algebraic structures part 1

7/27/2019 Algebraic Structures Part 1

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CHAPTER 1

Group Actions and Sylows Theorem

1.1. Group Actions

Definition 1.1.1. Let G be a group and T a set. An action of G on T is afunction : G T T such that, denoting (g, t) by g t,

(1) For all g, h G, t T, g (h t) = (gh) t.(2) For all t T, eG t = t.

Given such an action we may define maps g : T T for all g G byg(t) = g t. Denote by P(T) the set of bijections (or permutations) from T to T.This of course is a group under composition of functions.

Theorem 1.1.2. Let be an action of a group G on a set T. For g G defineg : T T by g(t) = g t. Then

(1) g P(T) for all g G(2) the map : G P(T) given by (g) = g is a homomorphism of groups.

Conversely, given any group homomorphism : G P(T), the map : G T T

given by (g, t) = (g)(t) is a group action.

Proof. First observe that for any g, h G and t T

g h(t) = g(h(t)) = g (h t)) = (gh) t = gh(t)

so (g h) = gh. This proves that g g1 = IT = g1 g. Hence g P(T).Obviously the same result now implies that is a homomorphism. The proof ofthe converse is similar.

Example 1.1.3. The classic example of a group action is the action of a matrixgroup on the corresponding vector space. For instance, for any field F, take G =GLn(F) and T = F

n. It is clear that the usual matrix multiplication satisfies theaxioms above.

Example 1.1.4. The group operation defines an action of G on itself by leftmultiplication. Let g P(G) be the associated permutation ofG given by g(h) =gh and let : G P(G) be the map (g) = g. The homomorphism is calledthe left regular representation of G.

Example 1.1.5. A group can also act on itself by conjugation: g h = ghg1.

Theorem 1.1.6 (Cayleys theorem). Any finite group is isomorphic to a sub-group of Sn.

Proof. Let G be a finite group. Consider the left regular representation : G P(G). Ifg ker , then g is the trivial permutation; that is gh = g(h) =

h for all h G. But this implies that g = e by cancellation. Thus ker = {e}1


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1.2. THE COUNTING FORMULA 2

and is injective. By the First Isomorphism theorem, G is isomorphic to its image

which is a subgroup of P(G). Since P(G)= S|G|, the theorem follows.

1.2. The Counting Formula

Definition 1.2.1. Let G be a group acting on a set T. Let t T.

(1) The stabilizer of t in G is defined to be StabG(t) = {g G | g t = t}(2) The orbit oft under G is defined to be O(t) = {g t | g G}.

We leave it as an exercise to the reader to verify that StabG(Y) is a subgroupof G. For instance, let T be the real plane and G the group of rigid motions of theplane. Let Y be a figure in the plane (such as a hexagon or the letter H). ThenStabG(Y) is the usual symmetry group of the figure Y. Thus, we see again how thetheory of groups captures and generalizes the notion of symmetry. One can think

of the stabilizer StabG(Y) of the subset Y as an abstract version of the concept ofsymmetry of an object.

Theorem 1.2.2 (The Counting Formula). LetG be a group acting on a set T.Let t T.

(1) StabG(t) is a subgroup of G(2) If G is a finite group, then |G| = |StabG(t)||O(t)|

Proof. Part (1) has already been discussed.(2) Let H = StabG(t). Then by Lagranges Theorem, |G| = |H|[G : H].

Hence it suffices to establish a bijection between G/H and O(t). Define a function : G/H O(t) by (gH) = gt. We claim that is a bijection. First we need

to check that is well-defined. Suppose that gH = gH. Since g gH, thereexists an h H such that g = gh. But then g t = (gh) t = g (h t) = g t, asrequired. If t O(t), then there exists a g G such that t = g t. In this caset = g t = (gH) so is surjective. Finally,

(g) = (g) = g t = g t = g1g t = t = g1g H = gH = gH

so is also one-to-one.

Notice that the G-orbits inside T form a partition of T (that is, T is thedisjoint union of the orbits). Recall that in this case the orbits must arise froman equivalence relation on T. We can recreate the partition of T by defining thisrelation.

Let G be a group acting on a set T. Define a relation on T by s t if andonly if there exists a g G such that t = gs. It is a routine exercise to verify thatthis defines an equivalence relation. The equivalence class of t T is clearly justthe orbit O(t).

Example 1.2.3. Consider the group D8 of symmetries of the octagon. Eachelement of D8 induces a permutation of the set V of vertices of the octagon. Letv be one such vertex. The stabilizer of v is the set of all symmetries that fix v.clearly this is just the identity and the reflection through v and the opposite vertex.On the other hand the orbit O(v) is just the set of different vertices to which vcan be sent by a symmetry. Since the octagon is regular, this is just the set of all8 vertices. Thus |D8| = 16. Naturally the same argument shows that the set of

symmetries Dn of a regular n-gon has order 2n.


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1.3 . ACTION OF A GROUP ON ITSELF - THE CLASS EQUATION 3

Example 1.2.4. More interesting examples involve the symmetries of the reg-

ular solids. Consider for example the dodecahedron and let D be its group ofsymmetries. Recall that the dodecahedron has 20 vertices, 30 edges and 12 pentag-onal faces. Three faces meet at each vertex. Again fix a vertex v. The stabilizer ofv is he set of symmetries that fix v. These consist of the cyclic group of order threeof rotations and the three reflections through the edges meeting at the vertex. SoStabD(v) = S3 and has order 6. On the other hand the orbit of v is just the setof 20 vertices. So |D| = 20 6 = 120. The reader will probably easily agree thatthis is an easier method of finding the order of D than trying to identify all 120symmetries.

Another way to approach the problem is to notice that the group D acts onthe set of faces. If we fix a particular face f, then StabD(f) = D5, the group ofsymmetries of the pentagon, which has order 10. Since the orbit of f has size 12

we again get 120 for the order of the group.

Example 1.2.5. We can also derive the usual formula for combinations usingthe counting formula. Recall that the number of ways of choosing a set of k elementsfrom a set of n elements is given by the binomial coefficient

nk

=

n!

k!(n k)!

We can derive this formula in the following way. Let T = {1, 2, . . . , n} and let P(T)be the power set of T - the set of all subsets. Then clearly Sn acts on P(T) in thenatural way and the set of subsets of size k is precisely the orbit of the element{1, 2, . . . , k} P(T). An element of Sn will be in the stabilizer if it separately

permutes the sets {1, 2, . . . , k} and {k, k + 1, . . . , n}. Thus

StabSn({1, 2, . . . , k}) = A({1, 2, . . . , k}) A({k, k + 1, . . . , n})= Sk Snk

hence the counting formula states that

n! = |Sk Snk||O({1, 2, . . . , k})| = k!(n k)!

nk

which yields the formula above.

1.3. Action of a group on itself - the class equation

Now we study in more detail the action of a group on itself via conjugation:

g h = ghg1. In this case the stabilizer of an element h is the set:

CG(h) = {g G | ghg1 = h}

= {g G | gh = hg}

called the centralizer ofh; it is the set of elements of G that commute with h. Notethat CG(h) contains the center Z(G) and the subgroup h, so that if h = e thenCG(h) is a non-trivial subgroup. Note also that CG(h) = G if and only ifh Z(G).

