551, Fall '02         A3a.    The Noether Isomorphism Theorems         p. 1

Math 551 – Algebra – Fall 2002

Richard Lyons Rutgers University New Brunswick, New Jersey, USA

A. Groups

3. The Meat-Axe: Noether Isomorphism Theorems and the Jordan-Hölder Theorem

If K ◁ G, then G is broken up (in some way) into the groups K and G/K. In general G cannot be recovered just from K and G/K (for example, both G = Z_{2} × Z_{2} and G = Z_{4} have subgroups K ≅ Z_{2} such that G/K ≅ Z_{2}, though Z_{4} ≅ Z̸_{2} × Z_{2}). Nevertheless the pair K and G/K contain a significant amount of information about the structure of G, and we pursue this idea in this section. For instance, if H ≤ G, how does the “cut” K cut through H? And if H ◁ G how does it cut through G/H? What happens if we cut up a group G “as much as possible”, and is there indeed a maximal way to do it?

3a. The Noether Isomorphism Theorems

We begin with the three fundamental isomorphism theorems often named for Emmy Noether, the cigar-smoking pioneer of “modern” algebra.

Theorem. (First Isomorphism Theorem) Let ϕ : G → H be a homomorphism of groups, and let K = ker ϕ. Then there exists a unique isomorphism \overline{ϕ} : G/K → ϕ(G) such that the following diagram commutes:

\begin{matrix}G & \underrightarrow{ϕ} & H \\ ↓ & & ↑ \\ G/K & \underrightarrow{ϕ̄} & ϕ(G) \end{matrix}

Here the mapping on the left is π_{K} and the mapping on the right is the inclusion of ϕ(G) in K.

Proof. In fact there is a unique mapping \overline{ϕ} such that the diagram commutes; the only possibility is to define \overline{ϕ}(gK) = ϕ(g). Since ϕ(g) = ϕ(g′) whenever gK = g′K, this is well-defined. Moreover \overline{ϕ}(gKg′K) = \overline{ϕ}(gg′K) = ϕ(gg′) = ϕ(g)ϕ(g′) = \overline{ϕ}(gK)\overline{ϕ}(g′K) so \overline{ϕ} is a homomorphism. Clearly ϕ(G) = \overline{ϕ}(G/K) so \overline{ϕ} is surjective; also gK ∈ ker \overline{ϕ} ⟺ ϕ(g) = 1 ⟺ g ∈ ker ϕ = K ⟺ gK = K, so \overline{ϕ} is injective. Hence \overline{ϕ} is an isomorphism.

QED

p. 2         A3.  Noether and Jordan-Hölder Theorems         551, Fall '02

Corollary. If ϕ is any homomorphism from G to H, then G/ ker ϕ ≅ ϕ(G). If ϕ is surjective, then G/ ker ϕ ≅ H.

1. Let H ≤ G. Then we obtain a homomorphism ϕ : G → Σ_{G/H} whose image is transitive on G/H and whose kernel is K = ∩_{g∈G}^{g}H. In particular K ≤ H and G/K is isomorphic to a transitive subgroup of Σ_{G/H}.

2. Applying this to G ≅ Σ_{4} and H ∈ Syl_{2}(G), we have |H| = 8 so |G : H| = 3. We get a homomorphism Σ_{4} → Σ_{3} whose image I is transitive and whose kernel K is a normal subgroup such that Σ_{4}/K ≅ I. The only possibility is K = V, and so Σ_{4}/V ≅ Σ_{3}.

Theorem. (Second Isomorphism Theorem, or Parallelogram Law) Let H and K be subgroups of G and suppose that H ≤ N_{G}(K). Then HK ≤ G, K ◁ HK and H ∩ K ◁ H, and

H/H ∩ K ≅ HK/K

via the isomorphism h(h ∩ K) ↦ hK (h ∈ H).

Proof. We are given that ^{h}K = K for all h ∈ H, so that hK = Kh for all such h. Therefore HK = KH, which implies as before that HK ≤ G: (HK)(HK) = H(KH)K = H(HK)K = HHKK = HK, etc. Moreover, N_{HK}(K) is a subgroup of HK containing K and H, so equals HK, and so K ◁ HK. Now consider the homomorphism which is the composite of the inclusion and the canonical projection:

ϕ : H → HK → HK/K.

For any h ∈ H and k ∈ K, we have hkK = hK = ϕ(h), so ϕ is surjective. Moreover for any h ∈ H, we have h ∈ ker ϕ ⟺ hK = K ⟺ h ∈ K ⟺ h ∈ H ∩ K, so ker ϕ = H ∩ K. The first isomorphism theorem then implies that the mapping

\overline{ϕ} : H/H ∩ K → HK/K taking h(H ∩ K) ↦ hK

is an isomorphism. QED

Corollary. Suppose that K ◁ G and H is a subgroup of G such that H ∩ K = 1. Then G/K has a subgroup isomorphic to H. Moreover if HK = G then G/K ≅ H.

However, if K ◁ G it is not necessarily the case that H ∩K = 1 for any nontrivial subgroup of G, let alone that H ∩ K = 1 and G = HK for some H (in which case we say that G splits over K). For instance, in G = Z_{4}, there is a unique subgroup K of order 2, but there is no subgroup H of G such that H ∩ K = 1, other than the trivial subgroup H = 1.

