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Part IB — Groups, Rings and Modules
Based on lectures by O. Randal-Williams
Notes taken by Dexter Chua
Lent 2016
These notes are not endorsed by the lecturers, and I have modified them (often significantly) after lectures. They are nowhere near accurate representations of what was actually lectured, and in particular, all errors are almost surely mine. Groups Basic concepts of group theory recalled from Part IA Groups. Normal subgroups, quotient groups and isomorphism theorems. Permutation groups. Groups acting on sets, permutation representations. Conjugacy classes, centralizers and normalizers. The centre of a group. Elementary properties of finite p-groups. Examples of finite linear groups and groups arising from geometry. Simplicity of An. Sylow subgroups and Sylow theorems. Applications, groups of small order. [8]
Rings Definition and examples of rings (commutative, with 1). Ideals, homomorphisms, quotient rings, isomorphism theorems. Prime and maximal ideals. Fields. The characteristic of a field. Field of fractions of an integral domain. Factorization in rings; units, primes and irreducibles. Unique factorization in principal ideal domains, and in polynomial rings. Gauss’ Lemma and Eisenstein’s irreducibility criterion. Rings Z[α] of algebraic integers as subsets of C and quotients of Z[x]. Examples of Euclidean domains and uniqueness and non-uniqueness of factorization. Factorization in the ring of Gaussian integers; representation of integers as sums of two squares. Ideals in polynomial rings. Hilbert basis theorem. [10]
Modules Definitions, examples of vector spaces, abelian groups and vector spaces with an endomorphism. Sub-modules, homomorphisms, quotient modules and direct sums. Equivalence of matrices, canonical form. Structure of finitely generated modules over Euclidean domains, applications to abelian groups and Jordan normal form. [6]
Contents IB Groups, Rings and Modules
- 0 Introduction Contents
- 1 Groups
- 1.1 Basic concepts
- 1.2 Normal subgroups, quotients, homomorphisms, isomorphisms
- 1.3 Actions of permutations
- 1.4 Conjugacy, centralizers and normalizers
- 1.5 Finite p-groups
- 1.6 Finite abelian groups
- 1.7 Sylow theorems
- 2 Rings
- 2.1 Definitions and examples
- 2.2 Homomorphisms, ideals, quotients and isomorphisms
- 2.3 Integral domains, field of factions, maximal and prime ideals
- 2.4 Factorization in integral domains
- 2.5 Factorization in polynomial rings
- 2.6 Gaussian integers
- 2.7 Algebraic integers
- 2.8 Noetherian rings
- 3 Modules
- 3.1 Definitions and examples
- 3.2 Direct sums and free modules
- 3.3 Matrices over Euclidean domains
- 3.4 Modules over F[X] and normal forms for matrices
- 3.5 Conjugacy of matrices*
- Index
1 Groups
1.1 Basic concepts
We will begin by quickly recapping some definitions and results from IA Groups.
Definition (Group). A group is a triple (G, · , e), where G is a set, · : G×G → G is a function and e ∈ G is an element such that
(i) For all a, b, c ∈ G, we have (a · b) · c = a · (b · c). (associativity)
(ii) For all a ∈ G, we have a · e = e · a = a. (identity)
(iii) For all a ∈ G, there exists a−^1 ∈ G such that a · a−^1 = a−^1 · a = e.(inverse)
Some people add a stupid axiom that says g · h ∈ G for all g, h ∈ G, but this is already implied by saying · is a function to G. You can write that down as well, and no one will say you are stupid. But they might secretly think so.
Lemma. The inverse of an element is unique.
Proof. Let a−^1 , b be inverses of a. Then
b = b · e = b · a · a−^1 = e · a−^1 = a−^1.
Definition (Subgroup). If (G, · , e) is a group and H ⊆ G is a subset, it is a subgroup if
(i) e ∈ H,
(ii) a, b ∈ H implies a · b ∈ H,
(iii) · : H × H → H makes (H, · , e) a group.