The orbit ofh is called the conjugacy class of h and is the set of elements of Gwhich are conjugate to h:

C(h) = {ghg1 | g G}

Interpreting the counting formula in this context yields:


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1.3 . ACTION OF A GROUP ON ITSELF - THE CLASS EQUATION 4

Corollary 1.3.1. Let G be a group and h G. Then

|C(g)| = |G|/|CG(h)| = [G : CG(h)]

In particular, the cardinality of a conjugacy class must divide the order of thegroup. Of course, the conjugacy class C(h) must contain h. We say the conjugacyclass is trivial if C(h) = {h}.

Theorem 1.3.2 (The class equation). LetG be a finite group and let g1, . . . , gtbe representatives of the non-trivial conjugacy classes of G. Then

|G| = |Z(G)| +t

i=1

[G : CG(gi)].

Proof. Since being conjugate to is an equivalence relation, the conjugacy

classes of G partition G. For elements in the center, the conjugacy classes are alltrivial. Thus these can be lumped together to yield the partition

G = Z(G) t

i=1

C(gi).

Hence

|G| = |Z(G)| +t

i=1

|C(gi)|

and the theorem follows from the corollary above.

The reader should pause to take in the significance of this theorem. the sum-mands on the right all divide |G| and at most one (the first) can be equal to 1.

Example 1.3.3. Consider the case when G = S3. The conjugacy classes are{e}, {(12), (13), (23)} and {(123), (132)}, so the class equation is 6 = 1 + 3 + 2.

Suppose that |G| = 9, then the class equation could only be 9 = 3+3+3 or 9 = 9(the last case of course is the class equation of an Abelian group). In particular, itis not possible to have a class equation that begins 1 + . . . , so the center of G mustbe non-trivial. This is a general phenomenon which occurs whenever the order ofG is a power of a prime.

Corollary 1.3.4. Let p be a prime and let G be a finite group of order pt for

some positive integer t. Then Z(G) = {e}.

Proof. All the summands of the form |C(gi)| on the right hand side are divisorsof pn and are not equal to 1. Therefore they are divisible by p. Since the left handside is divisible by p, the remaining summand, |Z(G)| must also be divisible by p.Hence Z(G) = {e}.

Corollary 1.3.5. Let p be a prime and let G be a finite group of order p2.Then G is Abelian.

Proof. Exercise.

Note that the example of S3 shows that a group of order pq with p = q does

not need to be Abelian.


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1.4. SYLOWS THEOREM 5

1.4. Sylows Theorem

Sylows theorem represents one of the high points of elementary group theory,combining simple ideas to prove a complex and surprising result. The theoremconsists of three parts, the first of which is a partial converse to Lagranges Theorem.Recall that Lagranges Theorem states that if G is a finite group of order n and ifH is a subgroup of order k, then k|n. We turn this around to ask whether for anyk dividing n whether there exists a subgroup of order k. In general the answer tothis question is No. Sylows Theorem answers this question when k is a largestpossible power of a prime.

Let G be a group of order n and let p be a prime divisor of n. Then we canwrite n = pem where gcd(p, m) = 1. In its simplest from Sylows Theorem hasthree parts:

(1) There exists subgroups ofG of order pe

.(2) All such subgroups are conjugate.(3) The number of such subgroups divides m and is congruent to 1 modulo p.

To prove Sylows Theorem we shall make use of some further group actions onsets connected to G. Let S be the set of subgroups of G. We know that if H isa subgroup, then so is any conjugate gH g1. Hence G acts on S by conjugation.That is, we may define an action of G on S by:

for any g G, g H = gH g1

Then the stabilizer of a subgroup under this action is the normalizer of the subgroup,

NG(H) = {g G | gH g1 = H}

The orbit is just the set of subgroups conjugate to H, so part (2) above can berestated as saying the subgroups of order pe from a single orbit under this action.

Definition 1.4.1. Let G be a finite group and let p be a prime dividing |G|.Suppose that pe is the highest power of p dividing n. A subgroup of G of order pe

is called a Sylow p-subgroup of G.

Lemma 1.4.2. Let G be a finite Abelian group of order n and let p be a primedividing n. Then G contains a subgroup of order p.

Proof. We work by induction on n = |G|. The case where n = 2 is trivial.Now suppose that the result is true for all Abelian groups of order less than n.Clearly it suffices to find an element of G of order p. We may certainly pick a

non-trivial element g G. Let o(g) = m. If p | m, then we may write m = pm

and o(gm

) = p.On the other hand suppose that p m. Let N = g. Then N is a (normal)

subgroup of order m and the quotient group G/N has order n/m, which is smallerthan n but still divisible by p (since p m). By induction G/N contains an elementof order p, say hN. This means that hpN = (hN)p = N and hence that hp N.Since h N, this implies that hp is strictly contained in h and hence thato(hp) < o(h). But we know that

o(hp) =o(h)

gcd(p, o(h))

This implies that gcd(p, o(h)) = 1 and hence, since p is prime that p|o(h). Returning

to the argument above, this yields an element in G of order p.


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Theorem 1.4.3. LetG be a finite group of order n and let p be a prime divisor

of n. Then Sylow p-subgroups exist.Proof. We prove the result by complete induction on n = |G|. Clearly the

result is true when n = 2. Assume that the result is true for all groups of orderk < n.

Let p be a prime dividing n. Suppose first that p Z(G). Looking at the classequation we see that p cannot divide all the remaining terms on the right. So theremust exist a g G such that [G : CG(g)] is not divisible by p. But in this case|CG(g)| = p

em where gcd(p, m) = 1 and m < m. By induction CG(g) has asubgroup H of order pe and H is clearly a Sylow p-subgroup of G.

Now assume that p|Z(G). By the lemma, Z(G) contains a subgroup N oforder p. Since N is contained in the center ofG it is a normal subgroup of G. Now|G/N| = pe1m, so by induction G/N contains a Sylow p-subgroup, K, which hasorder pe1. Let : G G/N be the natural surjection and let H = 1(K). ThenK = H/N by the first isomorphism theorem, so |H| = p.pe1 = pe. Hence H is aSylow p-subgroup of G.

The next lemma touches on one of the crucial properties of Sylow subgroups:no element of one Sylow p-subgroup can normalize non-trivially another one. Thatis if P and P are two Sylow p-subgroups and x P satisfies xPx1 = P, thenx P.

Lemma 1.4.4. Let P and P be two Sylow p-subgroups. Then P NG(P) =P P. In particular, P NG(P

) if and only if P = P.

Proof. Let N = NG(P) and let Q = P N. Then applying the second

isomorphism theorem to N and the subgroups Q and P, yieldsQP

P=

Q

Q P

This implies that the order of QP/P is of the form pf for some non-negativeinteger f and that |QP| = |QP/P||P| = pe+f, which contradicts LagrangesTheorem unless f = 0. Hence Q P = Q. So P N P which implies thatP N = P P.

