### 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 {I*_{j}*}*_{j∈J}*, one has a upper*
*bound I = ∪I*_{j}*. 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.

1

2

*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*

*A*_{1} *:= {a ∈ A|ahas parentless ancestor in A},*
*A*_{2} *:= {a ∈ A|ahas parentless ancestor in B},*

*A*3 *:= {a ∈ A|ahas infinite ancestor}.*

*And so does B.*

*Then we claim that f restricted to A*_{1}*, A*_{3} *are bijections to B*_{1}*, B*_{3}.
*And g restricted to B*_{2}*, B*_{3} *are bijections to A*_{2}*, A*_{3}. 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| = ℵ*_{0}.

*Proposition 1.9. If A is infinite, then ℵ*_{0} *≤ |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 A*_{0} *in A, one sees that |A*_{0}*| =*

*|A*_{0}*| + |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*
*p*^{n}*m (where (n, m) = 1). Then there are subgroups of order p*^{i}*for all*
*0 ≤ i ≤ n.*

*Furthermore, for each subgroup H*_{i}*of order p*^{i}*, there is a subgroup*
*H*_{i+1}*of order p*^{i+1}*such that H*_{i}*C H*_{i+1}*for 0 ≤ i ≤ n − 1.*

*In particular, there exist a subgroup of order p** ^{n}*, 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| ≡ |S*_{0}*| (mod p).*

*proof of the theorem. We will find subgroup of order p** ^{i}* inductively. By

*Cauchy’s theorem, there is a subgroup of order p. Suppose that H*

*is a subgroup of order p*

^{i}*. Consider the group action that H acts on*

*S = G/H by translation. One show that xH ∈ S*

_{0}if and only if

*x ∈ N*

_{G}*(H). Thus |S*

_{0}

*| = |N*

_{G}*(H)/H|. If i < n, then*

*|S*_{0}*| ∼= |S| ∼= 0 (mod p).*

*By Cauchy’s theorem, the group N*_{G}*(H)/H contains a subgroup of or-*
*der p. The subgroup is of the form H*_{1}*/H, hence |H*_{1}*| = p** ^{i+1}*. Moreover,

*H C H*

_{1}.

¤
*Example 0.3. If G is a finite p-group of order p*^{n}*, then one has a*
*series of subgroups {e} = H*_{0} *< H*_{1} *< ... < H*_{n}*= G such that |H*_{i}*| = p*^{i}*and H*_{i}*C H*_{i+1}*, H*_{i+1}*/H*_{i}*∼*= Z*p**. 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 p*^{n}*m then a subgroup P is a Sylow p-subgroup*
*if and only if |P | = p** ^{n}* by the proof of the first theorem.

*Theorem 0.5 (Second Sylow theorem). Let G be a finite group of order*
*p*^{n}*m. 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 .*

1

2

*Proof. Let S = G/P and H acts on S by translation. Thus by the*
*Lemma, one has |S*_{0}*| ≡ |S| = m(mod p). Therefore, S*_{0} *6= ∅. One has*

*xP ∈ S*_{0} *⇔ hxP = xP* *∀h ∈ H ⇔ x*^{−1}*hx < 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*
*p*^{n}*m. 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 ∈ S*0 *⇔ xQx*^{−1}*= Q ∀x ∈ P ⇔ P < N**G**(Q).*

*Both P, Q are Sylow p-subgroup of N*_{G}*(Q) and therefore conjugate in*
*N*_{G}*(Q). However, Q C N*_{G}*(Q), Q has no conjugate other than itself.*

*Thus one concludes that P = Q. In particular, S*_{0} *= {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 r*_{p}*:= number of Sylow p-subgroups. Then*
*r*_{5} *= 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,*
*r*_{p}*= 1, hence the subgroup < a > is normal. Then one notices that*
*G =< a >< b > for < a > ∩ < b >= {e}. Moreover, bab*^{−1}*= a*^{k}*for*
*some k. One has*