We write H ≤ G if H is a subgroup of G.
Note that the last condition in some sense encompasses the first two, but we need the first two conditions to hold before the last statement makes sense at all.
Lemma. H ⊆ G is a subgroup if H is non-empty and for any h 1 , h 2 ∈ H, we have h 1 h− 2 1 ∈ H.
Definition (Abelian group). A group G is abelian if a · b = b · a for all a, b ∈ G.
Example. We have the following familiar examples of groups
(i) (Z, +, 0), (Q, +, 0), (R, +, 0), (C, +, 0).
(ii) We also have groups of symmetries:
(a) The symmetric group Sn is the collection of all permutations of { 1 , 2 , · · · , n}. (b) The dihedral group D 2 n is the symmetries of a regular n-gon. (c) The group GLn(R) is the group of invertible n × n real matrices, which also is the group of invertible R-linear maps from the vector space Rn^ to itself.
(iii) The alternating group An ≤ Sn.
(iv) The cyclic group Cn ≤ D 2 n.
(v) The special linear group SLn(R) ≤ GLn(R), the subgroup of matrices of determinant 1.
(vi) The Klein-four group C 2 × C 2.
(vii) The quaternions Q 8 = {± 1 , ±i, ±j, ±k} with ij = k, ji = −k, i^2 = j^2 = k^2 = −1, (−1)^2 = 1.
With groups and subgroups, we can talk about cosets.
Definition (Coset). If H ≤ G, g ∈ G, the left coset gH is the set
gH = {x ∈ G : x = g · h for some h ∈ H}.
For example, since H is a subgroup, we know e ∈ H. So for any g ∈ G, we must have g ∈ gH. The collection of H-cosets in G forms a partition of G, and furthermore, all H-cosets gH are in bijection with H itself, via h 7 → gh. An immediate consequence is
Theorem (Lagrange’s theorem). Let G be a finite group, and H ≤ G. Then
|G| = |H||G : H|,
where |G : H| is the number of H-cosets in G.
We can do exactly the same thing with right cosets and get the same conclusion. We have implicitly used the following notation:
Definition (Order of group). The order of a group is the number of elements in G, written |G|.
Instead of order of the group, we can ask what the order of an element is.
Definition (Order of element). The order of an element g ∈ G is the smallest positive n such that gn^ = e. If there is no such n, we say g has infinite order. We write ord(g) = n.
A basic lemma is as follows:
Lemma. If G is a finite group and g ∈ G has order n, then n | |G|.
Proof. Consider the following subset:
H = {e, g, g^2 , · · · , gn−^1 }.
This is a subgroup of G, because it is non-empty and gr^ g−s^ = gr−s^ is on the list (we might have to add n to the power of g to make it positive, but this is fine since gn^ = e). Moreover, there are no repeats in the list: if gi^ = gj^ , with wlog i ≥ j, then gi−j^ = e. So i − j < n. By definition of n, we must have i − j = 0, i.e. i = j. Hence Lagrange’s theorem tells us n = |H| | |G|.
This is indeed a group. Normality was defined such that this is well-defined. Multiplication is associative since multiplication in G is associative. The inverse of gH is g−^1 H, and eH is easily seen to be the identity. So far, we’ve just been looking at groups themselves. We would also like to know how groups interact with each other. In other words, we want to study functions between groups. However, we don’t allow arbitrary functions, since groups have some structure, and we would like the functions to respect the group structures. These nice functions are known as homomorphisms.
Definition (Homomorphism). If (G, · , eG) and (H, ∗, eH ) are groups, a function φ : G → H is a homomorphism if φ(eG) = eH , and for g, g′^ ∈ G, we have
φ(g · g′) = φ(g) ∗ φ(g′).
If we think carefully, φ(eG) = eH can be derived from the second condition, but it doesn’t hurt to put it in as well.