Theorem 1.4.5. LetG be a finite group of order n and let p be a prime dividingn.

(1) All Sylow p-subgroups are conjugate to each other.

(2) Let np be the number of Sylow p-subgroups of G. Thennp|m and np 1 (mod p).

Proof. Let P be a Sylow p-subgroup. Let P = {gP g1 | g G} be theset of conjugates of P; lets denote these conjugates by P = P1, . . . , P s, so thatP = {P1, . . . , P s}. Then P itself acts on P by conjugation. An orbit will be asingleton set {Pi} if and only if P NG(Pi). But this can only happen if P = Piby the lemma. Hence {P} is the only trivial orbit; all the other orbits must havemore than one element and hence, because |P| = pe, the size of these orbits isdivisible by p. This proves that |P|, the number of conjugates of P, is congruentto 1 modulo p.

Now let P be any other Sylow p-subgroup. Again we let P act on P and

consider the orbit decomposition. The size of the orbits must be a power of p.


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Thus at least one orbit must be trivial; otherwise we would conclude that p |

P, contradicting the conclusion of the previous paragraph. Suppose this orbit is{Pi}. Then P NG(Pi). By the lemma, this implies that P = Pi. Henceevery Sylow p-subgroup is conjugate to P, proving (1). Thus np is equal to thenumber of conjugates of P; hence np 1 (mod p). Finally, applying the CountingTheorem to the action of G on subgroups, we see that np = [G : NG(P)]. But[G : NG(P)][NG(P) : P] = [G : P] = m, so np|m.

Example 1.4.6. As an easy illustrative example, consider the group D5 ofsymmetries of the regular pentagon. Recall that D5 has order 10 = 2.5 and itselements are the identity, four non-trivial rotations and five reflections. The theoremstates that the number of Sylow 5-subgroups must divide 2 and be congruent to1 modulo 5. Thus there is a single normal subgroup of order 5; of course this

is the group consisting of the identity and the four non-trivial rotations. On theother hand the theorem states that in a group of order 10 the number of Sylow2-subgroups must divide 5 and be congruent to 1 modulo 2. Hence it must be 1or 5. Clearly in the case of D5, the answer is 5 and these groups are the 5 cyclicgroups of order 2 generated by the reflections.

As a first application we see that most groups of order pq are Abelian (ofcourse S3 is an example that shows that this is not always true).

Theorem 1.4.7. Let G be a group of order pq where p and q are primes andp < q. If p does not divide q 1, then G is Abelian.

Proof. Consider the number nq of Sylow q-subgroups. Then nq|p and nq 1

(mod q). hence nq = 1 and there is only one Sylow q-subgroup, say H, which mustthen be normal.Now consider the number np of Sylow p-subgroups. We know by Sylows The-

orem that np|q and np 1 (mod p). If np = q, this would imply that p|(q 1),which is not possible. Hence np = 1 also. Let the unique Sylow p-subgroup beK. By Lagranges Theorem, H K = {e}. Now HK must be a subgroup that isproperly bigger than K, so its order is greater than q and by Lagranges Theoremdivides pq. Hence HK = G. Thus by Lemma ??, G = H K = Cq Cp and inparticular, G is Abelian.

If there is only Sylow p-subgroup for some p, then we know that this subgroupmust be normal. Thus the Sylow theorems provide us with a method of provingthat a group has a proper non-trivial normal subgroup. This turns out to be anextremely important question. Groups that do not have proper non-trivial normalsubgroups are called simple groups and form the fundamental building blocks offinite group theory as we shall see in the next chapter.

Definition 1.4.8. A group is said to be simple if it has no non-trivial propernormal subgroups.

Cyclic groups of prime order are simple because they have no non-trivial propersubgroups at all. Conversely, an Abelian simple group must be cyclic of prime order(why?). At this stage it is not at all obvious whether it is possible for a groupsto have non-trivial proper subgroups yet still be simple (because none of them arenormal). We shall answer this question in the next section. Our next result says

that if there are any such groups, they have to have at least 60 elements. Its proof


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1.5. THE ALTERNATING GROUPS 8

is one of the most satisfying exercises in elementary group theory, so I wont spoil

the fun by providing the proof.Theorem 1.4.9. There are no non-Abelian simple groups of order less than 60.

Proof. Exercise.

1.5. The Alternating Groups

Continuing with our theme of group actions, we consider a canonical action ofthe symmetric group Sn on the vector space R

n. this allows us easily to identifythe alternating group, the subgroups of Sn consisting of permutations that canbe represented as a product of an even number of transpositions. the alternatinggroups are among the most important families of finite groups and their structureplays a key role in explaining the non-existence of a formula for the solutions of aquintic equations.

Proposition 1.5.1. Let Sn and let

iei Rn, where i R. Define

: Rn Rn by

iei

=

ie(i)

Then is an invertible linear transformation and = . Define : Sn GLn(R) by () = . Then is an injective homomorphism.

Proof. We leave it to the reader to verify that is a linear transformation.Now

iei = ie(i)=

ie(i)

=

iei

Since e = IRn , the above proves that is invertible. It also proves that is ahomomorphism.

Let : Sn GLn(R) be the map () = . Then defines an action of Snon Rn. Note that since (ei) = e(i), is the matrix with a 1 in the ((i), i)-th position and zeros everywhere else. These matrices are known as permutationmatrices.

Example 1.5.2. Consider explicitly the case when n = 3. Then

((12)) =

0 1 01 0 0

0 0 1

, ((13)) =

0 0 10 1 0

1 0 0

, ((23)) =

1 0 00 0 1

0 1 0

and

((123)) =

0 0 11 0 0

0 1 0

, ((132)) =

0 1 00 0 1

1 0 0

.

Notice that the determinant of any permutation matrix is 1 since there isexactly one term in the usual expansion of the determinant that is non-zero forsuch matrices and this term is 1. The composition sgn = det : Sn {1} is

a homomorphism and is known as the sign representation.


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Definition 1.5.3. The kernel of the map sgn is called the alternating group

and is denoted by An.Note that An must have n!/2 elements by the first isomorphism theorem since

Sn has n! elements and Sn/An = C2.

Example 1.5.4. The alternating group A4 is the subgroup consisting of allthree cycles and products of disjoint transpositions:

{e, (123), (132), (134), (143), (124), (142), (234), (243), (12)(34), (13)(24), (14)(23)}

Sylows theorem predicts that the number of Sylow 3-subgroups must be 1 or 4;clearly in this case the answer is 4, the eight 3-cycles splitting into 4 mutuallyinverse pairs. On the other hand the number of Sylow 2-subgroups should be either1 or 3; clearly the answer here is 1. The group V = {e, (12)(34), (13)(24), (14)(23)}is the unique Sylow 2-subgroup.