Theorem. (The Third Isomorphism Theorem, or the Correspondence Theorem) Let ϕ : G → H be a surjective homomorphism, and let K = ker ϕ. Let S be the set of all subgroups of G containing K, and let T be the set of all subgroups of H. Then

1) For each S ∈ S, ϕ(S) ∈ T ; and for each T ∈ T , ϕ^{−1}(T) ∈ S.

551, Fall '02              A3b.    Groups with operators              p. 3

2) The mappings S ↦ ϕ(S) and T ↦ ϕ^{−1}(T) are mutually inverse bijections between S and T .

3) These bijections have the following further properties, for all S, S′ ∈ S and T, T′ ∈ T :

a) S ≤ S′ if and only if ϕ(S) ≤ ϕ(S′), and if these conditions hold then |S′ : S| = |ϕ(S′) : ϕ(S)|;

b) S ◁ S′ if and only if ϕ(S) ◁ ϕ(S′), and if these conditions hold then S′/S ≅ ϕ(S′)/ϕ(S);

c) ϕ(S ∩ S′) = ϕ(S) ∩ ϕ(S′).

d) ϕ(⟨S, S′⟩) = ⟨ϕ(S), ϕ(S′)⟩.

Proof. 1) is obvious (note that ϕ^{−1}(T) ⊇ ϕ^{−1}(1) = ker ϕ = K). To prove 2) we must show that ϕ(ϕ^{−1}(T)) = T and ϕ^{−1}(ϕ(S)) = S for all S ∈ S and T ∈ T . The first is a property of any surjective mapping of sets. As for the second, it is automatic from the definition of ϕ^{−1} that S ⊆ ϕ^{−1}(ϕ(S)). Let x ∈ ϕ^{−1}(ϕ(S)). Then ϕ(x) ∈ ϕ(S) so ϕ(x) = ϕ(y) for some y ∈ S. Then xy^{−1} ∈ ker ϕ = K ≤ S (recall S ∈ S!). Hence x = xy^{−1}y ∈ S, proving 2).

The first and third statements of 3) are left to the reader. (For the statement about indices, check that g_{1}S = g_{2}S ⟺ ϕ(g_{1})ϕ(S) = ϕ(g_{2})ϕ(S), the converse statement requiring S to contain ker ϕ. As for 3b), if S ◁ S′, then for every x ∈ S and g ∈ S′, an equation of the form ≅ gx = y holds for some y ∈ S′. Hence the image of this equation under ϕ is also true, whence ϕ(S) ◁ ϕ(S′). Conversely suppose that ϕ(S) ◁ ϕ(S′). Then the composite of ϕ|S′ and the canonical projection

ψ : S′ → ϕ(S′) → ϕ(S′)/ϕ(S)

is a homomorphism; both pieces are surjective so ψ is also surjective. Moreover x ∈ ker ψ ⟺ ϕ(x) ∈ ϕ(S) ⟺ x ∈ ϕ^{−1}ϕ(S) = S, so by the first isomorphism theorem S ◁ S′ and S′/S ≅ ϕ(S′)/ϕ(S). QED

Corollary. If H ≤ K ≤ G with H ◁ G and K ◁ G, then K/H ◁ G/H, and G/H/K/H ≅ G/K.

The proof is left to the reader.

These theorems answer the question above about how a normal subgroup K ◁ G cuts through a subgroup H of G or through a quotient G/H of G. Namely if H ≤ G, then H ∩ K ◁ H, and H ∩ K is a subgroup of K, while H/H ∩ K is isomorphic to a subgroup of G/K. On the other hand if H ◁ G, then HK ◁ G (easy to check), so HK/H ◁ G/H, with HK/H ≅ K/H ∩ K isomorphic to a quotient of K and G/H/HK/H ≅ G/HK ≅

G/K/HK/K isomorphic to a quotient of G/K.

3b. Groups with operators We pause to broaden the scope of this discussion, which has an almost trivial extension to a slightly more complex situation, and as a result the theory extends without essential change to vector spaces and modules, instead of just groups. The key notion is that of a group with operators, which consists of a group G and a set S which operates on G, but not in the sense of group action (for S itself need not be a group or indeed have any structure beyond that of a set). Instead, the only axiom is that the operators from S preserve the group structure on G.

Definition. Let G be a group and S a set. We say that G is an S-group (or a group with operators S) if there is defined a function

S × G → G, taking (s, g) ↦ sg,

such that s(gh) = (sg)(sh) for all s ∈ S and g, h ∈ G. Moreover, a subgroup H ≤ G is called an S-subgroup of G if and only if sg ∈ H for all s ∈ S and g ∈ H. If G and H are S-groups (for the same S) then a mapping ϕ : G → H is an S-homomorphism (resp. S-isomorphism) if and only if it is a homomorphism (resp. isomorphism) of groups and ϕ(sg) = sϕ(g) for all s ∈ S, g ∈ G. The two S-groups G and H are S-isomorphic if and only if there exists an S-isomorphism from one to the other.

The exponential notation g^{s} in place of sg is perhaps more suggestive, since the axiom for a group with operators then becomes

(gh)^{s} = g^{s}h^{s}.

