denoted N. This give rise to a group <
X|R >:= F (X)/N. We call it the group generated by X with relations R.
Example 0.3. Let X = {x, y} and R = {xyx−1y−1}, then < X|R >∼= Z2.
Example 0.4. Let X = {x, y} and R = {xn, y2, xyxy}, then < X|R >∼= Dn.
Remark 0.5. If the generators and relations are given explicitly as in the previous examples, we usually write it as < x, y|xn = y2 = 1, (xy)2 = 1 >.
It’s in general not easy to determine the structure of the groups of the type < X|R >. We present here the Todd-Coxeter algorithm if
< X|R > is finite. Please note that we don’t have good criterion for finiteness of < X|R > yet.
The idea behind the Todd-Coxeter algorithm is the following: if G :=< X|R > is finite, then one has an embedding G → S|G| by Cay- ley’s theorem. For elements in G and label it by 1, 2, 3.. and so on.
Make tables out of the relation which basically describe the multipli- cation table of G (or how they correspond to permutations). Once the tables are completed, then we are done.
Please notice that if G is infinite, then the procedure will never end.
For more detail, please see [Ar] p.223- or [Ro] p. 351-.
[Ar] M. Artin, Algebra, Prentice-Hall International edition.
[Ro] J. Rotman, An Introduction to the Theory of Groups, GTM 148.
Advanced Algebra I
representation of finite groups
Another interesting realization of the tetrahedral group T is done by choose coordinates such that ±ei are those midpoint of 6 edges. Then one can express T as a finite subgroup of GL(3, R). This is an example of a representation.
Definition 0.1. A n-dimensional matrix representation of a group G is a homomorphism
R : G → GL(n, F ),
where F is a field. A representation is faithful if R is injective. And we write Rg for R(g)
It’s essential to work without fixing a basis. Thus we introduce the concept of representation of a group on a finite dimensional vector space V .
Definition 0.2. By a representation of G on V , we mean a homomor- phism ρ : G → GL(V ), where GL(V ) denote the group of invertible linear transformations on V . We write ρg for ρ(g)
Remark 0.3. By fixing a basis β of V , one has β : GL(V ) → GL(n, F )
T 7→ matrix of T.
And one has a matrix representation R := β ◦ ρ.
Furthermore, if a change of basis is given by a matric P , then one has the conjugate representation R0 = P RP−1, that is R0g = P RgP−1 for all g ∈ G.
Remark 0.4. We would like to remark that the concept of a linear representation of G on V is equivalent to G acts on V linearly. More precisely, G acts on the vector space V and the action satisfying
g(v + v0) = gv + gv0, g(cv) = cg(v) for all g ∈ G, v, v0 ∈ V and c ∈ F .
Definition 0.5. Let ρ, ρ0 be two representations of G on V, V0. They are said to be isomorphic if there is an isomorphism of τ : V → V0 which is compatible with ρ and ρ0. That is,
τ ρs(v) = ρ0sτ (v), for all s ∈ G, v ∈ V .
Example 0.6. A representation of degree 1 is a homomorphism R : G → C∗. Since every element has finite order, Rg is a root of unity.
In particular, |Rg| = 1.
1
2
Example 0.7 (Regular representation). Let G be a finite group of order g and let V be a vector space with basis {et}t∈G. For s ∈ G, let Rs be the linear map of V to V which maps et to est. This is called the regular representation of G.
Note that es = Rs(e1) for all s ∈ G. Hence the image of e1 ∈ V form a basis. On the other hand, if τ : G → W is a representation with the property that there is a v ∈ W such that {τs(v)}s∈G forms a basis. Then W is isomorphic to the regular representation. This is the case by considering τ : V → W with τ (es) = ρs(v).
More generally, if G acts on a finite set X, the one can have a representation similarly on the vector space V with basis X. This is called the permutation representation associated to X.
Let ρ, ρ0 be two representations of G on V, V0, then one can define ρ⊕
ρ0, ρ ⊗ ρ0 naturally. Note that if degree of ρ and ρ0 are d, d0 respectively, then degree of ρ ⊕ ρ0 is d + d0 and degree of ρ ⊗ ρ0 is dd0.
Definition 0.8. V is irreducible representation if V is not a direct sum of two representation non-trivially.