We now look more closely at the group A5 and prove that it is simple. Up tothis point we know that the cyclic groups of prime order are simple but we knowof no other examples. We also know that no non-Abelian group of order less than60 can be simple. Thus A5 is the smallest non-Abelian simple group. Its simplicityis intimately connected with the Abel-Ruffini theorem that there is no algebraicformula for the roots of a quintic polynomial.

The group A5 has 60 elements, too many to write out explicitly. One canhowever look at the number of elements of different types. In addition to theidentity element we see, using some straightforward counting arguments, that wehave:

20 3-cycles such as (123).

24 5-cycles such as (12345). 15 products of two disjoint 2-cycles such as (12)(34).

Now the orders of elements of A5 that are possible from Lagranges Theoremare the divisors of 60, namely 1, 2, 3, 4, 5, 6, 10, 12, 15, 20 and 30. But only 1, 2, 3and 5 actually occur. This is an important illustration of the fact that k divides |G|does not necessarily imply that there exist an element of order k.

Lets look at the Sylow subgroups. Note that 60 = 22.3.5. The Sylow 3-subgroups have order 3, so they must be cyclic. The number of them should becongruent to 1 modulo 3 and must divide 20, so the options are 1, 4 and 10. Sinceeach of the 3-cycles above generates a cyclic subgroup of order 3 which containsit and one other 3-cycle, there must be 10 such Sylow 3-subgroups. Similarly the

Sylow 5-subgroups are cyclic of order 5, and the number of them must divide 12 andbe congruent to 1 module 5. So there must be six Sylow 5-subgroups. Finally theSylow 2-subgroups are of order 4 and the number of them is congruent to 1 modulo4 and divides 15. The possibilities are 1 or 5. But there are 5 subgroups similar tothe Klein group {e, (12)(34), (13)(24), (14)(23)}. so the answer is obviously five.

Now lets review the class equation for A5. Recall that the class equation comesfrom the partition of G into conjugacy classes. The equation itself describes theorder ofG as the sum of (a) the order of the center, which is the union of the trivialconjugacy classes; and (b) the orders of the non-trivial conjugacy classes.

First recall the following basic fact about conjugation inside Sn.

Lemma 1.5.5. Let Sn and let (a1a2 . . . at) be a t-cycle. Then

(a1a2 . . . at)1

= ((a1)(a2) . . . (at))


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hence all t-cycles are conjugate in Sn.

Thus the three cycles in S5 form a single conjugacy class. Of course its notcompletely clear that they are still conjugate inside the subgroup A5. Recall that|G| = |C(g)|.|CG(h)|. Since |CS5((123))| = 20, the size of the centralizer of (123)must be 6. But the centralizer contains (123) and (45) and together these generatethe group of order 6, so we must have:

CS5((123)) = {e, (123), (132), (45), (123)(45), (132)(45)}

But thenCA5((123)) = CS5((123)) A5 = {e, (123), (132)}

So the order of the conjugacy class of (123) in A5 must be 60/3 = 12 and thereforeall of the 3-cycles are still conjugate in A5.

A similar argument can be applied to the 5-cycles but with a different result.

the 24 5-cycles are conjugate in S5. therefore |CS5((12345))| = 120/24 = 5; henceCS5((12345)) = (12345). Thus,

CA5((12345)) = CS5((12345)) A5 = (12345)

Hence, the order of the conjugacy class of (12345) in A5 must be 60/5 = 12. Thusinside A5 the 5-cycles fall into two distinct conjugacy classes of size 12.

A similar argument applied to the products of disjoint transpositions yields asingle conjugacy class of size 15. hence the class equation of A5 reads:

60 = 1 + 12 + 12 + 15 + 20

Interestingly, the simplicity of A5 is an easy consequence of this.

Theorem 1.5.6. The group A5 is simple

Proof. The key observation is that any normal subgroup must be a union ofconjugacy classes; for the definition of normality can be interpreted as the statementthat a group is normal if and only if contains all conjugates of any of its elements.Thus if K is a proper non-trivial normal subgroup, its order must be a subsumof the right hand side of the class equation that includes the first term. The onlypossibilities less than 30 are:

13, 16, 21, 25, 28

and clearly none of these numbers divide 60. Hence no proper non-trivial normalsubgroup exists.

Theorem 1.5.7. The group An is simple for n 5.

Proof. We prove the result by induction on n. The base case n = 5 wasproved in Theorem 1.5.6. So suppose that n 6. To simplify notation, let G = An.Consider the groups Gi = StabG(i). The elements of Gi are all even permutationsof the set {1, 2, , . . . , i 1, i + 1, . . . , n} so Gi = An1. Assume that G is not simpleand let N be a proper, non-trivial normal subgroup of N. Then N Gi Gi. Sinceby induction Gi is simple, N Gi = Gi or {e}. Suppose that N Gi = Gi. Let j bedifferent from i. Pick k and l also distinct from i and j. Since (ij)(kl)(i) = j, weknow (from a homework exercise) that StabG(i) = (ij)(kl)StabG(j)(kl)(ij) (notethat we cant just use (ij) here because it does not belong to G.) Since N is normal,it must therefore contain all the groups G1, . . . , Gn. let G and write as aproduct of transpositions,

= (a1a2)(a3a4) . . . (ar1ar)


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1.6. FREE GROUPS AND RELATIONS 11

where the ai are not necessarily distinct. Note that the number of transposi-

tions must be even so we can split them into consecutive pairs beginning with(a1a2)(a3a4). Each of these products must belong to some Gi (since n 5), so theproduct belongs to N. We have proved that G = N contradicting the assumptionthat N was proper.

Hence N Gi = {e} for all i. This means that if N, then for all i, (i) = i.It also implies that if , N and (i) = (i) for any i, then = . Let Nand consider the decomposition of into disjoint cycles. We consider two cases.

(1) The decomposition involves at least one cycle of length at least three.So we can write = (a1a2a3 . . . )0. Choose such that (a1) = a1,(a2) = a2 and (a3) = a3. (Note that must involve at least threenumbers distinct from a1, a2, so we are using the fact that n 5 at thispoint.) Then

= 1 = (a1a2(a3) . . . )01

Hence (a1) = (a1) and = , a contradiction.

(2) Otherwise the decomposition must involve only transpositions. Since alln numbers must be involved and is even, we must have n 8 and

= (a1a2)(a3a4)(a5a6)(a7a8)0

Take = (a3a5)(a7a8). Then,

= 1 = (a1a2)(a5a4)(a3a6)(a7a8)01

So again, (a1) = (a1) and = , a contradiction.

Thus we contradicted the choice of a non-trivial element of N. Hence N = {e}.

This contradicts the assumption that N is non-trivial. We have contradicted theassumption that G is not simple. Hence G must be simple.