Ex. A. Let V be a vector space over C (or more generally over any field F). The axioms for a vector space prescribe that V is an abelian group with respect to addition, and also that several axioms hold concerning scalar multiplication, so that V is a C-group (or an F -group), with respect to the operations of addition and scalar multiplication. Then a C-subgroup (or F -subgroup) of V is nothing other than a subspace; and a C-homomorphism (C-isomorphism) between two vector spaces is just a linear transformation (isomorphism of vector spaces). We then immediately get from the first theorem that if T : V → W is a linear transformation of vector spaces, then V/ ker(T) ≅ im(T) (as vector spaces), which immediately gives e.g. the “rank plus nullity” theorem: dim ker(T) + dim im(T) = dim V. Likewise the second theorem gives (W + X)/W ≅ X/(W ∩ X) for all subspaces W, X of a vector space V, and the third theorem implies that given a subspace W of a vector space V, the set of subspaces of V containing W is in one-to-one correspondence with the set of subspaces of V/W; also for W ≤ X ≤ V, V/X ≅ (V/W)/(X/W).

Ex. A’. Let M be a module over a ring R (same axioms as for a vector space, except that the scalars come from a ring R instead of from a field).

Ex. B. Let G be a group and consider G as a G-group via the definition

g · h = ^{g}h ∀g, h ∈ G.

551, Fall '02        A3c.  Normal Series; Jordan-Hölder Theorem        p. 5

Then the G-subgroups of G are the normal subgroups of G. A G-homomorphism from G to G is a homomorphism ϕ such that ^{g}(ϕ(h)) = ^{ϕ(g)}(ϕ(h)) for all g, h ∈ G. In particular the image of such a homomorphism is a normal subgroup of G, which is ordinarily not the case for homomorphisms.

Ex. B’. Any group G can be considered an Aut(G)-group via α · g = α(g). The Aut(G)- subgroups of G are called the characteristic subgroups of G. E.g., Z(G) is a characteristic subgroup of G.

Notice that this is also an action of Aut(G) on G, i.e., αβ(g) = α(β(g)). Thus for any characteristic subgroup N of G we obtain a homomorphism

Aut(G) → Aut(N), α ↦ α|N.

Ex. C. Let Γ be a group and V a vector space over a field F. A representation of Γ on V is a

homomorphism

ϕ : Γ → GL(V).

Such a homomorphism amounts to the structure of a Γ-group on V, by which Γ acts on V; the connection being

g · v = ϕ(g)(v),

Thus we may consider V to be a Γ ∪ F -group. The usual notion of equivalence of representations is just the notion of Γ ∪ F -isomorphism. That is, representations of Γ on V and W if and only if there exists an invertible abelian group homomorphism V → W preserving the actions of both Γ and F; i.e., a nonsingular linear transformation preserving the action of Γ. The representation is irreducible if and only if V is a simple Γ ∪ F -group.

Notice that given a representation ϕ : Γ → V, if W is a Γ ∪ F -subgroup of V, then both W and V/W are Γ ∪ F -groups and so give representations of Γ as well.

3c. Normal series and the Theorem of Jordan and Hölder

Definition. A (S-)group G ≠ {1} is a simple (S-)group if and only if there exist no normal (S-)subgroups of G other than 1 and G itself.

In other words simplicity is equivalent to having no nontrivial quotients. We have seen that A_{n} is simple for n ≥ 5.

Now let G be a (S-)group. In the case G = 1 there is nothing to discuss, so assume that G ≠ 1. If G is not simple then there exists a proper normal (S-)subgroup H, and so

1 ◁_{≠} H ◁_{≠} G.

If H and G/H are both (S)-simple we can stop; but otherwise a further term may be inserted, either between 1 and H, or between H and G, by the Correspondence Theorem:

1 ◁_{≠} K ◁_{≠} H ◁_{≠} G or 1 ◁_{≠} H ◁_{≠} K ◁_{≠} G.

If the three corresponding quotients are (S-)simple, we stop; otherwise we can insert a further term somewhere, and so on.

If this process stops after finitely many steps we reach a series

1 = G_{n} ◁_{≠} G_{n−1} ◁_{≠} ⋯ ◁_{≠} G_{2} ◁_{≠} G_{1} ◁_{≠} G_{0} = G  (3A)

in which

the quotients G_{0}/G_{1}, G_{1}/G_{2}, ⋯, G_{n−1}/G_{n} are all (S-)simple.  (3B)

Equivalently it is a series in which no further terms may be inserted to produce another “normal” series. Such a series (3A) satisfying (3B) is called a (S-)composition series of G.

Definition. A normal series for a (S-)group G is a finite series as in (3A) (but not necessarily satisfying (3B)). The integer n is the length of the series, and the quotients in (3B) are the factors of the series. If the factors are all (S-)simple, then the normal series is called a (S-)composition series and its factors are called the composition factors of G.

The Jordan-Hölder Theorem will justify calling the factors “the” composition factors of G.

One obvious sufficient condition for G to possess a composition series is that G be finite. Another is that G possess both the maximum and minimum condition on subgroups†: that is, there exist no infinite ascending chains or descending chains of subgroups of G:

H_{1} <_{≠} H_{2} <_{≠} ⋯ <_{≠} H_{n} <_{≠} ⋯ ≤ G or G ≥ H_{1} >_{≠} H_{2} >_{≠} ⋯ >_{≠} H_{n} >_{≠} · · · .

For if G satisfies these conditions, then G must possess a maximal (S-)normal subgroup G_{1}. (Choose any normal subgroup M_{1}. If this is not maximal, we get M_{1} <_{≠} M_{2} for some normal M_{2}. If M_{2} is not maximal, we get M_{1} <_{≠} M_{2} <_{≠} M_{3} and this process must terminate by the maximum condition.)

