Chapter 9

Series of Subgroups

In the future, as in the past, the great ideas must be simplifying ideas.

—André Weil

If G is a finite abelian group, it can be shown that G is isomorphic to a direct product of cyclic groups (see Chapter 7). This result is an example of a structure theorem, that is, a theorem showing that every group in a suitable defined class may be constructed in a systematic way from well understood groups in the class. Such theorems are hard to come by, and the result for finite abelian groups is a stunning example. The structure of nonabelian finite groups is much more complicated.

Suppose that groups K and H are given. It is a very difficult problem to describe all groups G that have a normal subgroup K₁ isomorphic to K such that G/K₁ is isomorphic to H. If we could solve this extension problem, the solution would give an inductive method for constructing all finite groups. Direct and semidirect products solve this problem in very special cases. Although the general problem is far from being solved, the classes of groups that can be built up this way are of interest.

To illustrate, suppose that we use only abelian groups as building blocks. Starting with an abelian group G₀, we construct G₁ ⊇ G₀ such that G₀ G₁ and G₁/G₀ is abelian. Next, we extend G₁ to obtain G₂ ⊇ G₁ such that G₁ G₂ and G₂/G₁ is abelian. After n steps, we have a chain

G = 389G_n ⊇ G_n−1 ⊇ ⊇ G₁ ⊇ G₀,

where G_i G_i+1 and G_i+1/G_i is abelian for each i. Such a group G is called solvable, and the theory of these groups is successful in the following sense: The class of solvable groups is large (it contains all finite groups of odd order), but at the same time, many theorems are true for all solvable groups but do not hold in general. We investigate solvable groups in Section 9.2.

If we use simple groups as building blocks in this way, the resulting groups are those studied in Section 9.1. In this case, the above chain of subgroups is called a composition series for G, and the famous Jordan–Hö lder theorem asserts that G uniquely determines the series of groups G_n/G_n−1, G_n−1/G_n−2, . . ., G₁/G₀, G₀. This leads to the useful notion of the composition length of a group.

Section 9.3 deals with somewhat more specialized central series and begins the study of finite nilpotent groups. These groups are characterized as the groups that are the direct product of their Sylow subgroups, equivalently if every Sylow subgroup is normal. In addition, the Frattini subgroup is defined for every finite group and shown to be nilpotent.

9.1 The Jordan–Hölder Theorem

Much of what we do in this chapter is concerned with groups G that admit a chain of subgroups with certain nice properties. A subnormal series for G is a chain

G = G₀ ⊇ G₁ ⊇ G₂ ⊇ ⊇ G_n = {1}

of subgroups of G such that G_i+1 G_i for each i. The factor groups G_i/G_i+1 are called the factors of the subnormal series. Note that we do not insist that the subgroups G_i are normal in G. Moreover, by possibly deleting some of the groups G_i, we clearly may assume that G_i ≠ G_i+1 for each i.

Example 1. G ⊇ {1} is a subnormal series for any group G. The only factor is G.

Example 2. A₄ ⊃ K ⊃ H ⊃ ε is a subnormal series for A₄, we have K = {ε, (12)(34), (13)(24), (14)(23)} and H = {ε, (12)(34)}. The factors are C₃, C₂, and C₂ in order. Note that H is not normal in A₄.

The most interesting cases are the groups that admit a subnormal series in which every factor is abelian or every factor is simple. We investigate the first case in Section 9.2; the second calls for another definition.

If G is a group, a subnormal series G = G₀ ⊃ G₁ ⊃ ⊃ G_n = {1} is called a composition series for G if each factor G_i/G_i+1 is simple. In this case, the factors G_i/G_i+1 are called composition factors of G, and the integer n is called the length of the composition series. If G = {1}, we say that G has a composition series of length 0.

Example 3. The simple groups are those with a composition series of length 1.

Example 4. Every finite group G has a composition series. This holds by definition if G = {1}. If G ≠ {1}, write G = G₀ and choose a maximal normal subgroup G₁ of G₀ (it exists because G is finite). Then G₀/G₁ is simple by Theorem 6 §8.1. If G₁ ≠ {1}, choose a maximal normal subgroup G₂ of G₁ and continue in this way. The series G = G₀⊃ G₁ ⊃ G₂ ⊃ must reach {1} eventually because G is finite, so it is a composition series.

The converse of Example 4 is false: Any infinite simple group has a composition series of length 1. However, the converse does hold for abelian groups.

Example 5. An abelian group G has a composition series if and only if G is finite. Indeed, if G = G₀ ⊃ G₁ ⊃ ⊃ G_n = {1} is a composition series, each composition factor G_i/G_i+1 is a simple abelian group and so is finite. Hence (Exercise 11), |G| = |G/G₁||G₁/G₂| |G_n−1/G_n| is also finite.

The finite abelian groups are not the only ones having all composition factors abelian. Theorem 8 §8.2 shows that every finite p-group has this property:

Example 6. If p is a prime, each finite p -group has a composition series in which every composition factor is isomorphic to C_p.

A group may have several different composition series. For example, let G be a cyclic group of order 12 and, for each divisor d of 12, let H_d be the unique subgroup of G of order d. Then G has three composition series. These series, along with their composition factors, are

Note that the length is 3 in each case and the factors are also the same except for the order in which they appear. Hence, these series are all equivalent in the following sense: Two composition series for a group are said to be equivalent if they have the same length and the composition factors can be paired in such a way that corresponding factors are isomorphic. This is clearly an equivalence relation on the set of all composition series of a group G (assuming that there is one). The remarkable thing is that any two composition series for G are equivalent. This is the most important theorem in this section.

Theorem 1. Jordan–Hölder Theorem. If a group has a composition series, any two composition series are equivalent.

We will give the proof at the end of this section.

If a group G has a composition series, it uniquely determines the length of the series and the composition factors (including multiplicities). Hence, we can speak of the composition length of the group G, denoted lengthG, and of the composition factors of G.

Composition series were first discussed by Camille Jordan. In 1869, he showed (for groups of permutations) that the orders of the composition factors are the same for every composition series of the group. However, it was not until 20 years later, after the abstract definition of a group had been given, that Otto H ölder observed that the group uniquely determines the composition factors themselves and that they are the same in any composition series.

Example 7. If n ≥ 5, then S_n ⊃ A₅ ⊃ {ε} is a composition series because A_n is simple (Theorem 8 §2.8). Hence, S_n has length 2 and the composition factors are C₂ ≅ S_n/A_n and A_n. If n = 4, we get a composition series

S₄ ⊃ A₄ ⊃ K ⊃ H ⊃ {ε},

where K = {ε, (12)(34), (13)(24), (14)(23)} and H = {ε, (12)(34)}. Hence, S₄ has length 4 and the composition factors are C₂, C₃, C₂, and C₂.