1.6. Free Groups and Relations

Let S be a set and define W(S) to be the set of words in S; that is the set ofall finite ordered strings of the form

a1a2 . . . an

where ai S. For instance ifS = {a,b,c} then the following are examples of words:

abaac, ababa, aaa, bbbcbab

Note that we include in W(S) the empty word which we will denote by 1. Wecan define a binary operation on S by concatenation

(a1a2 . . . an).(b1 . . . bm) = a1a2 . . . anb1 . . . bm

This operation is associative and has an identity operation (hence it makes W(S)into a semi-group). We want to make W(S) into a group. In order to do this wefirst introduce elements that will serve as the inverses of the elements of S. LetS be a copy of S whose elements consists of {s | s S} and let S = S S. LetW = W(S). Inside W we want to define some additional equality relations suchas

aa = 1, bcbbccabc = bcbbabc

To do this we will define an equivalence relation on W so that these statements

are true in W/ . We define a reduction to be a pair of words of W (w, w) where


12/25


w = a1a2 . . . at and w = a1a2 . . . aibbai+1 . . . at for some a1, . . . at, b S. We say

that a word w can be reduced to another word w

if there is a series of reductions(w, w1), (w1, w2), . . . (ws, w). A reduced word is a word that cannot be reduced to

another word of shorter length. A word is reduced if and only if it contains nosubwords of the form aa.

Lemma 1.6.1. Every word has a unique reduced form.

Proof.

We define two words to be equivalent if they have the same reduced form.

Proposition 1.6.2. If w w and v v, then wv wv.

Proof. Let the reduced form ofw and w be w0 and that ofv and v be vo. By

a series of reductions we can reduce w to its reduced form w0. The same reductionswill reduce wv to w0v. Similarly we may reduce w0v to w0v0. Analogously we canreduce wv to w0v

and then to w0v0. Hence the reduced form of both wv and wv

is w0v0. Thus wv wv.

Proposition 1.6.3. The operation of concatenation induces a group structureon F(S) = W(S)/ .

Proof. By the previous proposition the operation factors through to an op-eration on F(S). It remains associative and the empty word is till an identity.However we now have that every element has an inverse.

The group F(S) is called thefree group on the set S. It has the followingimportant universal property.

Theorem 1.6.4. Let G be a group, let S be a set and let : S G be afunction. Then there is a group homomorphism : F(S) G such that |S = .

Proof. Extend to S by defining (s) = (s)1. Define : W(S) G by

(a1 . . . at) = (a1) . . . (at). If w = a1a2 . . . at and w = a1a2 . . . aibbai+1 . . . at

then

(w) = (a1a2 . . . aibbai+1 . . . at)

= (a1) . . . (ai)(b)(b)1ai+1 . . . (an)

= (a1) . . . (ai)(ai+1) . . . (an)

= (w)

So is invariant under reductions. Hence if w w then (w) = (w). Thus

induces a map on F(S) by ([w]) = (w).

Let R F(S), and denote by R the normal closure of R (the intersection ofall the normal subgroups containing R).

Definition 1.6.5. The group generated by a set S subject to the relations Ris defined to be

S | R = F(S)/ R

For instance S3 can be show to be isomorphic to

a, b | a2, b3, abab

.


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Theorem 1.6.6. Let G be a group and let S G which satisfy the relations

R. Then there exists a homomorphism : S | R G such that (s) = s for alls S.

Proof. let : F(S) G be the homomorphism guaranteed by theorem 1.6.4.Then R ker( and hence R ker() also. Thus induces a map onF(S)/ R = S | R such that (s) = s.


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CHAPTER 2

Composition series and solvable groups

2.1. Composition series

One of the fundamental theorems in number theory is the unique factorizationtheorem which states that every number can be expressed in a unique way as a

product of primes. Interestingly an analogous result holds for groups. If N G,then we can think of N and G/N as being factors of G. We can then try andfactorize N and G/N. If G is finite this process must stop and at this point wehave reduced down to simple groups, which play the role of primes in group theory.

Definition 2.1.1. A normal series for a group G is a finite sequence of sub-groups, G0 = {e} G1 G2 . . . Gt = G such that Gi Gi+1. The length of theseries is defined to be t.

Definition 2.1.2. A composition series is a normal series in which all thefactor groups are simple

Example2.1.3

.Let G be S4. Recall the definition of the Klein 4-group V andlet C = {e, (12)(34)}. Then

{e} C V A4 S4

is a composition series for S4. Note that the definition of normal series does notrequire that the terms of the series be normal in the group G. In the case above Vand A4 are normal in G but C is not.

The unique factorization theorem for integers has two parts. One is the exis-tence part which states that every integer can be factored as a product of primes.The second is the uniqueness part which states that there is only one way of fac-toring a number into a product of primes. The analogous result for groups has two

similar parts: (1) composition series always exist for finite groups; and (2) the setof composition factors is the same for any composition series.

Proposition 2.1.4. Any finite group has a composition series.

Proof. We use induction on n = |G|. The case n = 1 is trivial. Considerthe set of all proper normal subgroups of |G| and pick one, say N, whose orderis as large as possible. By the fourth isomorphism theorem G/N is simple. Now|N| < n, so by induction, N has a composition series,

N0 = {e} N1 N2 . . . N s = N

Then clearly N0 N1 N2 . . . N s G is a composition series for G.

14


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2.2. THE JORDAN HOLDER THEOREM 15

2.2. The Jordan Holder Theorem

Suppose that G is a finite Abelian group of order n and let G0 = {e} G1 G2 . . . Gt = G be a composition series for G. Then the composition factors must all beAbelian simple groups; that is, they are isomorphic to a cyclic group of prime order,say Gi/Gi1 = Cpi . By Lagranges Theorem, n = p1 . . . pt. Thus while there maybe many different composition series for G, the length of any composition series isequal to t, the number of primes in the prime factorization of n; and the collectionof composition factors is the same in the sense that the types of composition factorsthat occur and the number of times they occur is invariant. For instance, ifG = is an Abelian group of order 60 then

{e} 12 6 2 G

and

{e} 20 10 5 G

are both composition series with composition factors C5, C2, C3, C2 and C3, C2, C2, C5respectively. We can thus think of a composition series in some way as a primefactorization for a finite group. The Jordan Holder Theorem is then the analog ofthe unique factorization theorem for integers. Its states that composition series areunique in the sense that they have the same length and that the set of compositionfactors occurring, counted with multiplicity is the same for any such series.

Definition 2.2.1. Let G be a group and let G0 = {e} G1 G2 . . . Gs = Gand H0 = {e} H1 H2 . . . H t = G be two normal series for G. We say that theseries are equivalent if s = t and there exists a bijection between the sets of factor

groups for which corresponding factors are isomorphic.Theorem 2.2.2 (Jordan-Holder). Any two composition series of a finite group

are equivalent.