Then G/G_{1} is (S-)simple by the maximality of G_{1}. Moreover G_{1} inherits the maximum condition, so possesses a maximal normal subgroup G_{2}, and continuing we obtain a composition series

1 ◁_{≠} G_{n−1} ◁_{≠} ⋯ ◁_{≠} G_{2} ◁_{≠} G_{1} ◁_{≠} G,

the process terminating by the minimum condition.

Examples of infinite S-groups satisfying these conditions are finite-dimensional vector spaces (with S being the field of scalars).

The main thrust of the Jordan-Hölder Theorem is the uniqueness statement. We say that two normal series of G, say (3A) and

1 = H_{n′} ◁_{≠} H_{n′−1} ◁_{≠} ⋯ ◁_{≠} H_{2} ◁_{≠} H_{1} ◁_{≠} H_{0} = G,  (3C)

are equivalent if and only if n = n′ and the two lists

G_{0}/G_{1}, G_{1}/G_{2}, ⋯, G_{n−1}/G_{n} and H_{0}/H_{1}, H_{1}/H_{2}, ⋯, H_{n′−1}/H_{n′}

of composition factors can be reordered so that corresponding terms are (S)-isomorphic.

† Actually there is a weaker sufficient condition: that G possess the maximum and minimum condition on subnormal subgroups. A subgroup of G is subnormal in G if and only if it is a term in some normal series for G.

Theorem (Jordan and Hölder). Let G be an S-group. Then the following conditions hold.

E) If G is finite, or more generally possesses the maximum and minimum conditions on subgroups, then G possesses a (S-)composition series; moreover any (S-)normal series may be “refined” by the suitable addition of new terms to a (S-)composition series.

U) Any two (S-)composition series are equivalent.

Proof. We have already proved the first existence statement. The second follows by a similar argument.

A slightly stronger statement than the uniqueness statement is easier to prove. It is the following:

Theorem. Suppose that G has a composition series (3A) (of length n) and also has a normal series (3C) of length n′ ≥ n. Then the series (3C) is a composition series, n′ = n, and the two composition series are equivalent.

This implies the Jordan-Hölder uniqueness statement, since given two composition series, of length n and n′, we may assume without loss that n′ ≥ n, and then the theorem gives us what we want (a composition series is a certain kind of normal series).

Proof Let series (3A), (3C) be given as assumed in the theorem. We go by induction on n. We consider two cases.

Case 1. H_{1} ≤ G_{1}. In this case by inserting the term G_{1} in the series (3C) (unless it was there already, as H_{1}), we get a series, call it (3C′), which is like (3C) but of length n′ or n′ + 1, and with the next-to-top term G_{1}. From G_{1} down, the series (3A) and (3C′) give a composition series of G_{1} of length n − 1, and a normal series of length n′ − 1 or n′. Since n′ − 1 ≥ n − 1, induction implies that these two series for G_{1} are equivalent, and both have length n − 1. Therefore (3C′) had length n, which implies that (3C) had length n and H_{1} = G_{1}. Now the composition factors of (3A) and (3C) are obtained from those of our two composition series for G_{1} just by appending the one further group G/G_{1} = G/H_{1}, so we are done in this case.

Case 2. H_{1} ≰ G_{1}. Therefore G_{1}H_{1} > G_{1}. But G_{1} is a maximal normal subgroup of G since (3A) is a composition series. Therefore G_{1}H_{1} = G (as G_{1}H_{1} ◁ G). We set

K_{2} = G_{1} ∩ H_{1} ◁ G.

We get a parallelogram with G at the top, K_{2} at the bottom and G_{1} and H_{1} the other two vertices. By the second isomorphism theorem,

G/G_{1} ≅ H_{1}/K_{2} and G/H_{1} ≅ G_{1}/K_{2}.

We now construct a normal series (C) for K_{2} as follows. If possible, construct a composition series (C) for K_{2}. Otherwise following the procedure described at the beginning of this section we obtain normal series of arbitrary length; we choose (C) to have length n − 1 (anything larger would do just as well).

Then the two series

1 = G_{n} <_{≠} G_{n−1} <_{≠} ⋯ <_{≠} G_{1} and ⋯ (C) ⋯ K_{2} <_{≠} G_{1}

are a composition series for G_{1} of length n − 1 and either a composition series for G_{1} or a normal series of length n. By induction, they are both composition series and are equivalent. In particular (C) must be a composition series for K_{2}, and has length n − 2. Now

⋯ (C) ⋯ K_{2} <_{≠} H_{1} and 1 = H_{n′} <_{≠} H_{n′−1} <_{≠} ⋯ <_{≠} H_{1}

are, respectively, a composition series for H_{1} of length n − 1 and a normal series for H_{1} of length n′ − 1 ≥ n − 1. Again by induction the second series is a composition series equivalent to the first. Consequently the factors of series (3A) are those of (C), together with G_{1}/K_{2} and G/G_{1}. Likewise the factors of (3C) are those of (C), together with G/H_{1} and H_{1}/K_{2}. By the parallelogram law, we are finished. QED

Alternative proof. Lang uses the “Zassenhaus Butterfly Lemma” to prove the following theorem, from which the uniqueness part of Jordan-Hölder follows immediately.