*a = b*^{2}*ab*^{−2}*= a*^{k}^{2}*.*

*It follows that k = 1, −1. If k = 1, then G is abelian. Thus we assume*
*that k = −1. This gives the group D*_{p}*:=< a, b|a*^{p}*= b*^{2} *= 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.*

3

*(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*^{−1}*y.*

*Because x = ex = (g*^{−1}*g)x = g*^{−1}*(gx) = g*^{−1}*y.*

*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) ∼= S**n**.*

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*

*O**x**:= {gx|g ∈ G}.*

*And the stabilizer of x is defined as*

*G**x**:= {g ∈ G|gx = x}.*

Then one has the following Proposition 0.4.

*|G| = |O**x**| · |G**x**|.*

*Sketch. Consider S**y* *:= {g ∈ G|gx = y}. Then G is a disjoint union of*
*S*_{y}*for all y ∈ O*_{x}*. Furthermore, fix a g such that y = gx, then one has*
*S*_{y}*= gG** _{x}*. Thus

*|G| = | ∪**y∈O**x* *S**y**| =* X

*y∈O**x*

*|G**x**| = |O**x**| · |G**x**|.*

¤

1

2

*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 O*_{x}*= O*_{y}*if and only if y ∈ O** _{x}*.

*Thus for convenience, we pick a representative in each orbit and let I*be a set of representatives of orbits. We have:

*S = ∪*_{x∈I}*O*_{x}*.*
In particular,

*|S| =*X

*x∈I*

*|O*_{x}*|.*

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**|O*_{x}*| 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, G**x* *= C(x), the centralizer. And O**x* *= {x}*

*if and only if x ∈ Z(G), the center of G. So, for G finite, the equation*

*|S| =* P

*x∈I**|O*_{x}*| 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**|O*_{x}*|) is*
*very useful if the group is a finite p-group. By a finite p-group, we*
*mean a group G with |G| = p*^{n}*for some n > 0. Consider now G is a*
*finite p group acting on S. Let*

*S*_{0} *:= {x ∈ S|gx = x, ∀g ∈ G}.*

Then the class equation can be written as

*|S| = |S*_{0}*| +* X

*x∈I,x6∈S*0

*|O*_{x}*|.*

One has the following

*Lemma 0.7. Let G be a finite p-group. Keep the notation as above,*
*then*

*|S| ≡ |S*0*| (mod p).*

3

*Proof. If x 6∈ S*0*, then 1 6= |O**x**| = p** ^{k}*.

¤
*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 := {(a*1*, ..., a**p**)|a**i* *∈ G,*Y

*a**i* *= e}.*

*And consider a group action C*_{p}*×S → S by (1, (a*_{1}*, .., a*_{p}*)) 7→ (a*_{p}*, a*_{1}*, ..., a** _{p−1}*).

*One claims that S*_{0} *= {(a, a, ..., a)|a ∈ G}.*

*By the Lemma, one has |S| ≡ |S*_{0}*| (mod . It follows that p | |S*_{0}*|.*

*In particular, |S*_{0}*| > 1, hence there is (a, ..., a) ∈ S*_{0} *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 C*2*(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 C*_{2}*(G) is a nor-*
*mal subgroup of G. And then let C*_{3}*(G) to be the inverse image of*
*Z(G/C*_{2}*(G)). By doing this inductively, one has an ascending chain of*
normal subgroups

*{e} < C*_{1}*(G) := Z(G) < C*_{2}*(G) < ...*

*Definition 0.1. G is nilpotent if C*_{n}*(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 C**i* * C**i+1**. The group G has finite order thus the ascending*
*chain must terminates, say at C**n**. If C**n* *6= G, then G/C**n*has non-trivial
*center. One has C*_{n}* C*_{n+1}*which is impossible. Hence C*_{n}*= 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 C*_{i}*(H × K) = C*_{i}*(H) × C*_{i}*(K). If C*_{n}*(H) =*
*H, C*_{m}*(K) = K then C*_{l}*(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 N*_{G}*(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 N**G**(H).*

*By applying this Claim to N**G**(H), then it says that N**G**(H) can’t be*
*a proper subgroup of G since N*_{G}*(N*_{G}*(H)) = N*_{G}*(H). Thus it follows*

*that N*_{G}*(H) = G.* ¤

We have seen that we have a series of subgroup by taking centers.