Proof. We shall prove the theorem by induction on the s length of the shorterseries. If s = 1, then G is simple, so the result is trivially true. Now assume thatG0 = {e} G1 G2 .. . Gs = G and H0 = {e} H1 H2 .. . Ht = G are twocomposition series with s t. If Gs1 = Ht1 then the result from the inductionhypothesis applied to Gs1. Otherwise Gs1Ht1 is a normal subgroup ofG strictlybigger than Gs1. Since G/Gs1 is simple, the fourth isomorphism theorem tellsus that Gs1Ht1 = G. Therefore,

Gs1Gs1 Ht1 =

Gs1Ht1Ht1 =

G

Ht1

andHt1

Gs1 Ht1=

Gs1Ht1Gs1

=G

Gs1Now let

{e} K1 K2 . . . K u = Gs1 Ht1

be a composition series for Gs1 Ht1. Then

{e} K1 K2 . . . K u Gs1

is a composition series for Gs1, as is

{e} G1 G2 . . . Gs1


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2.3. SOLVABLE GROUPS 16

Thus, by induction, u = s2 and the two series are equivalent. A similar argument

shows that{e} K1 K2 . . . K u Ht1

is equivalent to

{e} H1 H2 . . . H t1

and u = t 2. Hence s = t and by combining all the above information we see thatthe two original series are equivalent.

2.3. Solvable groups

IfG is an Abelian group, its composition factors must also be Abelian. However,the converse is far from true, as the examples of S3 and S4 illustrate. It is quite

possible for a non-Abelian group to have Abelian composition factors. These kindof groups turn out to be extremely important.

Definition 2.3.1. A group G is said to be solvable if it has a normal series

G0 = {e} G1 G2 . . . Gt = G

for which the factors Gi/Gi1 are Abelian.

This definition works for both finite and infinite groups. For finite groups onecan equivalently define a solvable group to be one whose composition factors areall Abelian. The symmetric groups S3 and S4 are solvable but S5 is not (since itscomposition factors are A5 and C2). More generally, the fact that there are no

non-Abelian simple groups of order less than 60 implies that all groups of order lessthan 60 are solvable and that A5 is the smallest group that is not solvable.The following result about solvable groups is of fundamental importance.

Theorem 2.3.2. Let G be a group and H a normal subgroup. Then G issolvable if and only if both H and G/H are solvable.

Proof. Suppose that G is solvable. Then there exists a normal series

G0 = {e} G1 G2 . . . Gt = G

for which the factors Gi/Gi1 are Abelian. Let Hi = GiH. Then for any k Hi1and h Hi, then hkh

1 Gi1 H = Hi1, so the Hi form a normal series forH. Moreover, using the second isomorphism theorem,

HiHi1

=G1 H

Gi1 H=

Gi H

Gi1 (Gi H)=

Gi1(Gi H)

Gi1

GiGi1

Hence Hi/Hi1 is Abelian, and H is solvable. Similarly, let Ki = GiH/H. ThenK0 = {e} K1 K2 .. . Kt = G/H is a normal series for G/H. Using both thesecond and third isomorphism theorems we see that

KiKi1

=GiH/H

Gi1H/H=

GiH

Gi1H=

Gi(Gi1H)

Gi1H=

GiGi1H Gi

=Gi/Gi1

Gi1H Gi/Gi1

Thus Ki/Ki1 is a homomorphic image of Gi/Gi1 and hence is Abelian. ThusG/H is solvable.

The converse we leave as an exercise.


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2.4. P-GROUPS AND NILPOTENT GROUPS 17

2.4. p-Groups and Nilpotent Groups

Definition 2.4.1. Let p be a prime. A p-group is a group of order pr for somer.

Theorem 2.4.2. Any p-group is solvable.

Proof. Let G be a p-group. We know that G has a composition series. ByLagranges Theorem, the composition factors of a p-group must all be p-groupsthemselves. Thus the composition factors are simple p-groups. Since the center ofa p-group is a non-trivial normal subgroup by 1.3.4, a simple p-grop must be equalto its center and hence abelian.

In fact p-groups have many other very special properties.

Theorem 2.4.3. Let G be a group of order pr where p is prime.

(1) If N is a normal subgroup of G, then N Z(G) = {e}.(2) G contains a composition series consisting of normal subgroups. In par-

ticular G contains a normal subgroup of order ps for all s = 1, . . . , n.(3) If H is a proper subgroup of G, then H < NG(H).

Proof. (1) Notice that N is closed under conjugation by elements ofG. There-fore we have an action of G on N by conjugation. Just as in the proof of 1.3.4, welook at the orbits under this action and deduce that the number of trivial orbitsmust be divisible by p. The trivial orbits are precisely the intersection of N withZ(G).

(2) We prove that G contains a composition series consisting of normal sub-groups by induction on r. Ifr = 1, the assertion is trivial. Ifr > 1, then G contains

a non-trivial center and hence an element of order p by 1.4.2. Thus, G containsa normal subgroup of order p, say N. By induction, G/N contains a compositionseries consisting of normal subgroups, say

N/N = N1/N N2/N . . . N t/N = G/N

By the fourth isomorphism theorem, the Ni must be normal in G, so

{e} N = N1 N2 . . . N t = G

is a composition series for G consisting of normal subgroups ofG. Since Ni/Ni1 =Cp, we must have that |Ni| = pi, proving the second assertion.

(3) Again we use induction on the order of G, the base case being trivial.Consider HZ(G). IfHZ(G) = {e}, then the elements of the center are in NG(H)

but not in H. IfHZ(G) = {e}, then H contains a normal subgroup M of order p.Now NG(H)/M = NG(H/M) (exercise). By induction NG(H/M) strictly containsH/M. By the fourth isomorphism theorem, NG(H) strictly contains H.

Groups that have composition series consisting of normal subgroups are calledsupersolvable groups. However p-groups actually have an even stronger property they have a normal series where each Gi/Gi1 Z(G/Gi1). Such groups arecalled nilpotent. the standard defiition is slightly different.

Definition 2.4.4. Let G be a group. we define iteratively the upper centralseries by

Z0(G) = {e}, Z1(G) = Z(G), Zi(G)/Zi1(G) = Z(G/Gi1)

The group Zi(G is called the i-th center of the group G.


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2.4. P-GROUPS AND NILPOTENT GROUPS 18

Definition 2.4.5. The commutator of two elements x, y in a group G is the

element [x, y] = xyx1y1

Lemma 2.4.6. The i-th center Zi(G) is the set of elements {x G | [x, y] Zi1(G) for all y G}.

Proof. Let x G. Set Zi1 = Zi1(G). Then

x Zi xZi1 Z(G/Zi1)

xZi1yZi1 = yZi1xZi1 for all y G

xZi1yZi1[xZi1]1y[Zi1]

1 for all y G

xyx1y1Zi1 = Zi1 for all y G

[x, y] Zi1 for all y G

Definition 2.4.7. A group is called nilpotent if there exists an integer t suchthat Zt(G) = G.

Theorem 2.4.8. All p-groups are nilpotent.

Proof. It suffices to note that if Zi(G) = G, then Zi(G) Zi+1(G).

Recall that ifK and N are normal subgroups of a group G such that H K =

{e}, then HK= H K .

Lemma 2.4.9. LetG be a finite group and let p1, . . . pr be the prime divisors of|G|. If G has a unique Sylow pi-subgroup Pi, for each i, then G = P1 Pr.