Schreier Refinement Theorem. Any two normal (S)-series for a group G have equivalent refinements.

Proof. Take a normal S-series

1 = G_{n} ◁ G_{n−1} ◁ ⋯ ◁ G_{1} ◁ G_{0} = G.

This “filtration” of G allows us to filter any subgroup and any quotient. Thus if L ≤ G, then

1 = L ∩ G_{n} ◁ L ∩ G_{n−1} ◁ ⋯ ◁ L ∩ G_{1} ◁ L ∩ G_{0} = L

is a normal series of L, by the second isomorphism theorem applied to the subgroups H = L ∩ G_{i−1} and K = G_{i} of G_{i−1}.

To do an analogous thing with a quotient G/H, given H ◁ G, we must raise everything “above the level of H”; remember that the set of subgroups of G/H are in bijective correspondence with the set of subgroups of G containing H. From our normal series for G and the normal subgroup H we get a “partial” normal series (going down only to H)

H = G_{n}H ◁ G_{n−1}H ◁ ⋯ ◁ G_{1}H ◁ G_{0}H = G

which when reduced modulo H gives a normal series for G/H (with the same factors, up to isomorphism).

Exercise. G_{i−1}H/G_{i}H is a quotient of G_{i−1}/G_{i}. (Hint. Apply the parallelogram law to G_{i−1} and G_{i}H.)

Now if we have K ◁ H ≤ G (H/K is then called a “section” of G), we can apply both of the above to filter H/K by our given normal series for G. The result is a partial normal series from K to H:

K = (H ∩ G_{n})K ◁ (H ∩ G_{n−1})K ◁ ⋯ ◁ (H ∩ G_{1})K ◁ (H ∩ G_{0})K = H.

Now we are ready to consider a second given normal series for the same group G:

1 = H_{m} ◁ H_{m−1} ◁ ⋯ ◁ H_{1} ◁ H_{0} = G.

We may apply the filtering process to each H_{i−1}/H_{i}, thereby refining the normal series of H’s by replacing H_{i} ◁ H_{i−1} by the longer

H_{i} = (H_{i−1} ∩ G_{n})H_{i} ◁ (H_{i−1} ∩ G_{n−1})H_{i} ◁ ⋯ ◁ (H_{i−1} ∩ G_{1})H_{i} ◁ (H_{i−1} ∩ G_{0})H_{i} = H_{i−1}.

We have thus refined the series of H’s to a series of length mn whose factors are

(H_{i−1} ∩ G_{j−1})H_{i}/(H_{i−1} ∩ G_{j})H_{i}, 1 ≤ i ≤ m, 1 ≤ j ≤ n.

In the same way we can refine the series of G’s to another series of length mn whose factors are

(H_{i−1} ∩ G_{j−1})G_{j}/(H_{i} ∩ G_{j−1})G_{j}, 1 ≤ i ≤ m, 1 ≤ j ≤ n.

The proof is completed by the Zassenhaus Butterfly Lemma, which states that the two groups just displayed are isomorphic: QED

Lemma (Zassenhaus). (H_{i−1} ∩ G_{j−1})H_{i}/(H_{i−1} ∩ G_{j})H_{i} ≅ (H_{i−1} ∩ G_{j−1})G_{j}/(H_{i} ∩ G_{j−1})G_{j} for all i, j.

Proof. Apply the parallelogram law to (H_{i−1} ∩ G_{j})H_{i} and H_{i−1} ∩ G_{j−1}. Since H_{i} ◁ H_{i−1} and G_{j} ◁ G_{j−1}, the first of these is normalized by the second. We first need to simplify the intersection of these two groups:

[(H_{i−1}∩G_{j})H_{i}]∩(H_{i−1}∩G_{j−1})=(H_{i−1}∩G_{j})[H_{i}∩(H_{i−1}∩G_{j−1})]=(H_{i−1}∩G_{j})(H_{i}∩G_{j−1}),

the first step by the modular law * (as H_{i−1} ∩ G_{j} ≤ H_{i−1} ∩ G_{j−1}) and the second step since H_{i} ≤ H_{i−1}. Notice that the resulting expression is symmetric in G’s and H’s, so a similar application of the parallelogram law brings the other group in the statement of the lemma to the same form. QED

* Modular Law.

If A, B and C are subgroups of G, and A ≤ C, then AB ∩ C = A(B ∩ C).

Proof. Clearly the right side is contained in the left side as A ≤ C. Conversely if c ∈ AB ∩ C, we write c = ab, a ∈ A, b ∈ B and conclude that b = a^{−1}c ∈ C since A ≤ C. Thus b ∈ B ∩ C so c = ab ∈ A(B ∩ C).

Corollary. The dimension of a finite-dimensional vector space is uniquely determined.

Proof. A basis {v_{1}, ⋯, v_{n}} of the vector space V gives rise to the normal series

0 = V_{n} <_{≠} V_{n−1} <_{≠} ⋯ <_{≠} V_{1} <_{≠} V_{0} = V

where V_{i} is the span of v_{i+1}, ⋯, v_{n}. This is in fact a composition series as each factor is the span of a single vector. Now apply U) of Jordan-Hölder. QED

Corollary. Unique factorization holds in Z.

The proof is left to the reader, and is based on the fact that a factorization of n yields a composition series for the cyclic group Z_{n}.