One might ask whether a representation is irreducible or not. We threrefore introduce the G-invariant subspace as we did in linear alge- bra.
Definition 0.9. Let ρ : G → GL(V ) be a representation. A vector subspace W of V is said to be a G-invariant subspace if ρs(W ) ⊂ W for all s ∈ G. It’s clear that the restriction of G action on V to W give a representation of G on W , which is called the subrepresentation of V .
Theorem 0.10 (Maschke’s Theorem). Every representation of a finite is a direct sum of irreducible representations.
Proof. It suffices to prove that for any G-invariant subspace W ⊂ V . There is a G-invariant complement of W . By a complement of W , we mean a subspace W0 such that W ∩ W0 = {e}, and W + W0 = V .
We first pick any complement W0. Then V = W ⊕ W0. Let p : V → W be the projection. We are going to modify W0 to get a G-invariant complement.
To this end, we average p over G to get p0 := 1
g X
t∈G
ρtpρ−1t , where g = |G|.
One checks that p0 : V → W and p0(w) = w for all w ∈ W . That is, p0 : V → W is a projection.
Let W0 := ker(p0). We check that W0 is G-invariant since ρsp0ρ−1s = p0
3
for all s ∈ G. It follows that if x ∈ W0, p0ρs(x) = ρs(p0(x)) = 0, which shows that ρs(x) ∈ W0.
This proves that the representation on V is isomorphic to W ⊕ W0.
¤ Remark 0.11. A matrix over C of finite order is diagonalizable. Hence every matrix representation over the field C is diagonalizable. We therefore assume the field to be the complex number field.
Moreover, let λ be an eigenvalue of ρs for some s. Then |λ| = 1.
Definition 0.12. Let ρ : G → GL(V ) be a linear representation on the vector space V . We define the character as χρ := T r ◦ ρ : G → C.
Proposition 0.13. Let χ be the character of ρ : G → GL(V ).
(1) χ(1) = n := dimV ,
(2) χ(s−1) = χ(s) for all s ∈ G, (3) χ(tst−1) = χ(s) for all s, t ∈ G.
(4) if χ0 is the character of another representation ρ0, then the char- acter of ρ ⊕ ρ0 is χ + χ0.
One can define a hermitian dot product on characters as
< χ, χ0 >:= 1 g
X
s∈G
χ(s)χ0(s).
The main theorem for character is the following:
Theorem 0.14. Let G be a group of order g, and let ρ1, ... represent the isomorphism classes of irreducible representations of G. Let χi be the character of ρi for each i.
(1) Orthogonality Relations:
<< χi, χj >= 0if i 6= j,
< χi, χi >= 1 for each i.
(2) The number of isomorphism classes of irreducible representa- tions of G is the same as the number of conjugacy classes of G.
(denote it by r).
(3) Let di be the degree of ρi, then di|g and g =
Xr i=1
d2i.
Example 0.15. Consider G = D4. It’s clear that r = 5. Hence it’s only possible to have d1 = 2, d2 = ... = d5 = 1.
Advanced Algebra I
representation of finite groups,II Characters
Let ρ be a 1-dimensional representation of a group G. Then in this case ρ = χ : G → C∗. One sees that χ(st) = χ(s)χ(t) for all s, t ∈ G.
Such character is called an abelian character.
Let ˆG be the set of all 1-dimensional characters, it forms a group under the multiplication χχ0(g) := χ(g)χ0(g).
Exercise 0.1. Let G be an abelian group. Prove that G ∼= ˆG
Recall that a representation ρ : G → GL(V ) is the same as a linear action G × V → V . Suppose now that there are two representation ρ, ρ0 on V, V0 respectively. A linear transformation T : V → V0 is said to be G-invariant if it’s compatible with representations. That is,
T ρs(v) = ρ0s(T v), for all v ∈ V .
Thus an isomorphism of representation is nothing but a G-invariant bijective linear transformation.
Exercise 0.2. It’s easy to check that if T : V → V0 is G-invariant, then the ker(T ) ⊂ V and im(T ) ⊂ V0 are G-invariant subspaces.