Proof. We prove by induction that P1 . . . P i = P1 Pi. The case i = 1is trivial. By induction P1 . . . P i = P1 Pi, so |P1 . . . P i| = |P1| . . . |Pi| andpi+1 |P1 . . . P i|. Hence Pi+1 P1 . . . P i = {e}. So

P1 . . . P i+1 = (P1 . . . P i)Pi+1 = (P1 . . . P i) Pi+1 = P1 Pi+1

The result then follows.

Lemma 2.4.10. LetG be a finite nilpotent group. If H < G, then H < NG(H).

Proof. As for p-groups. Either H Z(G) in which case we can apply induc-tion, or the elements of the center not in H are in NG(H).

Theorem 2.4.11. A finite group is nilpotent if and only if it is a product ofp-groups.

Proof. It suffices to show from Lemma 2.4.9 that any Sylow p-subgroup isnormal. Let P be a Sylow p-subgroup and let N = NG(P). Since P N, it mustbe the unique Sylow p-subgroup of N. Consider M = NG(N). Let m M. ThenmP m1 mN m1 = N and |mP m1| = |P| so mP m1 is a Sylow p-subgroup ofN. Hence mP m1 = P. Thus P M. Since N is the maximal subgroup in whichP is normal we must have M = N. By 2.4.10, if N were a proper subgroup of G,

it would have to be a proper subgroup of M. hence N = G and P G.


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2.5. THE DERIVED SERIES 19

Definition 2.4.12. For two subgroups H and K of a group G, we define

[H, K] = [h, k] | h H, k K

This group is called the commutator of the groups H and K.

Definition 2.4.13. The lower central series is defined by

G0 = G, G1 = [G, G], Gi = [G, Gi1]

Lemma 2.4.14. Let G be a group and let N M be normal subgroups. ThenM/N ZG/N if and only if N [G, M]. In particular G/N is abelian if and onlyif N [G, G]

Proof.

M/N Z(G/N) xNyN = yNxN for all x G, y M

xyx1y1N = N for all x G, y M

[x, y] N for all x G, y M

[M, G] G

Theorem 2.4.15. The groups Gi are all normal in G. A group G is nilpotentif and only if Gt = {e} for some t.

Proof. Exercise.

2.5. The Derived Series

We now return to solvable groups and give an alternative characterization of asolvable group.

Definition 2.5.1. The derived series of a group G is defined by

G(0) = G, G(1) = [G, G], G(i) = [G(i1), G(i1)]

Theorem 2.5.2. A group G is solvable if and only if Gt = {e} for some positiveinteger t.

Proof. Suppose that G(t) = {e} for some t. The commutator of a pair ofnormal subgroups is again normal, so the derived series is a normal series for G:

G

(t)

G

(t1)

. . . G

(1)

GMoreover G(i1)/G(i) = G(i1)/[G(i1), G(i1)] is abelian by Lemma 2.4.14 so G issolvable.

Conversely, suppose that G is solvable and let

N0 = {e} N1 N2 . . . N t = G

be a normal series with abelian factors. Then by Lemma 2.4.14, [Nj , Nj ] Nj1 forall j = 1, . . . , t. In particular, G(1) = [Nt, Nt] Nt1. We prove that G

(i) Ntiby induction, the base case having just been proved. Suppose that G(i) Nti.Then

G(i+1) = [G(i), G(i)] [Nti, Nti] Nti1 = Nt(i+1)

Thus G(t)

N0 = {e}, as required.


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2.6. AUTOMORPHISMS AND THE SEMI-DIRECT PRODUCT 20

2.6. Automorphisms and the Semi-Direct product

Definition 2.6.1. An automorphism of a group G is a group isomorphismfrom G to G. The set of all automorphisms of G is a subgroup of P(G) called theautomorphism group and denoted by Aut(G).

We leave it to the reader to verify that Aut(G) is indeed a subgroup of P(G).

Theorem 2.6.2. Let g G. Denote by g : G G the conjugation mapg(h) = ghg

1. Then g Aut(G). Define : G G by (g) = g. Then is ahomomorphism with kernel ker = Z(G).

Proof. Exercise for the reader.

Definition 2.6.3. The automorphisms g are known as inner automorphisms

and the image of is called the inner automorphism group. It is denoted Inn(G) = = {g | g G}.

Proposition 2.6.4. let Aut(G) and let g G. Then

g1 = (g).

Hence Inn(G) is a normal subgroup of Aut(G).

Proof. Exercise for the reader.

Definition 2.6.5. Let H and K be groups and let : K Aut(H) be ahomomorphism. Denote the induced action of k K on h H by k h = (K)(h).Define an operation on H K by

(h1, k1) (h2, k2) = (h1(k1 h1), k1k2)

Theorem 2.6.6. The operation defined above endows the set H K with agroup structure.

Proof. Associativity: Let (h1, k1), (h2, k2), (h3, k3) H K. Then

(h1, k1)[(h2, k2)(h3, k3)] = (h1, k1)[(h2(k2 h3), k2k3)]

= (h1(k1 (h2(k2 h3))), k1(k2k3))

= (h1(k1 h2)(k1 (k2 h3)), (k1k2)k3)

= (h1(k1 h2)((k1k2) h3), (k1k2)k3)

= (h1(k1 h2), k1k2)(h3, k3)

= [(h1, k1)(h2, k2)](h3, k3)

It is easily seen that (eH, eK) is an identity element. Observe that

(k1 h1, k1)(h, k) = ((k1 h1)(k1 h), k1k) = (k1 eH, eK) = (eh, eK)

and that

(h, k)(k1 h1, k1) = (h(k (k1 h1)), kk1) = (h(eK h1), eK) = (eH, eK)

so (h, k)1 = (k1 h1, k1). Thus all the group axioms are satisfied for thisoperation.

Definition 2.6.7. The group described in the above theorem is called the

semidirect product of H and K with respect to . It is denoted by H K.


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2.7. LINEAR GROUPS 21

Example 2.6.8. It is easy to see that S3 is isomorphic to the semidirect product

of C3 C2 where is the (unique) isomorphism from C2 to Aut(C3).Theorem 2.6.9. LetG be a group and let H andK be subgroups. Suppose that

(1) H G(2) HK = G(3) H K = {e}

Then G = H K where : K Aut(H) is the map induced from conjugation onH.

Proof. Define f: H K G by f(h, k) = hk. Then

f((h1, k1))f((h2, k2)) = h1k1h2k2 = h1(k1h2k11 )k1k2

= h1(k1)(h2)k1k2 = f((h1(k1)(h2), k1k2))

= f((h1, k1)(h2, k2))

So f is a homomorphism. It is clearly surjective by assumption (2). To verifyinjectivity note that

f(h, k) = e = hk = e = h = k1 H K = h = k = e

by part (3).

Example 2.6.10. Groups with two Sylow subgroups. Suppose |G| = peqf for eand f distinct primes. Let P and Q be a Sylow p-subgroup and a Sylow q-subgrouprespectively. If P and Q are both normal, then G = P Q (as we saw earlier inthe discussion of nilpotent groups). If P is normal but Q isnt then we are in the

situation of the Theorem 2.6.9 and G = P Q. In particular this will alwayshappen when G is a non-abelian group of order pq where q < p.