The Jordan–Hölder theorem is a type of unique factorization theorem. In the following corollary, we use it to give another proof that the factorization of an integer n into primes is unique. Here, the composition factors play the role of primes.

Corollary. The factorization of an integer n ≥ 2 into primes is unique.

Proof. Let n = p₁p₂ p_r, where p_i are (not necessarily distinct) primes. If is a cyclic group of order n, then

is a composition series for G because the factors are (Indeed, for each k by Theorem 5 §2.4.) Since any other factorization of n into primes must yield the same composition factors, it follows that n uniquely determines the number r = lengthG and the primes p_i.

If K G and G/K ≅ H, the group G is called an extension of K by H. So if

G = G₀ ⊃ G₁ ⊃ ⊃ G_r = {1}

is a composition series for G, then each G_i is an extension of G_i+1 by a simple group. Thus, each finite group G is the result of a finite number of extensions by finite simple groups, and the Jordan–Hö lder theorem shows that G uniquely determines the simple groups used (up to order). Moreover, we know all the finite simple groups (see the discussion at the end of Section 2.8), so the complete description of all finite groups comes down to the extension problem: For a simple group H, describe all extensions of a given group G by H. This is a very difficult task.

We are going to prove that subgroups and homomorphic images of groups with a composition series again have composition series. The proof requires the follow-ing technical lemma that gives important information about subnormal series in general and will be referred to again later. For composition series, we use it to deduce some important properties of the length of a group and prove the Jordan–Hölder theorem itself.

Lemma 1. Let G = G₀ ⊇ G₁ ⊇ ⊇ G_n = {1} be a subnormal series for the group G and let K G.

Proof. We leave to the reader the verification that the series are subnormal.

(1) Define α: K ∩ G_i → G_i/G_i+1 by α(x) = xG_i+1. This is clearly a group homomorphism and ker α = {x K ∩ G_i x G_i+1} = K ∩ G_i+1. Hence, it remains to prove that α(K ∩ G_i) (G_i/G_i+1). But if x K ∩ G_i and y G_i, then

(yG_i+1)α(x)(yG_i+1)⁻¹ = (yxy⁻¹)G_i+1 = α(yxy⁻¹) α(K ∩ G_i)

because yxy⁻¹ (yKy⁻¹) ∩ G_i = K ∩ G_i.

(2) Since KG_i+1 KG_i (Exercise 15 §8.1), the third isomorphism theorem (Theorem 7 §8.1) shows that

To show that (KG_i)/(KG_i+1) is an image of G_i/G_i+1, define

by  α(xG_i+1) = xKG_i+1 for all x G_i.

This is well defined because xG_i+1 = yG_i+1 implies that y⁻¹x G_i+1 ⊆ KG_i+1. It is clearly a group homomorphism, and (as K G) it is onto because

kxKG_i+1 = x(x⁻¹kx)KG_i+1 = xKG_i+1 = α(xG_i+1)

holds for all k K and x G_i.

Now suppose that G = G₀ ⊇ G₁ ⊇ ⊇ G_n = {1} is a composition series for G. If K G, the subnormal series for K and G/K in Lemma 1 are also composition series. Indeed, in both cases, the factors are isomorphic to either normal subgroups or images of the simple groups G_i/G_i+1 and so are all either simple or {1}. Hence, after we eliminate equalities, these series become composition series for K and G/K, respectively, each with factors from G. This proves part of Theorem 2.

Theorem 2. Let G be a group and let K G. Then G has a composition series if and only if both K and G/K have composition series. Moreover, in this case,

Proof. If G has a composition series, we have already seen that this is true of K and G/K. Conversely, suppose

K = K₀ ⊃ K₁ ⊃ ⊃ K_m = {1}  and 

are composition series for K and G/K. Because G_iG_i+1 ≅ G_i/KG_i+1/K, the series

G = G₀ ⊃ G₁ ⊃ ⊃ G_r = K = K₀ ⊃ K₁ ⊃ ⊃ K_m = {1}

is a composition series for G. Now (1)–(3) are apparent.

Corollary 1. If θ: G → H is a homomorphism, then G has a composition series if and only if both ker θ and θ(G) have composition series. In this case,

lengthG = length ker θ + lengthθ(G).

Proof. Since G/ker θ ≅ θ(G), Theorem 2 applies with K = ker θ.

Corollary 2. If G₁, G₂, , G_n are groups, then G₁ × G₂ × × G_n has a composition series if and only if the same is true of each G_i. In this case,

length(G₁ × G₂ × × G_n) = lengthG₁ + lengthG₂ + + lengthG_n.

Solution. We prove it for n = 2, and then the general case follows by induction. Define θ: G₁ × G₂ → G₂ by θ(g₁, g₂) = g₂ for all g₁ G₁ and g₂ G₂. Then θ is a homomorphism, ker θ = G₁ × {1} ≅ G₁, and θ(G₁ × G₂) = G₂. Use Corollary 1.

Example 8. Let G be an abelian group of order where p_i are distinct primes. Show that lengthG = n₁ + n₂ + + n_r.

Solution. Proceed by induction on |G|. If |G| = 1, it is clear. In general, let K be a maximal (normal) subgroup of G. Then |G/K| is a prime divisor of |G|, say |G/K| = p₁. Hence, so induction and Theorem 2 give

We conclude this section with a proof of the Jordan–Hölder theorem. The proof requires the following lemma.

Lemma 2. Let G be a group and let H and K be distinct maximal normal subgroups of G. Then H ∩ K is maximal normal in both H and K. Moreover,

and   

Proof. We first claim that KH = G. Indeed, H ⊆ KH G and K ⊆ KH G, so if KH ≠ G, the fact that H and K are maximal normal in G implies that H = KH = K, contrary to assumption. Hence, KH = G, so the second isomorphism theorem (Theorem 1 §8.1) gives GK = KHK ≅ HH ∩ K. Because G/K is simple, this shows that K ∩ H is maximal normal in H. The rest is proved in the same way.