Lemma. If φ : G → H is a homomorphism, then
φ(g−^1 ) = φ(g)−^1.
Proof. We compute φ(g · g−^1 ) in two ways. On the one hand, we have
φ(g · g−^1 ) = φ(e) = e.
On the other hand, we have
φ(g · g−^1 ) = φ(g) ∗ φ(g−^1 ).
By the uniqueness of inverse, we must have
φ(g−^1 ) = φ(g)−^1.
Given any homomorphism, we can build two groups out of it:
Definition (Kernel). The kernel of a homomorphism φ : G → H is
ker(φ) = {g ∈ G : φ(g) = e}.
Definition (Image). The image of a homomorphism φ : G → H is
im(φ) = {h ∈ H : h = φ(g) for some g ∈ G}.
Lemma. For a homomorphism φ : G → H, the kernel ker (φ) is a normal subgroup, and the image im(φ) is a subgroup of H.
Proof. There is only one possible way we can prove this. To see ker(φ) is a subgroup, let g, h ∈ ker φ. Then
φ(g · h−^1 ) = φ(g) ∗ φ(h)−^1 = e ∗ e−^1 = e.
So gh−^1 ∈ ker φ. Also, φ(e) = e. So ker(φ) is non-empty. So it is a subgroup. To show it is normal, let g ∈ ker (φ). Let x ∈ G. We want to show x−^1 gx ∈ ker(φ). We have
φ(x−^1 gx) = φ(x−^1 ) ∗ φ(g) ∗ φ(x) = φ(x−^1 ) ∗ φ(x) = φ(x−^1 x) = φ(e) = e.
So x−^1 gx ∈ ker(φ). So ker(φ) is normal. Also, if φ(g), φ(h) ∈ im(φ), then
φ(g) ∗ φ(h)−^1 = φ(gh−^1 ) ∈ im(φ).
Also, e ∈ im(φ). So im(φ) is non-empty. So im(φ) is a subgroup.
Definition (Isomorphism). An isomorphism is a homomorphism that is also a bijection. Definition (Isomorphic group). Two groups G and H are isomorphic if there is an isomorphism between them. We write G ∼= H. Usually, we identify two isomorphic groups as being “the same”, and do not distinguish isomorphic groups. It is an exercise to show the following: Lemma. If φ is an isomorphism, then the inverse φ−^1 is also an isomorphism. When studying groups, it is often helpful to break the group apart into smaller groups, which are hopefully easier to study. We will have three isomorphism theorems to do so. These isomorphism theorems tell us what happens when we take quotients of different things. Then if a miracle happens, we can patch what we know about the quotients together to get information about the big group. Even if miracles do not happen, these are useful tools to have. The first isomorphism relates the kernel to the image.
Theorem (First isomorphism theorem). Let φ : G → H be a homomorphism. Then ker(φ) C G and G ker(φ)
∼= im(φ).
Proof. We have already proved that ker (φ) is a normal subgroup. We now have to construct a homomorphism f : G/ ker (φ) → im (φ), and prove it is an isomorphism. Define our function as follows:
f :
G
ker(φ)
→ im(φ)
g ker(φ) 7 → φ(g).
We first tackle the obvious problem that this might not be well-defined, since we are picking a representative for the coset. If g ker (φ) = g′^ ker (φ), then we know g−^1 · g′^ ∈ ker(φ). So φ(g−^1 · g′) = e. So we know
e = φ(g−^1 · g′) = φ(g)−^1 ∗ φ(g′).
Multiplying the whole thing by φ(g) gives φ(g) = φ(g′). Hence this function is well-defined. Next we show it is a homomorphism. To see f is a homomorphism, we have
f (g ker(φ) · g′^ ker(φ)) = f (gg′^ ker(φ)) = φ(gg′) = φ(g) ∗ φ(g′) = f (g ker(φ)) ∗ f (g′^ ker(φ)).