Theorem 0.3 (Schur’s Lemma). Let ρ, ρ0 be two irreducible represen- tation of G on V, V0 respectively. And let T : V → V0 be a G-invariant linear transformation. Then
(1) Either T is an isomorphism or T = 0.
(2) If V = V0,ρ = ρ0, then T is multiplication by a scalar.
Proof. (1) Since ker(T ) is a G-invariant subspace and V is irre- ducible. One has that either ker(T ) = 0 or ker(T ) = V . Hence T is injective or T = 0. If T is injective, by looking at im(T ), One must have im(T ) = V0. Therefore T is an isomorphism.
(2) Let λ be an eigenvalue of T . One sees that T1 := T − λI is also an G-invariant linear transformation. Since ker(T1) is non-zero, one has that ker(T1) = V . Thus T1 = 0 and hence T = λI.
¤ Suppose one has T : V → V0 not necessarily G-invariant. One can produce an G-invariant linear transformation by the ”averaging process”. For T (v) = s−1T (sv), we set
T (v) :=˜ 1 g
X
s∈G
s−1T (sv).
And it’s easy to check that this is G-invariant.
1
2
proof of the main theorem. (1) Let ρ, ρ0 be two irreducible repre- sentation of G on V, V0 with character χ, χ0 respectively.
Let T : V → V0 be any linear transformation. One can produce a G-invariant transformation ˜T .
Suppose first that ρ and ρ0 are not isomorphic. Then by Schur’s Lemma, ˜T = 0 for all T .
We fix bases of V, V0 and write everything in terms of matri- ces.
0 = ( ˜T )ij =X
t,k,l
(R0t−1)ik(T )kl(Rt)lj. Take T = Eij, then one has
0 =X
t,k,l
(R0t−1)ik(Eij)kl(Rt)lj =X
t
(R0t−1)ii(Rt)jj. Hence
< χ0, χ >=X
t,i,j
(R0t−1)ii(Rt)jj = 0.
Suppose now that ρ = ρ0, χ = χ0. The averaging process and Schur’s Lemma gives
λI = ˜T = 1 g
X
t
Rt−1T Rt. One notice that λd = tr( ˜T ) = tr(T ).
Now we set T = Eii, then 1
d = (λI)ii = 1 g
X
t
(Rt−1)ik(Eii)kl(Rt)li =X
t
(Rt−1)ii(Rt)ii. It follows that
< χ, χ >=X
t
X
i
(Rt−1)ii(Rt)ii =X
i
1 d = 1.
(2) A class function f on a group G is a complex value function such that f (s) = f (t) if s and t are conjugate. The space C of class function is clearly a vector space of dimension r, where r denotes the number of conjugacy classes of G. We claim that the set of character of irreducible representation form a orthonormal basis of C.
We remark that inner product can be defined on any class function.
Suppose now that φ is a class function which is orthogonal to every χi. For any character χ of an irreducible representation ρ, we can produce a linear transformation by averaging process T := 1gP
tφ(t)ρt. It’s clear that tr(T ) =< φ, χ >= 0. One sees that T : V → V is G-invariant. By Schur’s Lemma, T = λI.
But T r(T ) = 0. Thus T = 0 for any character χ.
3
We apply to the regular representation ρ : G → C[G], 0 = T (e1) = 1
g X
t
φ(t)ρt(e1) = 1 g
X
t
φ(t)et.
Since et forms a basis for C[G], it follows that φ(t) = 0 for all t ∈ G and hence φ = 0.
(3) We may assume that there are r irreducible representation. And suppose that the regular representation ρ is decomposed into n1ρ1⊕ ... ⊕ nrρr. One notice that ρ(1) = g and ρ(t) = 0 for all t 6= 1. By direct computation,
di =< χρ, χi >= ni, g =< χρ, χρ>=X
i
d2i.
To prove that di|g need some extra work on the group algebra C[G] which we will do later.
¤
Advanced Algebra I
Gruop algebra
Recall that by a regular representation of G, we consider a vector space with basis {es}s∈G. Let C[G] be the vector space with basis {es}s∈G. One can have a natural ring structure on C[G] as following:
X
s∈G
ases+X
s∈G
bses =X
s∈G
(as+ bs)es,
(X
s∈G
ases)(X
t∈G
btet) =X
s,t
asbtest =X
u∈G
(X
st=u
asbt)eu. We call C[G] the group algebra of G.