Corollary. Let Γ be any group and ϕ a representation of Γ on the finite-dimensional vector space V. Then V has a filtration by Γ-invariant subspaces

0 = V_{n} <_{≠} V_{n−1} <_{≠} ⋯ <_{≠} V_{1} <_{≠} V_{0}

such that each V_{i−1}/V_{i} affords an irreducible representation of Γ. Moreover any two such filtrations are equivalent in the sense of Jordan-Hölder.

The irreducible representations corresponding to the V_{i−1}/V_{i} are called the irreducible constituents of ϕ.

Problem. Determine all finite groups by determining a) the simple ones and b) all groups with a given set of composition factors.

Both these seem impossible, but a) has been solved and b) seems far too complex to leave any hope for a clear solution. However, 30 years ago the same was thought of a) !!

In the remaining parts of this section we address a) by giving some more examples of simple groups, and address b) by discussing solvable groups, which in the finite case are those built from the simplest composition factors–cyclic groups of prime order.

3d. Solvable Groups

The following result is easy:

Proposition. All subgroups and quotients of an abelian group are abelian.

However the “converse” is false; if N ◁ G and N and G/N are both abelian, then G need not be abelian. Example: G = Σ_{3}, N = A_{3}, or G = D_{2n}, N ≅ Z_{n}. To obtain a property which “persists under extensions” in the most economical way we are let to the following notion.

551, Fall '02                 A3d.    Solvable Groups                 p. 11

Definition. A finite group is solvable if and only if all its composition factors are abelian (hence cyclic of prime order). In general a group is solvable if and only if there exists a normal series (by definition of finite length!) all of whose factors are abelian.

The two definitions coincide for finite groups, since if a finite group has such a normal series then it can be refined to a composition series, whence all composition factors are abelian.

Example: Σ_{4} is solvable. For any v = (ab)(cd) ∈ V − {1}, the series

1 ◁ ⟨v⟩ ◁ V ◁ A_{4} ◁ Σ_{4}

is a composition series.

Proposition. All subgroups and quotients of solvable groups are solvable. If N ◁ G, and both N and G/N are solvable, then G is solvable.

Proof If K ◁ H ≤ G, and A ≤ G, then A ∩ K ◁ A ∩ H and A ∩ H/A ∩ K is isomorphic to a subgroup of H/K. This follows from the parallelogram law applied to K and A ∩ H. Now if

1 = G_{n} ◁_{≠} G_{n−1} ◁_{≠} ⋯ ◁_{≠} G_{2} ◁_{≠} G_{1} ◁_{≠} G_{0} = G

is a normal series of G with abelian factors, then

1 = G_{n} ∩ H ◁_{≠} G_{n−1} ∩ H ◁_{≠} ⋯ ◁_{≠} G_{2} ∩ H ◁_{≠} G_{1} ∩ H ◁_{≠} G_{0} ∩ H = H  (5A)

is a normal series of H, and G_{i−1} ∩ H/G_{i} ∩ H embeds in the abelian group G_{i−1}/G_{i} so is abelian. The statement for quotients similarly follows from the following fact: If K ◁ H ≤ G and N ◁ G, then KN ◁ HN and HN/KN is isomorphic to a quotient of H/K. The parallelogram law applid to KN and H yields in fact that HN/KN ≅ H/H ∩ KN; but K ≤ H ∩ KN so H/H ∩ KN is a quotient of H/K. In this situation the above series gives the series

N = NG_{n} ◁_{≠} NG_{n−1} ◁_{≠} ⋯ ◁_{≠} NG_{2} ◁_{≠} NG_{1} ◁_{≠} NG_{0} = G  (5B)

and so

1 = N/N ◁_{≠} NG_{n−1}/N ◁_{≠} ⋯ ◁_{≠} NG_{2}/N ◁_{≠} NG_{1}/N ◁_{≠} NG_{0}/N = G/N  (5C)

the factors of which are quotients of the original abelian factors so are again abelian.

Conversely suppose that N and G/N are solvable. Then G/N has a normal series with abelian factors, and replacing each term by its inverse image in G and using the correspondence theorem we get a series starting with N and going to G, with all factors abelian. We are assuming that N possesses such a series, and it can be attached to this one to give the desired series for G.

One useful way to analyze a solvable group is by its action on a minimal normal subgroup.

Theorem. Let G be a finite solvable group. Let 1 ≠ N ◁ G and suppose that no proper subgroup of N is normal in G. Then N ≅ Z_{p} × ⋯ × Z_{p} for some prime p.

A group is called elementary abelian if it is the direct product of groups of the same prime order. The structure theorem for finite abelian groups (see below) implies that if N is a finite abelian group and p a prime such that x^{p} = 1 for all x ∈ N, then N is elementary abelian. We shall use this fact in the proof, and aim to prove that N is abelian and has exponent p.

The proof uses the notion of characteristic subgroup:

Definition. A subgroup H ≤ G is characteristic in G if and only if α(H) = H for all α ∈ Aut(G). We write H char G.

If H char G, then Int(g)(H) = H for all g ∈ G, so gHg^{−1} = H, i.e., H ◁ G. The converse is false.

It also uses the notion of commutator subgroup.

Definition. Let x, y ∈ G. Then [x, y] = xyx^{−1}y^{−1}. Moreover, [G, G] = ⟨[x, y] | x, y ∈ G⟩.

A group is abelian if and only if [G, G] = 1.