Then the kernel of φ is
ker(φ) = {h ∈ H : hK = eK} = {h ∈ H : h ∈ K} = H ∩ K.
So the first isomorphism theorem says
H H ∩ K
∼= HK
K
Notice we did more work than we really had to. We could have started by writing down φ and checked it is a homomorphism. Then since H ∩ K is its kernel, it has to be a normal subgroup. Before we move on to the third isomorphism theorem, we notice that if K C G, then there is a bijection between subgroups of G/K and subgroups of G containing K, given by
{subgroups of G/K} ←→ {subgroups of G which contain K}
X ≤
G
K
−→ {g ∈ G : gK ∈ X} L K
G
K
←− K C L ≤ G.
This specializes to the bijection of normal subgroups:
{normal subgroups of G/K} ←→ {normal subgroups of G which contain K}
using the same bijection. It is an elementary exercise to show that these are inverses of each other. This correspondence will be useful in later times.
Theorem (Third isomorphism theorem). Let K ≤ L ≤ G be normal subgroups of G. Then G K
/ L
K
∼= G
L
Proof. Define the homomorphism
φ : G/K → G/L gK 7 → gL
As always, we have to check this is well-defined. If gK = g′K, then g−^1 g′^ ∈ K ⊆ L. So gL = g′L. This is also a homomorphism since
φ(gK · g′K) = φ(gg′K) = gg′L = (gL) · (g′L) = φ(gK) · φ(g′K).
This clearly is surjective, since any coset gL is the image φ(gK). So the image is G/L. The kernel is then
ker(φ) = {gK : gL = L} = {gK : g ∈ L} =
L
K
So the conclusion follows by the first isomorphism theorem.
The general idea of these theorems is to take a group, find a normal subgroup, and then quotient it out. Then hopefully the normal subgroup and the quotient group will be simpler. However, this doesn’t always work.
Definition (Simple group). A (non-trivial) group G is simple if it has no normal subgroups except {e} and G.
In general, simple groups are complicated. However, if we only look at abelian groups, then life is simpler. Note that by commutativity, the normality condition is always trivially satisfied. So any subgroup is normal. Hence an abelian group can be simple only if it has no non-trivial subgroups at all.
Lemma. An abelian group is simple if and only if it is isomorphic to the cyclic group Cp for some prime number p.
Proof. By Lagrange’s theorem, any subgroup of Cp has order dividing |Cp| = p. Hence if p is prime, then it has no such divisors, and any subgroup must have order 1 or p, i.e. it is either {e} or Cp itself. Hence in particular any normal subgroup must be {e} or Cp. So it is simple. Now suppose G is abelian and simple. Let e 6 = g ∈ G be a non-trivial element, and consider H = {· · · , g−^2 , g−^1 , e, g, g^2 , · · · }. Since G is abelian, conjugation does nothing, and every subgroup is normal. So H is a normal subgroup. As G is simple, H = {e} or H = G. Since it contains g 6 = e, it is non-trivial. So we must have H = G. So G is cyclic. If G is infinite cyclic, then it is isomorphic to Z. But Z is not simple, since 2 Z C Z. So G is a finite cyclic group, i.e. G ∼= Cm for some finite m. If n | m, then gm/n^ generates a subgroup of G of order n. So this is a normal subgroup. Therefore n must be m or 1. Hence G cannot be simple unless m has no divisors except 1 and m, i.e. m is a prime.
One reason why simple groups are important is the following:
Theorem. Let G be any finite group. Then there are subgroups
G = H 1 B H 2 B H 3 B H 4 B · · · B Hn = {e}.
such that Hi/Hi+1 is simple.
Note that here we only claim that Hi+1 is normal in Hi. This does not say that, say, H 3 is a normal subgroup of H 1.