Proof of the Jordan–Hölder Theorem. Suppose the group G has a composition series

(1)

of length n. We show by induction on n that every composition series

(2)

for G is equivalent to series (1). If n = 1, then G is simple, so G₁ = {1} = H₁ and the theorem holds. So assume that n ≥ 2 and that the theorem holds for all groups with a composition series of length less than n. In particular, it holds for G₁ because G₁ ⊃ G₂ ⊃ ⊃ G_n = {1} has length n − 1. If it happens that H₁ = G₁, then G₁ ⊃ H₂ ⊃ ⊃ H_m = {1} is another composition series for G₁ and so is equivalent to G₁ ⊃ G₂ ⊃ ⊃ G_n = {1} by induction and the theorem follows.

So assume that H₁ ≠ G₁ and let H₁ ∩ G₁ = L₀ ⊃ L₁ ⊃ ⊃ L_s = {1} be a composition series for H₁ ∩ G₁ by Theorem 2 as H₁ ∩ G₁ G. Now consider the following series for G:

(3)

(4)

As H₁ ≠ G₁, Lemma 2 asserts that H₁ ∩ G₁ is maximal normal in each of H₁ and G₁, so both (3) and (4) are composition series for G. Moreover, G/G₁ ≅ H₁/(H₁ ∩ G₁) and G/H₁ ≅ G₁/(H₁ ∩ G₁) by Lemma 2, so (3) and (4) are equivalent; denote this as (3) (4). Note that this equivalence holds even if s = 0, that is, if H₁ ∩ G₁ = {1}.

Now G₁ ⊃ G₂ ⊃ ⊃ G_n = {1} and G₁ ⊃ (H₁ ∩ G₁) ⊃ L₁ ⊃ ⊃ L_s = {1} are composition series for G₁ and so are equivalent by induction. This implies that (1) (3) and also that n − 1 = s + 1. But then the composition series H₁ ⊃ (H₁ ∩ G₁) ⊃ L₁ ⊃ ⊃ L_s = {1} has length n − 1 and so (again by induction) is equivalent to H₁ ⊃ H₂ ⊃ ⊃ H_m = {1}. This in turn implies that (2) (4). Piecing these equivalent series together gives (1) (3) (4) (2), which proves the Jordan–Hölder theorem.

Exercises 9.1

9.2 Solvable Groups

In Section 9.1, we were concerned with groups that admit a composition series, that is, a subnormal series in which all the factors are simple. Although those groups are of interest, we obtain an even more important class of groups when the factors are required to be abelian rather than simple.

A group G is called a solvable group¹⁰² if there exists a subnormal series

G = G₀ ⊇ G₁ ⊇ ⊇ G_n = {1}

such that each factor G_i/G_i+1 is abelian. Such a series is called a solvable series for G. Note that {1} is solvable.

Example 1. Every abelian group G is solvable because G ⊇ {1} is a solvable series.

Example 2. If p is a prime, every finite p -group is solvable. In fact, it has a composition series in which each factor is isomorphic to C_p (Theorem 8 §8.2).

Example 3. D_n is solvable for each n. Indeed, D_n has a cyclic subgroup H of index 2, so H D_n and D_n ⊇ H ⊇ {1} is a solvable series.

Example 4. S₄ is solvable because S₄ ⊇ A₄ ⊇ K ⊇ {ε} is a solvable series, where K = {ε, (12)(34), (13)(24), (14)(23)}.

Example 5. If p and q are primes, show that any group of order pq is solvable.

Solution. Because G is not simple by the Sylow theorems (Example 5 §8.4), let K ≠ {1} be a proper normal subgroup. Then |K| = p or |K| = q; either way G ⊃ K ⊃ {1} is a solvable series with factors C_p and C_q.

Suppose that G₁ is an abelian group. It is difficult to describe how to construct all groups G₂ such that G₁ G₂ and G₂/G₁ is abelian. Nonetheless, suppose we carry out this construction and then repeat it to construct a group G₃ such that G₂ G₃ and G₃/G₂ is abelian. If we continue this procedure, each group constructed is clearly solvable, and we can obtain every solvable group in this way—constructed from the bottom up, as it were. Viewing solvable groups in this way is useful, but an analogous top–down construction is actually more important. It is based on the derived subgroup introduced in Section 2.9.

Recall that an element of the form aba⁻¹b⁻¹ in a group G is called a commutator, denoted [a, b]. The set G′ of all products of commutators is a subgroup, called the derived subgroup of G, and has the following properties (Theorem 3 §2.9):

For a group G, repeatedly taking the derived subgroup leads to a subnormal series of subgroups G⊇ G′ ⊇ G^′′ ⊇ G^′′′ ⊇ in which each factor is abelian. There is a standard notation for these subgroups. Given a group G, construct subgroups G⁽⁰⁾, G⁽¹⁾, G⁽²⁾, of G as follows:

Thus, G⁽¹⁾ = G′, G⁽²⁾ = G^′′, G⁽³⁾ = G^′′′, and so on. Furthermore, G⁽ⁱ⁺¹⁾ G⁽ⁱ⁾ for each i and the subnormal series

G = G⁽⁰⁾⊇ G⁽¹⁾ ⊇ G⁽²⁾ ⊇

is called the derived series for G. Note that G⁽ⁱ⁾/G⁽ⁱ⁺¹⁾ is abelian for each i by Theorem 3 §2.9. The groups G⁽ⁱ⁾ are called the higher derived subgroups of G, and they are actually normal in G as the next theorem shows. The proof requires the following lemma; we leave the proof as Exercise 8.

Lemma 1. Let G denote a group and let H be a subgroup.

Theorem 1. If G is a group, we have G⁽ⁱ⁾ G for all i ≥ 0.

Proof. We use induction on i ≥ 0. It is clear if i = 0, so assume inductively that G⁽ⁱ⁾ G for some i. If a G then σ_a(G⁽ⁱ⁾) = G⁽ⁱ⁾, where σ_a is the inner automorphism of G determined by a. But then Lemma 1 gives

σ_a[G⁽ⁱ⁺¹⁾] = σ_a[(G⁽ⁱ⁾)′] ⊆ (G⁽ⁱ⁾)′ = G⁽ⁱ⁺¹⁾.

This shows that G⁽ⁱ⁺¹⁾ G and so completes the induction.

The solvable groups G are just those for which the derived series reaches {1}.

Theorem 2. A group G is solvable if and only if G⁽ⁿ⁾ = {1} for some n ≥ 1.