Another similar construction is to take commutators.

1

2

*Definition 0.5. Let G be a group. The commutator of G, denoted G*^{0}*is the subgroup generated by the subset {aba*^{−1}*b*^{−1}*}.*

*Roughly speaking, the subgroup G** ^{0}* measures measure the commu-

*tativity of a group. The smaller G*

*, the more commutative it is.*

^{0}*Theorem 0.6. G*^{0}*C G, and G/G*^{0}*is ableian. Moreover, if N C G, then*
*G/N is abelian if and only if G*^{0}*< N .*

*Proof.* *(1) for all g ∈ G, g(aba*^{−1}*b*^{−1}*g*^{−1}*∈ G*^{0}*, hence gG*^{0}*g < G** ^{0}*. So

*G*

^{0}*C G.*

(2)

*aG*^{0}*bG*^{0}*= abG*^{0}*= ab(b*^{−1}*a*^{−1}*ba)G*^{0}*= baG*^{0}*= bG*^{0}*aG*^{0}*.*

*(3) Consider π : G → G/N. If G/N is abelian, then π(aba*^{−1}*b** ^{−1}* =

*e, hence G*

^{0}*< N. Conversely, if G*

^{0}*< N, we have a surjective*

*homomorphism G/G*

^{0}*→ G/N. G/G*

*is abelian, hence so is it*

^{0}*homomorphic image G/N.*

¤
*Definition 0.7. We can define the the commutator inductively, i.e.*

*G*^{(2)} *:= (G** ^{0}*)

^{0}*, etc. The G*

^{(}

*i) is called the i-th derived subgroup of G.*

*It’s clear that G > G*^{0}*> G*^{(2)} *> ....*

*A group is solvable is G*^{(}*n) = {e} for some n.*

*Example 0.8. Take G = S*4*. The commutator is the smallest subgroup*
*that G/G*^{0}*is abelian. Since the only non-trivial normal subgroups of*
*S*4 *are V, A*4*. It’s clear that G*^{0}*= A*4 *(Or one can prove this by hand).*

*Similarly, one finds that G*^{(2)} *= A*^{0}_{4} *= V , and G*^{(3)} *= {e}. Hence S*_{4} *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 = H*_{0} *> H*_{1} *> H*_{2} *> ... > H*_{n}*,*
*such that H*_{i}*C H*_{i−1}*for all 1 ≤ i ≤ n.*

*A subnormal series is a solvable series if H*_{n}*= {e} and H*_{i−1}*/H*_{i}*is*
*abelian for all 1 ≤ i ≤ n.*

*A subnormal series is a normal series if all H*_{i}*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 > G*^{0}*> ...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} = H**n* *< ... < H*0 *= G. First*
*observe that G*^{0}*< H*_{1} *since G/H*_{1} *is abelian. We claim that G*^{(}*i) < H*_{i}

3

*for all i inductively. Which can be prove by the observation that the*
*intersection of the series {e} = H**n* *< ... < H*0 *= 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. A*_{5} *is not solvable, hence so is S*_{n}*for n ≥ 5.*

*0.1. simplicity of A*_{5}*. An element in S** _{n}*is said to be have cycle struc-

*ture (m*

_{1}

*, .., m*

_{r}*) with m*

_{1}

*≥ m*

_{2}

*≥ ... ≥ m*

_{r}*, m*

_{1}

*+ ... + m*

_{r}*= n if its*

*cycle decomposition is of length m*

_{1}

*, ..., m*

*respectively. For example,*

_{r}*(1, 2)(3, 4) ∈ S*

_{4}

*has cycle structure (2, 2) and (1, 2) ∈ S*

_{4}has cycle

*structure (2, 1, 1).*

*Remark 0.14. There is a one-to-one correspondence between cycle*
*structures of S*_{n}*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 (m*1*, ..., m**r*) the cycle class of
*(m*1*, ...m**r**). A consequence of this fact is that a subgroup N < S**n* 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 = S** _{n}*, then orbits are cycle classes.