Proof. If G is simple, let H 2 = {e}. Then we are done. If G is not simple, let H 2 be a maximal proper normal subgroup of G. We now claim that G/H 2 is simple. If G/H 2 is not simple, it contains a proper non-trivial normal subgroup L C G/H 2 such that L 6 = {e}, G/H 2. However, there is a correspondence between normal subgroups of G/H 2 and normal subgroups of G containing H 2. So L must be K/H 2 for some K C G such that K ≥ H 2. Moreover, since L is non-trivial and not G/H 2 , we know K is not G or H 2. So K is a larger normal subgroup. Contradiction. So we have found an H 2 C G such that G/H 2 is simple. Iterating this process on H 2 gives the desired result. Note that this process eventually stops, as Hi+1 < Hi, and hence |Hi+1| < |Hi|, and all these numbers are finite.
However, we don’t always want the whole symmetric group. Sometimes, we just want some subgroups of symmetric groups, as in our initial motivation. So we make the following definition.
Definition (Permutation group). A group G is called a permutation group if it is a subgroup of Sym(X) for some X, i.e. it is given by some, but not necessarily all, permutations of some set. We say G is a permutation group of order n if in addition |X| = n.
This is not really a too interesting definition, since, as we will soon see, every group is (isomorphic to) a permutation group. However, in some cases, thinking of a group as a permutation group of some object gives us better intuition on what the group is about.
Example. Sn and An are obviously permutation groups. Also, the dihedral group D 2 n is a permutation group of order n, viewing it as a permutation of the vertices of a regular n-gon.
We would next want to recover the idea of a group being a “permutation”. If G ≤ Sym(X), then each g ∈ G should be able to give us a permutation of X, in a way that is consistent with the group structure. We say the group G acts on X. In general, we make the following definition:
Definition (Group action). An action of a group (G, · ) on a set X is a function
∗ : G × X → X
such that
(i) g 1 ∗ (g 2 ∗ x) = (g 1 · g 2 ) ∗ x for all g 1 , g 2 ∈ G and x ∈ X.
(ii) e ∗ x = x for all x ∈ X.
There is another way of defining group actions, which is arguably a better way of thinking about group actions.
Lemma. An action of G on X is equivalent to a homomorphism φ : G → Sym(X).
Note that the statement by itself is useless, since it doesn’t tell us how to translate between the homomorphism and a group action. The important part is the proof.
Proof. Let ∗ : G × X → X be an action. Define φ : G → Sym(X) by sending g to the function φ(g) = (g ∗ · : X → X). This is indeed a permutation — g−^1 ∗ · is an inverse since
φ(g−^1 )(φ(g)(x)) = g−^1 ∗ (g ∗ x) = (g−^1 · g) ∗ x = e ∗ x = x,
and a similar argument shows φ(g) ◦ φ(g−^1 ) = idX. So φ is at least a well-defined function. To show it is a homomorphism, just note that
φ(g 1 )(φ(g 2 )(x)) = g 1 ∗ (g 2 ∗ x) = (g 1 · g 2 ) ∗ x = φ(g 1 · g 2 )(x).
Since this is true for all x ∈ X, we know φ(g 1 ) ◦ φ(g 2 ) = φ(g 1 · g 2 ). Also, φ(e)(x) = e ∗ x = x. So φ(e) is indeed the identity. Hence φ is a homomorphism. We now do the same thing backwards. Given a homomorphism φ : G → Sym(X), define a function by g ∗ x = φ(g)(x). We now check it is indeed a group action. Using the definition of a homomorphism, we know
(i) g 1 ∗ (g 2 ∗ x) = φ(g 1 )(φ(g 2 )(x)) = (φ(g 1 ) ◦ φ(g 2 ))(x) = φ(g 1 · g 2 )(x) = (g 1 · g 2 ) ∗ x.
(ii) e ∗ x = φ(e)(x) = idX (x) = x.
So this homomorphism gives a group action. These two operations are clearly in- verses to each other. So group actions of G on X are the same as homomorphisms G → Sym(X).