Proof. If G⁽ⁿ⁾ = {1}, then

G = G⁽⁰⁾ ⊇ G⁽¹⁾ ⊇ G⁽²⁾ ⊇ ⊇ G⁽ⁿ⁾ = {1}

is a solvable series for G because G⁽ⁱ⁾/G⁽ⁱ⁺¹⁾ = G⁽ⁱ⁾/[G⁽ⁱ⁾]′ is abelian for each i. Conversely, let G = G₀ ⊇ G₁ ⊇ ⊇ G_n = {1} be a solvable series for G. It suffices to show that G⁽ⁱ⁾ ⊆ G_i holds for each i. This is clear if i = 0, so assume that G⁽ⁱ⁾ ⊆ G_i for some i ≥ 0. As G_i/G_i+1 is abelian, we have Hence,

by Lemma 1, which completes the induction.

If G is solvable, then G⁽ⁱ⁾ ⊃ G⁽ⁱ⁺¹⁾ is strictly for all i by Corollary 1 of Theorem 3.

Theorem 2 provides a quick method of establishing several basic properties of solvable groups. We begin with the following result.

Theorem 3. Every subgroup and image of a solvable group is again solvable.

Proof. Suppose that G is solvable and let G⁽ⁿ⁾ = {1}. If H is a subgroup of G, it suffices to show that H⁽ⁱ⁾ ⊆ G⁽ⁱ⁾ for each i. This follows by induction: It is clear when i = 0, and if H⁽ⁱ⁾ ⊆ G⁽ⁱ⁾, then Lemma 1 gives

H⁽ⁱ⁺¹⁾ = [H⁽ⁱ⁾]′ ⊆ [G⁽ⁱ⁾]′ = G⁽ⁱ⁺¹⁾.

Now let α: G → K be an onto group homomorphism. It suffices to show that K⁽ⁱ⁾ ⊆ α[G⁽ⁱ⁾] for each i. This is clear if i = 0 because α is onto, so assume that K⁽ⁱ⁾ ⊆ α(G⁽ⁱ⁾). Then, given x and y in K⁽ⁱ⁾, write x = α(a) and y = α(b), where a, b G⁽ⁱ⁾. Hence,

[x, y] = [α(a), α(b)] = α([a, b]) α(G⁽ⁱ⁺¹⁾),

so K⁽ⁱ⁺¹⁾ ⊆ α(G⁽ⁱ⁺¹⁾) because K⁽ⁱ⁺¹⁾ = (K⁽ⁱ⁾)′.

Corollary 1. If H ≠ {1} is a subgroup of a solvable group G, then H′ ≠ H.

Proof. If H′ = H, then H⁽²⁾ = [H′]′ = H′ = H and an induction shows that H⁽ⁱ⁾ = H ≠ {1} holds for each i. As H is solvable by Theorem 3, this result contradicts Theorem 2.

Corollary 2. A simple group is solvable if and only if it is abelian (of prime order).

Proof. Let G be solvable. If G is simple, then either G′ = {1} (so G is abelian) or G′ = G, (contradicting Corollary 1). The converse is clear.

Example 6. If n ≥ 5, the symmetric group S_n is not solvable. For if S_n were solvable, A_n would be solvable by Theorem 3 so, as A_n is simple, Corollary 2 would imply that A_n is abelian, which is not the case. Hence, S_n is not solvable.

Example 6 explains the origin of the term solvable. A classical problem in the theory of equations was to find a formula for the roots of a real polynomial xⁿ + a_n−1xⁿ⁻¹ + + a₁x + a₀ in terms of the coefficients a_i. If n = 2, the solution is the famous quadratic formula: In general, such a formula should give the roots in terms of the coefficients a_i using only arithmetic operations and the extraction of roots. Such formulas were found for n = 3 and n = 4, but the case n = 5 proved to be difficult. Call a polynomial f solvable if such a formula exists. It will be shown in Chapter 10 that f is solvable if and only if a certain group (called the Galois group of f) is a solvable group. For example, the polynomial x⁵ − 6x + 2 has Galois group S₅ (Example 1 §10.3) and so cannot be solvable. Incidentally, the first proof that a nonsolvable polynomial exists was given in 1824 by the young Norwegian mathematician Niels Henrik Abel, building on the work of Paolo Ruffini.

Theorem 4 gives a useful way to show that a group is solvable.

Theorem 4. If K G, then G is solvable if and only if both K and G/K are solvable.

Proof. Assume that K and G/K are solvable and let

and

be solvable series. Then

G = G₀ ⊇ G₁ ⊇ ⊇ G_r = K = K₀ ⊇ K₁ ⊇ ⊇ K_m = {1}

is a subnormal series for G and the factors are abelian because G_iG_i+1 ≅ G_i/KG_i+1/K for each i. Hence, G is solvable. The converse follows by Theorem 3.

Example 7. If G₁, G₂, , G_n are groups, then G₁ × G₂ × × G_n is solvable if and only if the same is true of each G_i.

Solution. By induction, it suffices to prove it for n = 2. Let θ: G₁ × G₂ → G₂ be the projection given by θ(g₁, g₂) = g₂. Then (G₁ × G₂)/ker θ ≅ θ(G₁ × G₂) = G₂ and ker θ = G₁ × {1} ≅ G₁ are both solvable, so Theorem 4 applies.

The above theorems are valid for arbitrary groups. We now give some conditions equivalent to solvability in a finite group.

Theorem 5. The following conditions are equivalent for a finite group G.

Proof. Note first that G has a composition series because it is finite.

(1) ⇒ (2). Each composition factor is simple and solvable (Theorem 3), and so is abelian by Corollary 2 of Theorem 3.

(2) ⇒ (3). Any composition series for G is a solvable series by (2).

(3) ⇒ (1). The derived series G = G⁽⁰⁾⊇ G⁽¹⁾ ⊇ G⁽²⁾ ⊇ reaches {1} because G is finite and G⁽ⁱ⁾ ⊃ (G⁽ⁱ⁾)′ = G⁽ⁱ⁺¹⁾ for each i by (3).

Example 8. Let R be any ring. If n ≥ 3, show that the group G of all invertible n × n matrices over R is not solvable.

Solution. Let E_ij denote the n × n matrix with (i, j)-entry 1 and zeros elsewhere. Then E_ijE_jk = E_ik, whereas E_ijE_lk = 0 if j ≠ l. If I is the n × n identity matrix, this shows that I + E_ij is in G whenever i ≠ j and that (I + E_ij)⁻¹ = I − E_ij. Now let H be the subgroup of G generated by the matrices I + E_ij, i ≠ j. If i, j, and k are distinct indices (they exist because n ≥ 3), compute

This shows that every generator of H is a commutator from H and hence H′ = H. Thus, G is not solvable by Corollary 1 of Theorem 3.