*Example 0.15. In S*_{4}*, V is the union of class (1, 1, 1, 1) and (2, 2).*

*A*_{4} *is the union of V and the class (3, 1).*

*The purpose of this subsection is to show that A*_{5} is a simple non-
abelian group, hence a non-solvable group.

*Theorem 0.16. A*_{5} *is a simple non-abelian group.*

*Proof. One note that in S*_{5}*, 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 A*5 *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 : S*_{5} *× A*_{5} *→ A*_{5} and its
*restriction β : A*_{5}*× A*_{5} *→ A*_{5}*. For σ ∈ A*_{5}*, let O** _{α,σ}* be the orbit of the

*α and O*

_{β,σ}*be the orbit of the β. And let G*

_{α,σ}*, G*

*be the stabilizer.*

_{β,σ}*It’s clear that G*_{α,σ}*= C*_{S}_{5}*(σ) and G*_{β,σ}*= C*_{A}_{5}*(σ) = C*_{S}_{5}*(σ) ∩ A*_{5}.
*Thus we have either |G*_{β,σ}*| =* ^{1}_{2}*|G*_{α,σ}*| or |G*_{β,σ}*| = |G*_{α,σ}*|. Hence |O*_{β,σ}*| =*

*|O*_{α,σ}*| or |O*_{β,σ}*| =* ^{1}_{2}*|O*_{α,σ}*|.*

4

*case 1. If σ has cycle structure (5), then |O**α,σ**| = 24, |G**α,σ**| = 5. It*
*follows that |G**β,σ**| = 5 and hence |O**β,σ**| = 12.*

*case 2. If σ has cycle structure (3, 1, 1), then |O*_{α,σ}*| = 20, |G*_{α,σ}*| = 6.*

*However, one notice that there is an element τ ∈ C*_{S}_{5}*(σ) − C*_{A}_{5}*(σ) (e.g.*

*(45)(123) = (123)(45)). Hence |G*_{β,σ}*| 6= |G*_{α,σ}*| and must be* ^{1}_{2}*|G*_{α,σ}*| = 3.*

*Therefore |O*_{β,σ}*| = 20.*

*case 3. If σ has cycle structure (2, 2, 1),then |O*_{α,σ}*| = 15, |G*_{α,σ}*| = 8.*

*It follows that |O*_{β,σ}*| = 15.*

*Combining all this, if H < A*_{5} *is a normal subgroup, then |H| =*
*1 + 12r*_{1}*+ 20r*_{2}*+ 15r*_{3}*, where r*_{i}*are integers. Moreover |H| | |A*_{5}*| = 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} = H**n* *< ... < H*0 *= G,*
*the factors of the series are the quotient groups H*_{i−1}*/H*_{i}*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} = H*_{n}*< . < K <*

*.. < H*_{0} *= 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. ¤

5

*proof of Schreier’s theorem. Let {e} = G**n+1* *< ... < G*0 *= G and*
*{e} = H**m+1* *< ... < H*0 *= G be two subnormal series. Let G(i, j) =:=*

*G*_{i+1}*(G*_{i}*∩ H*_{j}*) (resp. H(i, j) := H*_{j+1}*(G*_{i}*∩ H** _{j}*)). 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 G*_{i+1}*, G*_{i}*, H*_{j+1}*, H** _{j}*, 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 x*1*x*2*... 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*

*X*0 *:= (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 ∈ X*0

*− {1} by*

*a*

*.*

^{−1}*A word on X is a sequence (a*_{1}*, a*_{2}*, ...) such that a*_{i}*∈ X*_{0} and there
*is n*_{0} *such that a*_{i}*= 1 for all i > n*_{0}*. (We call it a word of length n*_{0}.)