Lemma. The following subgroups of any group G are characteristic subgroups:

a) [G, G];
b) Z(G);
c) G^{n} = ⟨x^{n} | x ∈ G⟩, for an integer n.

Moreover, if G is finite and has a normal Sylow p-subgroup P for some prime p, then P char G.

Proof ^{g}[x, y] = [^{g}x,^{g} y] so conjugation by an element of G leaves the set of commutators invariant; therefore it leaves invariant the subgroup generated by them. The proof for c) is similar, and b) is left to the reader. For the final statement, P is the only Sylow psubgroup, being normal, so is characteristic since automorphisms carry Sylow p-subgroups to Sylow p-subgroups. QED

Lemma. If H char N ◁ G, then H ◁ G.

Proof Let g ∈ G. Then Int(g) : x ↦ gxg^{−1} is an automorphism of G, and it leaves N invariant since N ◁ G. Therefore Int(g)|N is an automorphism of N, so it leaves the characteristic subgroup H invariant. Therefore gHg^{−1} = H. QED

Here is the universal property of [G, G], or really the quotient G/[G, G], or really the projection G → [G, G].

Proposition. If N ≤ G, then N ≥ [G, G] if and only if N ◁ G and G/N is abelian.

551, Fall '02               A3e.    More Simple Groups               p. 13

Proof Suppose [G, G] ≤ N ≤ G. Then for any g ∈ G and n ∈ N, gng^{−1} = [g, n]n ∈ N, so N ◁ G. Now under the projection G → G/N, [x, y] maps to [xN, yN]. But [x, y] = 1 for all x, y since [G, G] ≤ N. Therefore [G/N, G/N] = 1. Conversely if N is a normal subgroup and G/N is abelian, the same reasoning shows that [x, y] ∈ N for all x, y ∈ G. Therefore [G, G] ≤ N.

Corollary. [G, G] < G if and only if G has a nontrivial abelian quotient. If G is a solvable group and G ≠ 1, then [G, G] < G.

Proof of Theorem Let G and N be as in the theorem. Then N is solvable since G is solvable. So [N, N] < N. But [N, N] char N ◁ G so [N, N] ◁ G. Therefore [N, N] = 1 and N is abelian. Let p be a prime divisor of |N|. The mapping N → N defined by x ↦ x^{p} is then a homomorphism, with image N^{p}. Its kernel is nontrivial by Sylow, and so N^{p} < N. As with the commutator subgroup we get N^{p} = 1, that is, x^{p} = 1 for all x ∈ N. Now we quote the structure theorem for finite abelian groups (see below) to complete the proof.

3e. More Simple Groups

So far the simple groups which we have encountered are all finite: Z_{p} for p prime, and A_{n} for n ≥ 5.

Exercise. Let Ω be an infinite set and let Σ_{Ω}^{o} be the subgroup of Σ_{Ω} consisting of all elements with finite support, i.e. all g such that gω = ω for all but finitely many ω ∈ Ω. Define A_{Ω}^{o} in a natural way, and prove that A_{Ω}^{o} is simple. (Hint. Let N ◁ A_{Ω}^{o}. Prove and use the fact that for any finite subset Ψ ⊆ Ω, N ∩ A_{Ψ} = 1 or A_{Ψ}, where A_{Ψ} is considered as a subgroup of A_{Ω}^{o} in the obvious way.)

Another source for simple groups, finite or infinite, is matrix groups, i.e. certain subgroups and quotients (and subgroups of quotients) of GL_{n}(K) for various fields K. We define

P GL_{n}(K) = GL_{n}(K)/Z and P SL_{n}(K) = SL_{n}(K)/Z ∩ SL_{n}(K)

where Z is the group of all scalar matrices (scalar multiples of the identity matrix) in GL_{n}(K). Notice that Z ≤ Z(GL_{n}(K)).

Exercise. Z = Z(GL_{n}(K)).

Also by the parallelogram law, P SL_{n}(K) ≅ ZSL_{n}(K)/Z ◁ P GL_{n}(K). So P GL_{n}(K) has a copy of P SL_{n}(K) as a normal subgroup.

Theorem. Let K be a field and n ≥ 2 an integer. If n = 2 assume that |K| ≥ 4. Then P SL_{2}(K) is simple.

The “P ” in P GL and P SL is for “projective”. Another source for simple groups are matrix groups, i.e. certain subgroups and quotients (and subgroups of quotients) of GL_{n}(K) for various fields K. We define

P GL_{n}(K) = GL_{n}(K)/Z and P SL_{n}(K) = SL_{n}(K)/Z ∩ SL_{n}(K)

where Z is the group of all scalar matrices (scalar multiples of the identity matrix) in GL_{n}(K). Notice that Z ≤ Z(GL_{n}(K)).

Exercise. Z = Z(GL_{n}(K)).

Also by the parallelogram law, P SL_{n}(K) ≅ ZSL_{n}(K)/Z ◁ P GL_{n}(K). So P GL_{n}(K) has a copy of P SL_{n}(K) as a normal subgroup.

Theorem. Let K be a field and n ≥ 2 an integer. If n = 2 assume that |K| ≥ 4. Then P SL_{2}(K) is simple.

The “P ” in P GL and P SL is for “projective”.