If F is a field, Example 8 shows that the general linear group GL_n(F) of all n × n invertible matrices over F is not solvable if n ≥ 3. If F is finite, Theorem 5 shows that a nonabelian simple group is lurking among the composition factors of GL_n(F). In fact, such a group exists even if F is infinite. The mapping A det A is an onto homomorphism GL_n(F) → F^* and the kernel is the special linear group SL_n(F) of all matrices with determinant 1. It is not difficult to verify that the center of SL_n(F) consists of all scalar matrices aI, where a F satisfies aⁿ = 1. The factor group

is called the projective special linear group (of degree n) over F. These groups comprise another infinite family of finite, simple, nonabelian groups (in addition to the alternating groups A_n, n ≥ 5). The theorem was proved in 1870 by Camille Jordan for a prime and, in early 1900, Leonard Eugene Dickson proved it for all finite fields.

Theorem 6. Jordan–Dickson Theorem. If F is a finite field, then PSL_n(F) is a finite nonabelian simple group for all n ≥ 2, except for and

The proof is beyond the scope of this book.¹⁰³

The class of solvable groups is large. Of course, it contains all abelian groups, and a celebrated theorem of William Burnside asserts that every group of order pⁿq^m is solvable, where p and q are primes. In a different direction, Georg Frobenius showed that every group of square-free order is solvable. In 1911, Burnside conjectured that every nonabelian finite simple group has even order, equivalently (Exercise 13) that every group of odd order is solvable. This conjecture remained an open question until 1963 when two American algebraists Walter Feit and John Thompson proved that it is true. The proof is 254 pages long and fills an entire issue of the Pacific Journal of Mathematics,¹⁰⁴ and it is widely regarded as the best single paper in finite group theory. Thompson went on to classify all minimal finite simple groups, that is, those in which every proper subgroup is solvable, and played an important role in the classification of all finite simple groups. He was awarded the Fields Medal in 1970, the highest honor a mathematician can attain.

Even though the class of solvable groups is very large, many theorems are true for solvable groups that are not true of groups in general. One such theorem, a fundamental strengthening of the Sylow theorems in any solvable group, was first proved in 1928 by the British mathematician Philip Hall.

Theorem 7. Hall's Theorem. Let G be a group of order nm, where n and m are relatively prime. If G is solvable, then

We omit the proof.¹⁰⁵ Hall went on to develop the theory of finite solvable groups and influenced an entire generation of group theorists.

Exercises 9.2

9.3 Nilpotent Groups

If G is a group, the definition of the derived subgroup G′ guarantees that G is abelian if and only if G′ = {1}. If the process of taking the derived subgroup is iterated, the derived series G = G⁽⁰⁾⊇ G⁽¹⁾ ⊇ G⁽²⁾ ⊇ is obtained, and G is solvable if and only if this series (of normal subgroups of G) reaches {1} in a finite number of steps (Theorem 2 §9.2). Note that G⁽¹⁾ = G′. Now the center Z(G) plays an analogous role to G′ in the sense that G is abelian if and only if Z(G) = G. In view of this, an irresistible question arises: Is there a way to iterate the formation of the center so as to create a series {1} = Z₀⊆ Z₁ ⊆ Z₂ ⊆ of normal subgroups of G (with Z(G) = Z₁) such that G is solvable if and only if this series reaches G in a finite number of steps? The answer is yes and no. Yes, there is a natural way to define such a series. No, it does not characterize the solvable groups in this way. Rather, it characterizes a smaller class of groups called the nilpotent groups. In this section, we define these groups and show that the finite ones are precisely the finite groups that are isomorphic to the direct product of their Sylow subgroups.

Central Series

If G is a group, define a series Z₀(G), Z₁(G), Z₂(G), . . . of normal subgroups of G inductively as follows:

Then Z_i+1(G) ⊇ Z_i(G) and Z_i(G) Z_i+1(G) for each i ≥ 0, and the series

{1} = Z₀(G)⊆ Z₁(G) ⊆ Z₂(G) ⊆

is called the ascending central series of G. Note that

The ascending central series may never reach G, even if G is solvable:

Example 1. Suppose Z(G) = {1} where G ≠ {1}—for example the solvable group S₃. Then Z₁(G) = Z(G) = {1}, so we have Hence, Z₂(G) = {1}, and this process continues inductively to show that Z_k(G) = {1} for all k.

To characterize the groups G for which the ascending central series reaches G, it is useful to define a related descending central series. This requires a new notion. Recall that the derived subgroup G′ is generated by the commutators [a, b] = aba⁻¹b⁻¹ in G. We extend this idea as follows. If H and K are subgroups of a group G, define

to be the subgroup generated by the commutators [h, k], with h H and k K. Note that [h, k]⁻¹ = [k, h] for all h H and k K. Hence, [H, K] = [K, H], and this group consists of all products of commutators of the form [h, k] or [k, h], where h H and k K. In particular, [G, G] = G′. Lemma 1 collects several other useful facts about these subgroups.

Lemma 1. Let H, K, H₁, and K₁ be subgroups of a group G.

Proof. We prove (5) and leave the rest as Exercise 2. If , then we have hKgK = gKhK for all g G and h H; that is, [h, g] K. Hence, [H, G] ⊆ K. Since this argument works in reverse, we have proved (5).

Now, given a group G, define a series Γ₀(G), Γ₁(G), Γ₂(G), . . . of subgroups of G inductively as follows:

Then Γ_i(G) G for all i ≥ 0 by induction on i (using (3) of Lemma 1), whence Γ_i+1(G) = [Γ_i(G), G] ⊆ Γ_i(G) for each i by (4) of Lemma 1. Thus, we obtain a series of normal subgroups

G = Γ₀(G) ⊇ Γ₁(G) ⊇ Γ₂(G) ⊇ .

This is called the descending central series of G. The name comes from the fact that, using (5) of Lemma 1,

holds for each i ≥ 0. Note that

Γ₁(G) = [G, G] = G′ is the derived subgroup of G.

If G is an abelian group, then Z₁(G) = G and Γ₁(G) = {1}. On the other hand, there are groups (even solvable ones by Example 1) for which the ascending central series does not reach G and the descending central series does not reach 1. However, if either possibility occurs, so does the other.

Lemma 2. The following are equivalent for a group G and an integer n ≥ 0:

Proof. Write Γ_i(G) = Γ_i and Z_i(G) = Z_i for each i.