*Given two words w*_{1} *:= (a*_{1}*, a*_{2}*, ..., a*_{n}*, 1, ...), w*_{2} *:= (b*_{1}*, b*_{2}*, ..., b*_{m}*, 1...)*
*of length n, m respectively, one can compose them to obtain a word*

*w*_{1} *◦ w*_{2} *:= (a*_{1}*, ..., a*_{n}*, b*_{1}*, ..., b*_{m}*, 1, ...).*

*A word w (of length n) is said to be reduced if*
*(1) a*_{i}*6= 1 for i ≤ n.*

*(2) a*_{i+1}*6= a*^{−1}_{i}*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 (a*1*, .., a**n**, 1, ...), we denote it by*
*a*_{1}*a*_{2}*...a** _{n}*.

*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}*...a*_{n}*) := f (a*_{1}*)...f (a** _{n}*). One checks that this is a group homomor-
phism and also ¯

*f ◦ ı = f .*

1

2

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*
*i*_{G}*: 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*^{−1}*y*^{−1}*}, then < X|R >∼*=
Z^{2}*.*

*Example 0.4. Let X = {x, y} and R = {x*^{n}*, y*^{2}*, xyxy}, then < X|R >∼*=
*D*_{n}*.*

*Remark 0.5. If the generators and relations are given explicitly as*
*in the previous examples, we usually write it as < x, y|x*^{n}*= y*^{2} =
*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 ±e** _{i}* 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 R*_{g}*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 ρ**g* *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 R*^{0}*= P RP*^{−1}*, that is R*^{0}_{g}*= P R*_{g}*P*^{−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 + v*^{0}*) = gv + gv*^{0}*,* *g(cv) = cg(v)*
*for all g ∈ G, v, v*^{0}*∈ V and c ∈ F .*

*Definition 0.5. Let ρ, ρ*^{0}*be two representations of G on V, V*^{0}*. They*
*are said to be isomorphic if there is an isomorphism of τ : V → V*^{0}*which is compatible with ρ and ρ*^{0}*. That is,*

*τ ρ**s**(v) = ρ*^{0}_{s}*τ (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, R*_{g}*is a root of unity.*

*In particular, |R*_{g}*| = 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 {e*_{t}*}*_{t∈G}*. For s ∈ G, let*
*R*_{s}*be the linear map of V to V which maps e*_{t}*to e*_{st}*. This is called the*
*regular representation of G.*

*Note that e*_{s}*= R*_{s}*(e*_{1}*) for all s ∈ G. Hence the image of e*_{1} *∈ 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 {τ*_{s}*(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 (e**s**) = ρ**s**(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, ρ*^{0}*be two representations of G on V, V*^{0}*, then one can define ρ⊕*

*ρ*^{0}*, ρ ⊗ ρ*^{0}*naturally. Note that if degree of ρ and ρ*^{0}*are d, d** ^{0}* respectively,

*then degree of ρ ⊕ ρ*

^{0}*is d + d*

^{0}*and degree of ρ ⊗ ρ*

^{0}*is dd*

*.*

^{0}*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 ρ**s**(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 W*^{0}*such that W ∩ W*^{0}*= {e}, and W + W*^{0}*= V .*

*We first pick any complement W*^{0}*. Then V = W ⊕ W*^{0}*. Let p : V →*
*W be the projection. We are going to modify W*^{0}*to get a G-invariant*
complement.