Definition (Permutation representation). A permutation representation of a group G is a homomorphism G → Sym(X).
We have thus shown that a permutation representation is the same as a group action. The good thing about thinking of group actions as homomorphisms is that we can use all we know about homomorphisms on them.
Notation. For an action of G on X given by φ : G → Sym(X), we write GX^ = im(φ) and GX = ker(φ).
The first isomorphism theorem immediately gives
Proposition. GX C G and G/GX ∼= GX^.
In particular, if GX = {e} is trivial, then G ∼= GX^ ≤ Sym(X).
Example. Let G be the group of symmetries of a cube. Let X be the set of diagonals of the cube.
Then G acts on X, and so we get φ : G → Sym(X). What is its kernel? To pre- serve the diagonals, it either does nothing to the diagonal, or flips the two vertices. So GX = ker(φ) = {id, symmetry that sends each vertex to its opposite} ∼= C 2. How about the image? We have GX^ = im (φ) ≤ Sym(X) ∼= S 4. It is an exercise to show that im (φ) = Sym (X), i.e. that φ is surjective. We are not proving this because this is an exercise in geometry, not group theory. Then the first isomorphism theorem tells us
GX^ ∼= G/GX.
So |G| = |GX^ ||GX | = 4! · 2 = 48.
Proof. We apply the previous example, giving φ : G → Sym(G/H), and let K be the kernel of this homomorphism. We have already shown that K ≤ H. Then the first isomorphism theorem gives
G/K ∼= im φ ≤ Sym(G/H) ∼= Sn.
Then by Lagrange’s theorem, we know |G/K| | |Sn| = n!, and we also have |G/K| ≥ |G/H| = n.
Corollary. Let G be a non-abelian simple group. Let H ≤ G be a proper subgroup of index n. Then G is isomorphic to a subgroup of An. Moreover, we must have n ≥ 5, i.e. G cannot have a subgroup of index less than 5. Proof. The action of G on X = G/H gives a homomorphism φ : G → Sym (X). Then ker (φ) C G. Since G is simple, ker (φ) is either G or {e}. We first show that it cannot be G. If ker (φ) = G, then every element of G acts trivially on X = G/H. But if g ∈ G \ H, which exists since the index of H is not 1, then g ∗ H = gH 6 = H. So g does not act trivially. So the kernel cannot be the whole of G. Hence ker(φ) = {e}. Thus by the first isomorphism theorem, we get
G ∼= im(φ) ≤ Sym(X) ∼= Sn.
We now need to show that G is in fact a subgroup of An. We know An C Sn. So im (φ) ∩ An C im (φ) ∼= G. As G is simple, im(φ) ∩ An is either {e} or G = im (φ). We want to show that the second thing happens, i.e. the intersection is not the trivial group. We use the second isomorphism theorem. If im(φ) ∩ An = {e}, then
im(φ) ∼=
im(φ) im(φ) ∩ An
∼= im(φ)An An
Sn An
∼= C 2.
So G ∼= im(φ) is a subgroup of C 2 , i.e. either {e} or C 2 itself. Neither of these are non-abelian. So this cannot be the case. So we must have im(φ) ∩ An = im(φ), i.e. im(φ) ≤ An. The last part follows from the fact that S 1 , S 2 , S 3 , S 4 have no non-abelian simple subgroups, which you can check by going to a quiet room and listing out all their subgroups.
Let’s recall some old definitions from IA Groups.
Definition (Orbit). If G acts on a set X, the orbit of x ∈ X is
G · x = {g ∗ x ∈ X : g ∈ G}.
Definition (Stabilizer). If G acts on a set X, the stabilizer of x ∈ X is
Gx = {g ∈ G : g ∗ x = x}.
The main theorem about these concepts is the orbit-stabilizer theorem.