(1) ⇒ (2). If Γ_n = {1}, we show that Γ_n−i ⊆ Z_i for each i = 0, 1, 2, . . . (so Z_n = G). This is clear if i = 0 by (1), so assume Γ_n−i ⊆ Z_i, where i > 0. If a Γ_n−i−1, then, for all g G, [a, g] [Γ_n−i−1, G] = Γ_n−i ⊆ Z_i. Thus, aZ_i is in the center of G/Z_i and so a Z_i+1. Hence, Γ_n−i−1 ⊆ Z_i+1 as required.

(2) ⇒ (3). Given (2), use G = Z_n ⊇ Z_n−1 ⊇ ⊇ Z₀ = {1} in (3).

(3) ⇒ (1). Given (3), we show that Γ_i ⊆ G_i for each i = 0, 1, 2, . . . (so Γ_n = {1}). This is clear if i = 0, so assume that Γ_i ⊆ G_i for some i > 0. We must show that [Γ_i, G] = Γ_i+1 ⊆ G_i+1, so we show that [a, g] G_i+1 for all a Γ_i, g G. But by (3), so a Γ_i ⊆ G_i implies that aG_i+1 commutes with gG_i+1 for all g G. This in turn implies that [a, g] G_i+1, as required.

A group G is called a nilpotent group if the conditions in Lemma 2 are satisfied for some n ≥ 0. The smallest integer n for which Γ_n(G) = {1}, equivalently Z_n(G) = G, is called the nilpotency class of G. Thus, if G is nilpotent, then G has class 1 if and only if it is abelian, and (Exercise 11) G has class 2 if and only if it is nonabelian and G′ ⊆ Z(G).

A series as in (3) of Lemma 2 is called a central series for G. Suppose that G = G₀ ⊇ G₁ ⊇ ⊇ G_n = {1} is any central series for a nilpotent group G. Then the proof that (3) ⇒ (1) in Lemma 2 derives the first of the following inclusions:

We leave the second inclusion for the reader (Exercise 7). Hence, we often call the series G =Γ₀(G) ⊇ Γ₁(G) ⊇ Γ₂(G) ⊇ and 1 = Z₀(G)⊆ Z₁(G) ⊆ Z₂(G) ⊆ the lower and upper central series, respectively.

Example 2. Every abelian group is nilpotent.

Example 3. If p is a prime, every finite p -group is nilpotent. In fact, if we write Z_i(G) = Z_i for each i, Theorem 6 §8.2 shows that Z_i+1/Z_i = Z(G/Z_i) is not trivial if Z_i ≠ G because G/Z_i is a p-group. Hence, 1 ⊂ Z₁ ⊂ Z₂ ⊂ , which eventually reaches G because G is finite.

Example 4. Show that every nilpotent group G is solvable, but not conversely.

Solution. If G is nilpotent, the series {1} = Z₀(G) ⊆ ⊆ Z_n(G) = G is a solvable series. However, S₃ is solvable but not nilpotent by Example 1.

Theorem 1. Every subgroup and image of a nilpotent group is again nilpotent.

Proof. Let G be nilpotent. To show that a subgroup K ⊆ G is nilpotent, it suffices to show that Γ_i(K) ⊆ Γ_i(G) for each i. This is clear if i = 0. If Γ_i(K) ⊆ Γ_i(G) for some i, then the induction goes through because

Γ_i+1(K) = [Γ_i(K), K] ⊆ [Γ_i(G), G] = Γ_i+1(G).

Now let α: G → H be an onto homomorphism; we show that Γ_i(H) ⊆ α[Γ_i(G)] for each i. If i = 0, it is because α is onto. In general, let Γ_i(H) ⊆ α(Γ_i(G)), and let y Γ_i(H) and h H. Write y = α(x), where x Γ_i(G), and (as α is onto) write h = α(g), g G. Then

[y, h] = [α(x), α(g)] = α[x, g] α[Γ_i(G), G] = α(Γ_i+1(G)).

Hence, Γ_i+1(H) = [Γ_i(H), H] ⊆ α(Γ_i+1(G)), as required.

Corollary. A group G is nilpotent if and only if G′ is nilpotent.

Proof. Write D = G′. If D⁽ⁿ⁾ = {1}, then G⁽ⁿ⁺¹⁾ = D⁽ⁿ⁾ = {1}.

By Theorem 1, if K G and G is nilpotent, then both K and G/K are nilpotent. The converse is false (S₃ is again a counterexample) in contrast to the situation for solvable groups. However, the converse does hold when K ⊆ Z(G).

Theorem 2. If G is a group, K ⊆ Z(G), and G/K is nilpotent, then G is nilpotent.

Proof. Assume that G/K is nilpotent, and let θ: G → G/K be the coset map. By hypothesis, let θ(G) = X₀ ⊇ X₁ ⊇ ⊇ X_n = {K} be a central series for θ(G), where for 0 ≤ i ≤ n − 1. Write X_i = G_i/K = θ(G_i) for 0 ≤ i ≤ n. Then we obtain the series

G = G₀ ⊇ G₁ ⊇ ⊇ G_n = K ⊇ G_n+1 = {1}.

We show this is a central series for G. First, because K ⊆ Z(G). To see that for 0 ≤ i < n, let a G_i. Then θ(a) θ(G_i) = X_i, so θ(a) X_i+1 commutes with θ(g) X_i+1 for all g G. Thus, θ[a, g] = [θ(a), θ(g)] X_i+1 = θ(G_i+1), say θ[a, g] = θ(b), b G_i+1. Thus, [a, g]b⁻¹ ker θ = K ⊆ G_i+1, so [a, g] G_i+1b = G_i+1. This means aG_i+1 commutes with gG_i+1, that is, , as required.

The next result will be needed in Theorem 4.

Theorem 3. If G₁, G₂, , G_n are nilpotent, so also is G₁ × G₂ × × G_n.

Proof. This follows because Γ_i(G₁ × G₂ × × G_n) ⊆ Γ_i(G₁) × × Γ_i(G_n) for each i, a fact that we leave as Exercise 6.

Theorem 3 and Example 3 combine to show that any finite direct product of finite p-groups (for various primes p) is nilpotent. In fact, every finite nilpotent group is isomorphic to such a direct product. We need the following notion.

A subgroup M of a group G is said to be maximal in G if M ≠ G and the only subgroups H such that M ⊆ H ⊆ G are H = M and H = G. Clearly, every proper subgroup K of a finite group is contained in a maximal subgroup—one of maximal order containing K. If G is finite, every subgroup of prime index is maximal by Example 6 §2.6.¹⁰⁶ The converse is not necessarily true (any subgroup of index 4 in A₄ is maximal), but it does hold in a finite p-group. Moreover, in this case, the maximal subgroups (of index p) are necessarily normal (see the corollary to Theorem 1 §8.3). This property characterizes the finite nilpotent groups.