*To this end, we average p over G to get*
*p*_{0} := 1

*g*
X

*t∈G*

*ρ*_{t}*pρ*^{−1}_{t}*,*
*where g = |G|.*

*One checks that p*_{0} *: V → W and p*_{0}*(w) = w for all w ∈ W . That*
*is, p*_{0} *: V → W is a projection.*

*Let W*_{0} *:= ker(p*_{0}*). We check that W*_{0} *is G-invariant since*
*ρ*_{s}*p*_{0}*ρ*^{−1}_{s}*= p*_{0}

3

*for all s ∈ G. It follows that if x ∈ W*_{0}*, p*_{0}*ρ*_{s}*(x) = ρ*_{s}*(p*_{0}*(x)) = 0, which*
*shows that ρ*_{s}*(x) ∈ W*_{0}.

*This proves that the representation on V is isomorphic to W ⊕ W*_{0}.

¤
*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 ρ*_{s}*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 χ**ρ* *:= 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 χ*^{0}*is the character of another representation ρ*^{0}*, then the char-*
*acter of ρ ⊕ ρ*^{0}*is χ + χ*^{0}*.*

One can define a hermitian dot product on characters as

*< χ, χ*^{0}*>:=* 1
*g*

X

*s∈G*

*χ(s)χ*^{0}*(s).*

The main theorem for character is the following:

*Theorem 0.14. Let G be a group of order g, and let ρ*_{1}*, ... represent*
*the isomorphism classes of irreducible representations of G. Let χ*_{i}*be*
*the character of ρ**i* *for each i.*

*(1) Orthogonality Relations:*

*<< χ**i**, χ**j* *>= 0if i 6= j,*

*< χ*_{i}*, χ*_{i}*>= 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 d*_{i}*be the degree of ρ*_{i}*, then d*_{i}*|g and*
*g =*

X*r*
*i=1*

*d*^{2}_{i}*.*

*Example 0.15. Consider G = D*_{4}*. It’s clear that r = 5. Hence it’s*
*only possible to have d*1 *= 2, d*2 *= ... = d*5 *= 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 ˆ*G be the set of all 1-dimensional characters, it forms a group*
*under the multiplication χχ*^{0}*(g) := χ(g)χ*^{0}*(g).*

*Exercise 0.1. Let G be an abelian group. Prove that G ∼*= ˆ*G*

*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, ρ*^{0}*on V, V*^{0}*respectively. A linear transformation T : V → V** ^{0}* is said

*to be G-invariant if it’s compatible with representations. That is,*

*T ρ**s**(v) = ρ*^{0}_{s}*(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 → V*^{0}*is G-invariant,*
*then the ker(T ) ⊂ V and im(T ) ⊂ V*^{0}*are G-invariant subspaces.*

*Theorem 0.3 (Schur’s Lemma). Let ρ, ρ*^{0}*be two irreducible represen-*
*tation of G on V, V*^{0}*respectively. And let T : V → V*^{0}*be a G-invariant*
*linear transformation. Then*

*(1) Either T is an isomorphism or T = 0.*

*(2) If V = V*^{0}*,ρ = ρ*^{0}*, 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 ) = V*^{0}*. Therefore T is an isomorphism.*

*(2) Let λ be an eigenvalue of T . One sees that T*_{1} *:= T − λI is also*
*an G-invariant linear transformation. Since ker(T*_{1}) is non-zero,
*one has that ker(T*_{1}*) = V . Thus T*_{1} *= 0 and hence T = λI.*

¤
*Suppose one has T : V → V*^{0}*not necessarily G-invariant. One*
*can produce an G-invariant linear transformation by the ”averaging*
*process”. For T (v) = s*^{−1}*T (sv), we set*

*T (v) :=*˜ 1
*g*

X

*s∈G*

*s*^{−1}*T (sv).*

*And it’s easy to check that this is G-invariant.*

1

2

*proof of the main theorem.* *(1) Let ρ, ρ** ^{0}* be two irreducible repre-

*sentation of G on V, V*

^{0}*with character χ, χ*

*respectively.*

^{0}*Let T : V → V** ^{0}* be any linear transformation. One can

*produce a G-invariant transformation ˜T .*

*Suppose first that ρ and ρ** ^{0}* are not isomorphic. Then by
Schur’s Lemma, ˜

*T = 0 for all T .*

*We fix bases of V, V** ^{0}* and write everything in terms of matri-
ces.