We claim that
C[G] ∼= Yr
i=1
Mdi(C).
Where r is the number of conjugacy classes of G and ni is the degree of each irreducible representation.
First of all, the irreducible representation ρi : G → GL(Wi) in- duces an algebra homomorphism ˜ρi : C[G] → End(Wi) ∼= Mni(C) by
˜ ρi(P
s∈Gases) =P
sasρi(g). Hence one has
˜
ρ : C[G] → Yr i=1
End(Ei) ∼= Yr i=1
Mdi(C).
We first claim the ˜ρ is surjective. Suppose not, then there is a lin- ear relation on the images. It follows that there is a relation on the coefficients of ρi. In particular, there is alinear erlation on χi. By the orthogonal property, this is impossible. Hence ˜ρ is surjective. However, they have the same dimension. Hence ˜ρ is an isomorphism.
Remark 0.1. C[G] is abelian if and only if G is abelian.
Our next goal it to determine the center Z(C[G]). In order to check x =P
ases is in center or not, we need to check for all t ∈ G, x =X
s∈G
ases= e−1t xet=X
s∈G
aset−1st =X
s∈G
atst−1es.
Note that t−1st is conjugate to s. Thus, it’s equivalent to have as = as0 for all s0 conjugate to s.
A special case is that the above equation holds for ec := P
σ∈ceσ, where c is a conjugacy class. Moreover, by our computation above, it’s indeed that
Z(C[G]) = { Xr
i=1
aieci|ai ∈ C, ci runs through all conjugacy classes}.
1
2
Example 0.2. Let G = S3. Then the center has a basis e(1), e(12)+ e(13)+ e(23), e(123)+ e(132)
By viewing the isomorphism ˜ρ, one sees that if u = P
ases ∈ Z(C[G]), then ˜ρi(u) is of the form λiI on the irreducible represen- tation Vi. The value λ can be computed. Note that the coefficient as actually gives a class function on G because as = as0 for s, s0 in the same conjugate class. We write it as a : G → C. By averaging process, one has a G-invariant ˜rhoi =P
s∈Gasρi(s) linear transformation on Vi. Thus one has
λi = 1 di
tr(X
s∈G
asρi(s)) = 1 di
Xasχi(s).
Theorem 0.3. Keep notation as before, then one has di|g.
To prove this result, we need some facts on integral extension and algebraic integers.
Remark 0.4. (1) Let R be a commutative ring, one can view it as a Z-module. An element x ∈ R is said to be integral over Z if x satisfies a monic integral polynomial in Z[x].
(2) In a commutative ring R, the elements which are integral over Z forms a subring of R.
(3) If R = C, then subring of elements which are integral over Z is called ring of algebraic integers, denoted A.
(4) A ∩ Q = Z.
Remark 0.5. The character χ(s) is an algebraic integer for all χ and all s ∈ G. This is because χ is sum of eigenvalues of a representation ρ.
However, the eigenvalues are root of unity which are algebraic integers.
Proposition 0.6. Let u = P
ases∈ Z(C[G]) such that as ∈ A. Then u is integral over Z.
Proof. Since Z(C[G]) is generated by ec. Let c1, ..., cr be all the con- jugacy classes and let ai := asi for some si ∈ ci. We first consider R := ⊕ri=1Zeci. It’s clear that R is a subring of Z(C[G]) which is finitely generated over Z. Now let M = Z[a1, ..., ar]. Since ai is inte- gral over Z, it follows that M is a finite Z-module. One checks that Z[u] ⊂ ⊕ri=1Meci and ⊕ri=1Meci is a finite Z-module. Hence u is inte-
gral over Z. ¤
proof of the theorem. For each i, take u = P
s∈Gχ(s−1)es. It’s clear that u ∈ Z(C[G]). By the previous Proposition, one has that u is integral over Z.
Note that one has natural ring homomorphism ωi : Z(C[G]) → C,
3
by sending u to λi the multiple of its i-component. It follows that the homomorphic image λi is an algebraic integer. One has now
λi := 1 di
Xχi(s−1)χi(s) = g
di < χi, χi >= g di.
Hence λi ∈ A ∩ Q = Z. It follows that di|g. ¤