Theorem (Orbit-stabilizer theorem). Let G act on X. Then for any x ∈ X, there is a bijection between G · x and G/Gx, given by g · x ↔ g · Gx. In particular, if G is finite, it follows that
|G| = |Gx||G · x|.
It takes some work to show this is well-defined and a bijection, but you’ve done it in IA Groups. In IA Groups, you probably learnt the second statement instead, but this result is more generally true for infinite groups.
1.4 Conjugacy, centralizers and normalizers
We have seen that every group acts on itself by multiplying on the left. A group G can also act on itself in a different way, by conjugation:
g ∗ g 1 = gg 1 g−^1.
Let φ : G → Sym(G) be the associated permutation representation. We know, by definition, that φ(g) is a bijection from G to G as sets. However, here G is not an arbitrary set, but is a group. A natural question to ask is whether φ(g) is a homomorphism or not. Indeed, we have
φ(g)(g 1 · g 2 ) = gg 1 g 2 g−^1 = (gg 1 g−^1 )(gg 2 g−^1 ) = φ(g)(g 1 )φ(g)(g 2 ).
So φ(g) is a homomorphism from G to G. Since φ(g) is bijective (as in any group action), it is in fact an isomorphism. Thus, for any group G, there are many isomorphisms from G to itself, one for every g ∈ G, and can be obtained from a group action of G on itself. We can, of course, take the collection of all isomorphisms of G, and form a new group out of it. Definition (Automorphism group). The automorphism group of G is
Aut(G) = {f : G → G : f is a group isomorphism}.
This is a group under composition, with the identity map as the identity.
This is a subgroup of Sym(G), and the homomorphism φ : G → Sym(G) by conjugation lands in Aut(G). This is pretty fun — we can use this to cook up some more groups, by taking a group and looking at its automorphism group. We can also take a group, take its automorphism group, and then take its automorphism group again, and do it again, and see if this process stabilizes, or becomes periodic, or something. This is left as an exercise for the reader. Definition (Conjugacy class). The conjugacy class of g ∈ G is
cclG(g) = {hgh−^1 : h ∈ G},
i.e. the orbit of g ∈ G under the conjugation action.
Definition (Centralizer). The centralizer of g ∈ G is
CG(g) = {h ∈ G : hgh−^1 = g},
i.e. the stabilizer of g under the conjugation action. This is alternatively the set of all h ∈ G that commute with g. Definition (Center). The center of a group G is
Z(G) = {h ∈ G : hgh−^1 = g for all g ∈ G} =
g∈G
CG(g) = ker(φ).
We show that if H contains a 3-cycle, then every 3-cycle is in H. Then we are done since An is generated by 3-cycles. For concreteness, suppose we know (a b c) ∈ H, and we want to show (1 2 3) ∈ H. Since they have the same cycle type, so we have σ ∈ Sn such that (a b c) = σ(1 2 3)σ−^1. If σ is even, i.e. σ ∈ An, then we have that (1 2 3) ∈ σ−^1 Hσ = H, by the normality of H and we are trivially done. If σ is odd, replace it by σ¯ = σ · (4 5). Here is where we use the fact that n ≥ 5 (we will use it again later). Then we have
¯σ(1 2 3)¯σ−^1 = σ(4 5)(1 2 3)(4 5)σ−^1 = σ(1 2 3)σ−^1 = (a b c),
using the fact that (1 2 3) and (4 5) commute. Now ¯σ is even. So (1 2 3) ∈ H as above. What we’ve got so far is that if H C An contains any 3-cycle, then it is An. Finally, we have to show that every normal subgroup must contain at least one 3-cycle. Claim. Let H C An be non-trivial. Then H contains a 3-cycle. We separate this into many cases
(i) Suppose H contains an element which can be written in disjoint cycle notation σ = (1 2 3 · · · r)τ, for r ≥ 4. We now let δ = (1 2 3) ∈ An. Then by normality of H, we know δ−^1 σδ ∈ H. Then σ−^1 δ−^1 σδ ∈ H. Also, we notice that τ does not contain 1 , 2 , 3. So it commutes with δ, and also trivially with (1 2 3 · · · r). We can expand this mess to obtain
σ−^1 δ−^1 σδ = (r · · · 2 1)(1 3 2)(1 2 3 · · · r)(1 2 3) = (2 3 r),
which is a 3-cycle. So done. The same argument goes through if σ = (a 1 a 2 · · · ar )τ for any a 1 , · · · , an.