0 = ( ˜*T )** _{ij}* =X

*t,k,l*

*(R*^{0}_{t}*−1*)_{ik}*(T )*_{kl}*(R** _{t}*)

_{lj}*.*

*Take T = E*

*ij*, then one has

0 =X

*t,k,l*

*(R*^{0}_{t}*−1*)_{ik}*(E** _{ij}*)

_{kl}*(R*

*)*

_{t}*=X*

_{lj}*t*

*(R*^{0}_{t}*−1*)_{ii}*(R** _{t}*)

_{jj}*.*Hence

*< χ*^{0}*, χ >=*X

*t,i,j*

*(R*^{0}_{t}*−1*)_{ii}*(R** _{t}*)

_{jj}*= 0.*

*Suppose now that ρ = ρ*^{0}*, χ = χ** ^{0}*. The averaging process and
Schur’s Lemma gives

*λI = ˜T =* 1
*g*

X

*t*

*R*_{t}^{−1}*T R*_{t}*.*
*One notice that λd = tr( ˜T ) = tr(T ).*

*Now we set T = E** _{ii}*, then
1

*d* *= (λI)**ii* = 1
*g*

X

*t*

*(R*_{t}* ^{−1}*)

*ik*

*(E*

*ii*)

*kl*

*(R*

*t*)

*li*=X

*t*

*(R*_{t}* ^{−1}*)

*ii*

*(R*

*t*)

*ii*

*.*It follows that

*< χ, χ >=*X

*t*

X

*i*

*(R*_{t}* ^{−1}*)

_{ii}*(R*

*)*

_{t}*=X*

_{ii}*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 χ**i**. For any character χ of an irreducible representation*
*ρ, we can produce a linear transformation by averaging process*
*T :=* ^{1}* _{g}*P

*t**φ(t)ρ*_{t}*. 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 (e*_{1}) = 1

*g*
X

*t*

*φ(t)ρ*_{t}*(e*_{1}) = 1
*g*

X

*t*

*φ(t)e*_{t}*.*

*Since e*_{t}*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*
*n*1*ρ*1*⊕ ... ⊕ n**r**ρ**r**. One notice that ρ(1) = g and ρ(t) = 0 for all*
*t 6= 1. By direct computation,*

*d*_{i}*=< χ*_{ρ}*, χ*_{i}*>= n*_{i}*,*
*g =< χ**ρ**, χ**ρ**>=*X

*i*

*d*^{2}_{i}*.*

*To prove that d**i**|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 {e**s**}**s∈G**. Let C[G] be the vector space with basis*
*{e*_{s}*}*_{s∈G}*. One can have a natural ring structure on C[G] as following:*

X

*s∈G*

*a*_{s}*e** _{s}*+X

*s∈G*

*b*_{s}*e** _{s}* =X

*s∈G*

*(a*_{s}*+ b*_{s}*)e*_{s}*,*

(X

*s∈G*

*a*_{s}*e** _{s}*)(X

*t∈G*

*b*_{t}*e** _{t}*) =X

*s,t*

*a*_{s}*b*_{t}*e** _{st}* =X

*u∈G*

(X

*st=u*

*a*_{s}*b*_{t}*)e*_{u}*.*
*We call C[G] the group algebra of G.*

We claim that

*C[G] ∼*=
Y*r*

*i=1*

*M*_{d}_{i}*(C).*

*Where r is the number of conjugacy classes of G and n** _{i}* is the degree
of each irreducible representation.