Theorem 4. Burnside–Wielandt Theorem. ¹⁰⁷ The following conditions are equivalent for a finite group G ≠ {1}:

Proof. (1) ⇒ (2). Write Z_i = Z_i(G) for each i and assume that Z_n = G. If H ≠ G is a subgroup of G, then Z₀ ⊆ H but Z_n H, so an integer k ≥ 0 exists such that Z_k ⊆ H but Z_k+1 H. Choose a Z_k+1, a ∉ H. Then aZ_k is in the center of G/Z_k, so if h H, aZ_k and hZ_k commute. Hence, hah⁻¹a⁻¹ Z_k ⊆ H, from which aHa⁻¹ ⊆ H. Thus, a N(H), and so N(H) ≠ H.

(2) ⇒ (3). Let M be a maximal subgroup of G. Since M ⊆ N(M) ⊆ G, (2) implies that N(M) = G. Hence, M G.

(3) ⇒ (4). Suppose P is a nonnormal Sylow p-subgroup of G. Then N(P) ≠ G, so let N(P) ⊆ M, where M is a maximal subgroup of G. Because P ⊆ M, (3) gives aPa⁻¹ ⊆ aMa⁻¹ = M for all a G. Hence, both P and aPa⁻¹ are Sylow p-subgroups of M and so are conjugate in M, say P = m(aPa⁻¹)m⁻¹ for some m M. But then ma N(P), so a M. Because a G was arbitrary, this means G ⊆ M, a contradiction. This proves (4).

(4) ⇒ (5). Let P₁, P₂, , P_r denote the distinct Sylow subgroups of G.

Claim 1. P₁P₂ P_k ≅ P₁ × P₂ × × P_k for each k = 2, 3, , r.

Proof. It is clear if k = 1. Assume inductively that P₁P₂ P_k ≅ P₁ × P₂ × × P_k for some k > 1. Then (P₁P₂ P_k) ∩ P_k+1 = {1} because elements in the two subgroups have relatively prime orders. By Theorem 6 §2.8,

(P₁P₂ P_k)P_k+1 ≅ (P₁ × P₂ × × P_k) × P_k+1 ≅ P₁ × P₂ × × P_k+1

because P₁P₂ P_k G and P_k+1 G. This proves the claim.

The claim gives |P₁P₂ P_r| ≅ |P₁||P₂| |P_r| = |G|. Hence G = P₁P₂ P_r and (5) follows, again by the Claim.

(5) ⇒ (1). This follows from Theorem 3 and Example 3.

It is interesting to compare (2) in Theorem 4 with the result (Theorem 5 §9.2) that a finite group G is solvable if and only if H′ ≠ H for every subgroup H ≠ {1}.

Since every finite abelian group is nilpotent, the implication (1) ⇒ (5) in Theorem 4 gives another proof of the primary decomposition theorem for finite abelian groups (Corollary 2 of Theorem 3 §7.2). We reformulate (1)(5) as

Corollary 1. A finite group G is nilpotent if and only if G is isomorphic to a finite direct product of p-groups for various primes p.

Frattini and Fitting Subgroups

One of the most important aspects of the study of nilpotent groups is that every finite group G contains a nilpotent subgroup Φ, which is characteristic in G (that is, σ(Φ) = Φ for every automorphism σ of G—these subgroups are discussed in Corollary 3 of Theorem 3 §2.8). We now turn to a discussion of this.

If G ≠ {1} is a finite group, define the Frattini subgroup

Φ(G) = {M ⊆ G M is a maximal subgroup of G}.

Define Φ{1} = {1}. This was introduced in 1885 by Giovanni Frattini.

Example 5. Φ(A₄) = {ε}. Indeed, K = {ε, (12)(34), (13)(24), (14)(23)} is maximal, being of index 3, and M = {ε, (123), (132)} is maximal (it has index 4, but A₄ has no subgroup of index 2). Hence, we have Φ(A₄) ⊆ K ∩ M = {ε}.

Example 6. If Q = { ± 1, ± i, ± j, ± k} is the quaternion group, then Φ(Q) = {1, − 1} because and are the only maximal subgroups.

Example 7. If and o(a) = pⁿ, where p is a prime, because is the unique maximal subgroup (of index p).

Theorem 5. Let G be a group and write Φ = Φ(G). Then the following hold:

Proof. (1) If U ⊆ H is a maximal subgroup, we must show that α(Φ) ⊆ U. If we define M = {m G α(m) U}, then it suffices to show that M is a maximal subgroup of G (since then Φ ⊆ M). But if M ⊆ K ⊆ G are subgroups of G then α(M) ⊆ α(K) ⊆ α(G). Since α is onto, this is U ⊆ α(K) ⊆ H, so α(K) = U or σ(K) = H. These imply that K = M or K = G.

(2) This follows from (1) if H = G and α is an automorphism of G.

(3) G′ ⊆ Φ if and only if G′ ⊆ M for each maximal subgroup M of G, if and only if M G for each M, and if and only if G is nilpotent by Theorem 4.

Corollary 1. The following are equivalent for a finite group G:

Proof. (1)(2) restates (3) of Theorem 5 and (2)⇒(3) is obvious.

(3)⇒(1). Given (3), we show that every maximal subgroup M of G is normal. Write Φ = Φ(G). Since Φ ⊆ M, the subgroup M/Φ is maximal in G/Φ by the correspondence theorem, so M/Φ G/Φ by (3) and Theorem 4. But then M G, again by the correspondence theorem.

To see that Φ(G) is a nilpotent group, we first characterize it in terms of the following concept. An element t G is called a nongenerator in G if it can be omitted from any generating set X of G; that is, if , then

Theorem 6. Let G denote any finite group. Then the following hold:

Proof. For convenience, write Φ = Φ(G).

(1) Write N = {t t is a nongenerator of G} and let a Φ. If a ∉ N, then X ⊆ G exists such that but So let where M is a maximal subgroup of G. Then a M because Φ ⊆ M, whence a contradiction. Hence, Φ ⊆ N. Conversely, if t N and M is a maximal subgroup of G, then g M (otherwise ). Hence, N ⊆ Φ.