(ii) Suppose H contains an element consisting of at least two 3-cycles in disjoint cycle notation, say σ = (1 2 3)(4 5 6)τ We now let δ = (1 2 4), and again calculate
σ−^1 δ−^1 σδ = (1 3 2)(4 6 5)(1 4 2)(1 2 3)(4 5 6)(1 2 4) = (1 2 4 3 6).
This is a 5-cycle, which is necessarily in H. By the previous case, we get a 3-cycle in H too, and hence H = An.
(iii) Suppose H contains σ = (1 2 3)τ , with τ a product of 2-cycles (if τ contains anything longer, then it would fit in one of the previous two cases). Then σ^2 = (1 2 3)^2 = (1 3 2) is a three-cycle.
(iv) Suppose H contains σ = (1 2)(3 4)τ , where τ is a product of 2-cycles. We first let δ = (1 2 3) and calculate
u = σ−^1 δ−^1 σδ = (1 2)(3 4)(1 3 2)(1 2)(3 4)(1 2 3) = (1 4)(2 3),
which is again in u. We landed in the same case, but instead of two transpositions times a mess, we just have two transpositions, which is nicer. Now let v = (1 5 2)u(1 2 5) = (1 3)(4 5) ∈ H. Note that we used n ≥ 5 again. We have yet again landed in the same case. Notice however, that these are not the same transpositions. We multiply
uv = (1 4)(2 3)(1 3)(4 5) = (1 2 3 4 5) ∈ H.
This is then covered by the first case, and we are done.
So done. Phew.
1.5 Finite p-groups
Note that when studying the orders of groups and subgroups, we always talk about divisibility, since that is what Lagrange’s theorem tells us about. We never talk about things like the sum of the orders of two subgroups. When it comes to divisibility, the simplest case would be when the order is a prime, and we have done that already. The next best thing we can hope for is that the order is a power of a prime.
Definition (p-group). A finite group G is a p-group if |G| = pn^ for some prime number p and n ≥ 1.
Theorem. If G is a finite p-group, then Z(G) = {x ∈ G : xg = gx for all g ∈ G} is non-trivial.
This immediately tells us that for n ≥ 2, a p group is never simple.
Proof. Let G act on itself by conjugation. The orbits of this action (i.e. the conjugacy classes) have order dividing |G| = pn. So it is either a singleton, or its size is divisible by p. Since the conjugacy classes partition G, we know the total size of the conjugacy classes is |G|. In particular,
|G| = number of conjugacy class of size 1
order of all other conjugacy classes.
We know the second term is divisible by p. Also |G| = pn^ is divisible by p. Hence the number of conjugacy classes of size 1 is divisible by p. We know {e} is a conjugacy class of size 1. So there must be at least p conjugacy classes of size 1. Since the smallest prime number is 2, there is a conjugacy class {x} 6 = {e}. But if {x} is a conjugacy class on its own, then by definition g−^1 xg = x for all g ∈ G, i.e. xg = gx for all g ∈ G. So x ∈ Z(G). So Z(G) is non-trivial.
The theorem allows us to prove interesting things about p-groups by induction — we can quotient G by Z(G), and get a smaller p-group. One way to do this is via the below lemma.
Lemma. For any group G, if G/Z(G) is cyclic, then G is abelian. In other words, if G/Z(G) is cyclic, then it is in fact trivial, since the center of an abelian group is the abelian group itself.