*First of all, the irreducible representation ρ**i* *: G → GL(W**i*) in-
duces an algebra homomorphism ˜*ρ**i* *: C[G] → End(W**i**) ∼= M**n**i*(C) by

˜
*ρ** _{i}*(P

*s∈G**a*_{s}*e** _{s}*) =P

*s**a*_{s}*ρ*_{i}*(g). Hence one has*

˜

*ρ : C[G] →*
Y*r*
*i=1*

*End(E*_{i}*) ∼*=
Y*r*
*i=1*

*M*_{d}_{i}*(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 ρ*_{i}*. In particular, there is alinear erlation on χ** _{i}*. 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

*a**s**e**s* *is in center or not, we need to check for all t ∈ G,*
*x =*X

*s∈G*

*a*_{s}*e*_{s}*= e*^{−1}_{t}*xe** _{t}*=X

*s∈G*

*a*_{s}*e*_{t}^{−1}* _{st}* =X

*s∈G*

*a*_{tst}^{−1}*e*_{s}*.*

*Note that t*^{−1}*st is conjugate to s. Thus, it’s equivalent to have a*_{s}*= a*_{s}^{0}*for all s*^{0}*conjugate to s.*

*A special case is that the above equation holds for e** _{c}* := P

*σ∈c**e** _{σ}*,

*where c is a conjugacy class. Moreover, by our computation above, it’s*indeed that

*Z(C[G]) = {*
X*r*

*i=1*

*a**i**e**c**i**|a**i* *∈ C, c**i* *runs through all conjugacy classes}.*

1

2

*Example 0.2. Let G = S*3*. 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

*a**s**e**s* *∈*
*Z(C[G]), then ˜ρ*_{i}*(u) is of the form λ*_{i}*I on the irreducible represen-*
*tation V*_{i}*. The value λ can be computed. Note that the coefficient a*_{s}*actually gives a class function on G because a*_{s}*= a*_{s}^{0}*for s, s** ^{0}* in the

*same conjugate class. We write it as a : G → C. By averaging process,*

*one has a G-invariant ˜rho*

*=P*

_{i}*s∈G**a*_{s}*ρ*_{i}*(s) linear transformation on V** _{i}*.
Thus one has

*λ** _{i}* = 1

*d*

*i*

*tr(*X

*s∈G*

*a*_{s}*ρ*_{i}*(s)) =* 1
*d**i*

X*a*_{s}*χ*_{i}*(s).*

*Theorem 0.3. Keep notation as before, then one has*
*d*_{i}*|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

*a*_{s}*e*_{s}*∈ Z(C[G]) such that a*_{s}*∈ A. Then*
*u is integral over Z.*

*Proof. Since Z(C[G]) is generated by e*_{c}*. Let c*_{1}*, ..., c** _{r}* be all the con-

*jugacy classes and let a*

_{i}*:= a*

_{s}

_{i}*for some s*

_{i}*∈ c*

*. We first consider*

_{i}*R := ⊕*

^{r}

_{i=1}*Ze*

_{c}

_{i}*. It’s clear that R is a subring of Z(C[G]) which is*

*finitely generated over Z. Now let M = Z[a*

_{1}

*, ..., a*

_{r}*]. Since a*

*is inte-*

_{i}*gral over Z, it follows that M is a finite Z-module. One checks that*

*Z[u] ⊂ ⊕*

^{r}

_{i=1}*Me*

_{c}

_{i}*and ⊕*

^{r}

_{i=1}*Me*

_{c}

_{i}*is a finite Z-module. Hence u is inte-*

gral over Z. ¤

*proof of the theorem. For each i, take u =* P

*s∈G**χ(s*^{−1}*)e** _{s}*. 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
*ω*_{i}*: Z(C[G]) → C,*

3

*by sending u to λ**i* *the multiple of its i-component. It follows that the*
*homomorphic image λ**i* is an algebraic integer. One has now

*λ**i* := 1
*d*_{i}

X*χ**i**(s*^{−1}*)χ**i**(s) =* *g*

*d*_{i}*< χ**i**, χ**i* *>=* *g*
*d*_{i}*.*

*Hence λ*_{i}*∈ A ∩ Q = Z. It follows that d*_{i}*|g.* ¤