We sketch a proof. Let V be a 2-dimensional vector space over the field K. Define GL(V) to be the group of nonsingular linear transformations: V → V; SL(V) to be the subgroup of GL(V) which is the kernel of the determinant homomorphism SL(V) → K^{×}; Z to be the subgroup of GL(V) consisting of all scalar mappings (for each α ∈ K^{×} there is unique corresponding scalar mapping s_{α} ∈ GL(V), namely s_{α}(v) = αv for all v ∈ V). Also define P GL(V) = GL(V)/Z and P SL(V) = SL(V)/Z ∩ SL(V).

Choosing a (ordered) basis for V leads to an isomorphism P SL(V) ≅ P SL_{2}(K). We may then think of P SL(V) acting on P(V), or equivalently, we may think of P SL_{2}(K) acting on the set of 1-dimensional spaces of 2 × 1 column vectors.

The group GL(V) acts naturally on the set P(V) of 1-dimensional subspaces of V, so SL(V) does as well. Moreover Z acts trivially on P(V), so the action GL(V) → Σ_{P(V)} lifts to an action GL(V)/Z → Σ_{P(V)}, i.e., P GL(V) acts on P(V). Likewise P SL(V) acts on P(V).

Exercise. The action of P SL(V) on P(V) is 2-transitive.

We also can show:

Lemma. The action of P SL(V) on P(V) is faithful.

This amounts to showing that if T : V → V is a linear transformation and for each v ∈ V there exists a scalar α_{v} (depending on v, perhaps) such that T v = α_{v}v, then α_{v} = α_{w} for all v, w ∈ V − {0}. (Look at T (v + w) to see that this is true, if v and w are linearly independent.)

The structure of a point stabilizer is also important, and this is easily seen concretely in P SL_{2}(K).

Lemma. The stabilizer in P SL_{2}(K) of the subspace ω spanned by [ \begin{matrix}1 \\ 0 \end{matrix} ] is the image \overline{B}

in P SL_{2}(K) of the group

B = \{[ \begin{matrix}α & β \\ 0 & α^{−1} \end{matrix} ] | α ∈ K^{×}, β ∈ K\} .

under the canonical homomorphism SL_{2}(K) → P SL_{2}(K). (B is a subgroup of SL_{2}(K); check it.)

This merely states that

[ \begin{matrix}α & β \\ γ & δ \end{matrix} ] [ \begin{matrix}1 \\ 0 \end{matrix} ] = [ \begin{matrix}ζ \\ 0 \end{matrix} ]

for some ζ if and only if γ = 0.

Lemma. B and \overline{B} are solvable.

Proof. Define ϕ : B → K^{×} by

ϕ ([ \begin{matrix}α & β \\ 0 & α^{−1} \end{matrix} ]) = α.

It is easily checked that ϕ is a homomorphism. Let u(β) = [ \begin{matrix}1 & β \\ 0 & 1 \end{matrix} ] for any β ∈ K, and

U = ker ϕ = {u(β) | β ∈ K} .

Then U is abelian and B/U is isomorphic to a subgroup of K^{×} so is abelian. Therefore the normal series B > U > 1 has abelian factors and so B is solvable. Since \overline{B} is the image of B, it is isomorphic to a quotient of B and so is solvable.

We need one additional fact about P SL_{2}(K); to motivate it we now give the main logic of the proof.

Main Logic Let G = P SL_{2}(K), which acts on Ω = P(V). Suppose that N ◁ G, but 1 < N < G. We derive a contradiction.

Since G acts faithfully and 2-transitively on Ω, and N ≠ 1, N is transitive on Ω (the same fact was used in the proof that A_{n} is simple, n ≥ 5). Therefore G = NG_{ω} = \overline{B}N. (For any g ∈ G there is n ∈ N with gω = nω and so g = gn^{−1}n ∈ G_{ω}N. Then G/N = \overline{B}N/N ≅ \overline{B}/\overline{B} ∩N is a quotient of the solvable group \overline{B} so is solvable. Since N ≠ G, G/N is a nontrivial (solvable) group. Therefore it has a nontrivial abelian quotient. Therefore G has a nontrivial abelian quotient. Therefore SL_{2}(K) has a nontrivial abelian quotient. Therefore SL_{2}(K) ≠ [SL_{2}(K), SL_{2}(K)]. This completes the proof as it contradicts the following fact:

Lemma. If |K| ≥ 4, and H = SL_{2}(K), then H = [H, H].

The lemma is proved by

a) checking that H = SL_{2}(K) is generated by all the elements u(β) defined above and all their transposes v(β) = u(β)^{T}. (This amounts to showing that any matrix of determinant 1 can be reduced to I by certain types of row and column operations.) b) checking that each u(β) and v(β) lies in [H, H]. (It is here that the hypothesis |K| ≥ 4 is used, to find an element α ∈ K such that α ≠ 0 and α ≠ ±1. Fix α and let h(α) be the diagonal matrix with diagonal entries α, α^{−1}; one computes

[h(α), u(β)] = u((α^{2} − 1)β).

As β varies over K, so does (α^{2} − 1)β, since α^{2} ≠ 1. Therefore every u(β) is in [H, H], and similarly so is every v(β).)

Then as the u’s and v’s generate H, [H, H] = H. QED

A similar argument, slightly more complicated, can be used to prove:

Theorem. For any field K and any n ≥ 3, P SL_{n}(K) is simple.

There are no exceptions here; the larger size of matrices makes it possible to express certain critical matrices as commutators, regardless of the size of the field.