Example 2.6.11. Dihedral groups. Consider the dihedral groups D2n, thesymmetries of the regular n-gon. let H be the subgroup of rotations through2k/n and let K be the subgroup generated by a reflection . If H, then1 = 1. Again the hypotheses of the theorem are satisfied and G = H K.

Example 2.6.12. Affine transformations. let V be a vector space. An affinetransformation is function f: V V of the form f(v) = (v) + w where islinear and w V. Such a transformation is invertible if and only if is. theset of invertible affine transformations forms a group A. Let L be the subgroupof invertible linear transformations and T the group of translations (v v + w).Then T is a normal subgroup and A = T L.

2.7. Linear Groups

2.7.1. General and Special Linear Groups. Let F be a field. The generallinear group is defined to be the set of n n matrices over F,

GLn(F) = {A Mn(F) | det(A) = 0}

It is easily seen to be a group under matrix multiplication. The special linear groupis the subgroup of matrices of determinant 1.

SLn(F) = {A GLn(F) | det(A) = 1}

IfF is a finite field then these groups are both of course finite.


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Proposition 2.7.1. LetF be a field of order q. Then

|GLn(F)| = (qn 1)(qn q) . . . (qn qn1)

Proof. A matrix is invertible if and only if its columns are independent.We can form such a matrix by choosing successive columns so that they are notcontained in the vector subspace of Fn spanned by the previous columns. Since|Fn| = qn and an i1-dimensional subspace contains qi1 vectors, there are qnqi1

choices for the i-th column.

The special linear groups provide us with one of the simplest infinite families ofsimple finite groups of Lie type. The center of SLn(F) is just the scalar matricesof determinant 1 and so is isomorphic to the subgroup of n-th roots of unity in F.The quotient is called the projective special linear group,

PSLn(F) = SLn(F)/(Z(SLn(F))

IfF is a finite field then PSLn(F) is simple with two exceptions: the groups PSL2(F)when |F| = 2 or 3.

2.7.2. Orthogonal groups. The orthogonal group is the group

On(F) = {A GLn(F) | AA = I}

It is easily verified that On(F) is a subgroup of GLn(F). Note that since det A =

det A, we must have (det A)2 = 1. Moreover there exist orthogonal matrices ofdeterminant 1 (note that if characteristic ofF is 2, then 1 = 1 so these statementsare true but vacuous). The determinant map det: On(F) F is a homomorphism

with image {1, 1} and kernel

SOn(F) = On(F) SLn(F).

Hence by the first isomorphism theorem, if the characteristic ofF is not 2,

On(F)/SOn(F) = Z2.

2.7.3. Symplectic groups. Let J be the 2n 2n matrix:

J =

0 I

I 0

.

We define the symplectic group Sp2n(R) by

Sp2n(R) = {A GL2n(R) | AJ A = J}.

2.7.4. Lorentz group. Another interesting group arises as the symmetries ofspace-time. Note that the definition of the symplectic group in terms of the matrixJ would define a subgroup of GLn(R) for any matrix J. Define

J =

1 0 0 00 1 0 00 0 1 00 0 0 1

and set

L = {A GL4(R) | A

J A = J}


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2.7.5. Unitary groups. When working over the complex numbers, the natu-

ral bilinear form is now the Hermitian form x, y = x

y where the bar denotes thecomplex conjugate. the symmetries of Hermitian space are the unitary matrices

Un = {A GLn(C) | AA = I}

Again we can look at the determinant map det: Un R. IfA Un and det A = z,

then it follows from the multiplicativity of the determinant that ||z|| = zz = 1.Hence the image of the determinant is contained inside the unit circle S1 of complexnumbers of modulus one. Again one can find suitable matrices A of any suchdeterminant, so that the image of det is the whole of S1. We define

SUn = Un SLn(C)

Thus the kernel of det is SUn and the first isomorphism gives

Un/SUn= S

1


24/25

APPENDIX A

The Isomorphism Theorems

Theorem A.0.2 (The First Isomorphism Theorem). Let G be a group and let : G H be a homomorphism. then G/ ker = (G).

Proof. Let K = ker . Define a map : G/K (G) by (gK) = (g).

We must first verify that this map is well-defined: that is, if gK = gK, then(g) = (g). But if gK = gK, then g = gk for some k K and so (g) =(gk) = (g)(k) = (g)e = (g). Thus is well-defined.

For any gK,gK G/K, (gK gK) = (ggK) = (gg) = (g)(g) =(gK)(gK). So is a homomorphism.

Suppose that (gK) = eH, then (g) = eH and g K; hence gK = K = eG/K,and so is injective.

Finally, let h (G). Then h = (g) for some g G. Hence h = (gK),proving that is surjective. Thus is an isomorphism, as required.

Lemma A.0.3. Let G be a group, let K be a normal subgroup of G and let Hbe a subgroup of G. Then

(1) HK = KH and this is a subgroup of G(2) H K is a normal subgroup of H

Proof. Exercise

Theorem A.0.4 (The Second Isomorphism Theorem). Let G be a group, letK be a normal subgroup of G and let H be a subgroup of G. Then HK/K =H/(H K).

Proof. Define a map : H HK/K by (h) = hK. Then is clearlya homomorphism because (hh) = hhK = hKhK = (h)(h). Now (h) =eHK/K (h) = K hK = K h K. So ker = H K. Finally

is clearly surjective since an arbitrary element of HK/K is of the form hkK forsome h H, k K. Since kK = K, hkK = hK and we see that every element ofHK/K is of the form hK for some h H. The result then follows from the firstisomorphism theorem.

Theorem A.0.5 (The Third Isomorphism Theorem). Let G be a group, letH, K be normal subgroups of G with H K. Then K/H is a normal subgroup ofG/H and

(G/H)/(K/H) = G/K.

Proof. Define : G/H G/K by (gH) = gK. One shows that is a well-defined homomorphism with ker = K/H. The result then follows again from the

first isomorphism theorem. The details are left as an exercise for the reader.

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A. THE ISOMORPHISM THEOREMS 25

Theorem A.0.6 (The Fourth Isomorphism Theorem). LetG be a group and let

N be a normal subgroup of G. Then there is a one-to-one correspondence betweensubgroups of G containing N and subgroups of G/N given by K K/N. Thecorrespondence also is a one-to-one correspondence between normal subgroups of Gcontaining N and normal subgroups of G/N. The correspondence also preservesinclusions and intersections.

Proof. Let K be a subgroup of G/N and let K = {g G | gN K/N}.Clearly K N. We claim it is a subgroup. let g, g K. Then ggN =(gN)(gN) K since K is a subgroup of G/N. Similarly if g K, thengN G/N. But (gN)1 = g1N G/N so g1 K. In the other direction, if Kis a subgroup of G containing N, then it is easily seen that K/N = {kN | k K}is as subgroup of G/N.

We leave it as an exercise to prove that the correspondence takes normal sub-groups to normal subgroups and that it preserves inclusions and intersections.

algebraic structures part 1

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