(2) By Theorem 4 we show that every Sylow p-subgroup P of Φ is normal in G. If g G, then gPg⁻¹ ⊆ gΦg⁻¹ = Φ by (2) of Theorem 5. Hence, both gPg⁻¹ and P are Sylow p-subgroups of Φ and so are conjugate in Φ, say t(gPg⁻¹)t⁻¹ = P, where t Φ. Thus, tg N(P), which yields G = ΦN(P). But then so, as Φ is finite, by (1). Hence, P G as required.

Note the proof of (2) shows that Sylow p-subgroups of Φ(G) are normal in G.

Corollary 1. Assume that Φ(G) is finitely generated (for example, if G is finite). If HΦ(G) = G, where H is a subgroup, then H = G.

Proof. Write Φ(G) = Φ. If then and the nongenerators t_i can be removed one by one.

The next result extends a useful theorem about finite p-groups.

Theorem 7. If G is a nilpotent group and {1} ≠ H G, then H ∩ Z(G) ≠ {1}.

Proof. Write Γ_i = Γ_i(G) for each i. We have G = Γ₀ ⊇ Γ₁ ⊇ ⊇ Γ_n = {1} for some n, so H ∩ Γ_n = {1} while H ∩ Γ₀ ≠ {1}. So there exists k such that H ∩ Γ_k ≠ {1} while H ∩ Γ_k+1 = {1}. Choose 1 ≠ h H ∩ Γ_k. If g G, then [h, g] [Γ_k, G] = Γ_k+1. Also, [h, g] = h⁻¹g⁻¹hg H(g⁻¹Hg) = H because H G. So [h, g] H ∩ Γ_k+1 = {1}, whence hg = gh. Thus, h Z(G) ∩ H.

In 1938, Hans Fitting identified a largest nilpotent normal subgroup in every finite group. His key result was

Theorem 8. Fitting's Theorem. If H and K are nilpotent, normal subgroups of a finite group G, so also is HK.

Proof. We proceed by induction on We have HK G, so we may assume (by induction) that G = HK and that H ≠ {1} ≠ K. Write W = Z(K). Then W ≠ {1} by Theorem 7. Also, W G being characteristic in K G. If we write N = [W, H], the proof falls into two cases.

Case 1. N = {1}. Then W centralizes H (and K), so W ⊆ Z(HK) = Z(G). But Moreover, and are both nilpotent by Theorem 1, so is nilpotent by induction. Hence, G is nilpotent by Theorem 2.

Case 2. N ≠ {1}. We have N ⊆ W ∩ H because W G and H G. In particular, V = N ∩ Z(H) ≠ {1} again by Theorem 7. As before, and both and are nilpotent, so is nilpotent by induction. But V centralizes H, and it also centralizes K because V ⊆ N ⊆ W = Z(K). Hence, V ⊆ Z(HK) = Z(G), so G is nilpotent by Theorem 2 and we are done in this case too.

Now let G be any finite group. If N₁ = {1}, N₂, . . ., N_k denote all the nilpotent, normal subgroups of G, define the Fitting subgroup F(G) of G by

Then F(G) G, and it is nilpotent by Theorem 8 and induction on k. This proves

Theorem 9. If G is a finite group, then F(G) is the largest nilpotent, normal subgroup of G in the sense that it contains every such subgroup.

Corollary 1. If α: G → H is an onto homomorphism, then α[F(G)] ⊆ F(H).

Proof. α[F(G)] H because α is onto, and it is nilpotent by Theorem 1.

Hence, a finite group G is nilpotent if and only if F(G) = G. Clearly, Z(G) ⊆ F(G), and Φ(G) ⊆ F(G) because it is nilpotent (Theorem 6) and normal in G (Theorem 5). Moreover, F(G) is a characteristic subgroup of G by the above corollary.

Lemma 3. If G is finite and N G, then N is nilpotent if and only if N′ ⊆ Φ(G).

Proof. Write Φ(G) = Φ. If N′ ⊆ Φ, then N′ is nilpotent by Theorem 6, and so N is nilpotent by the corollary to Theorem 1. Conversely, if N is nilpotent, then N′ ⊆ Φ(N) by Theorem 5, and it remains to show that Φ(N) ⊆ Φ. Write Φ(N) = H and suppose H6 ⊆ Φ. Then H6 ⊆ M for some maximal subgroup M of G, so HM = G. Since H ⊆ N, the modular law (Lemma 1 §8.1) gives

N = G ∩ N = HM ∩ N = H(M ∩ N).

Thus, and H = Φ(N) consist of (a finite number of) non-generators of N. Hence, N = M ∩ N, whence N ⊆ M, a contradiction.

We can now describe the relationship between the Frattini and Fitting subgroups in a finite group.

Theorem 10. Let G be a finite group and write Φ = Φ(G) and F = F(G).

Proof. (1) We have F′ G because it is characteristic in F G, so F′ ⊆ Φ by Lemma 3 because F is nilpotent. On the other hand, Φ ⊆ F by Theorem 9 because Φ is nilpotent. This proves (1).

(2) This follows from a more general result (Theorem 11).¹⁰⁸

Theorem 11. If G is a finite group and Φ(G) ⊆ N G, then N is nilpotent if and only if N/Φ(G) is nilpotent.

Proof. Write Φ(G) = Φ. If N is nilpotent, so is its image For the converse, assume that is nilpotent. To show that N is nilpotent, we show that every Sylow p -subgroup P of N is normal in N (and invoke Theorem 4). First, is a Sylow p-subgroup of (by Example 4 §8.4), whence by hypothesis. But then is characteristic in and it follows that Thus, ΦP G. But P is a Sylow p-subgroup of ΦP (because ΦP ⊆ N), so G = (ΦP)N_G(P) by Lemma 1 §8.4. Since G is finite and Φ consists of nongenerators, it follows that G = P N_G(P) = N_G(P). Hence, P G, so certainly P N as required.

There is much more information available on nilpotent groups in books on group theory.¹⁰⁹

Exercises 9.3

Notes

100. Due to Hans Zassenhaus.

101. Due to Otto Schreier.

102. These groups are also called soluble groups.

103. See Kargapolov, M.I. and Merzljakov, J.I., Fundamentals of the Theory of Groups, Springer-Verlag, 1979; Rotman, J.J., The Theory of Groups: An Introduction, 2nd ed., Boston: Allyn & Bacon, 1973; Artin, E., Geometric Algebra, New York: Interscience, 1957. For an elementary proof when n = 2, see Lang, S., Undergraduate Algebra, Berlin: Springer-Verlag, 1987.

104. Feit, W. and Thomson, J.G., Solvability of groups of odd order, Pacific Journal of Mathematics, 13 (1963), 775–1029.

105. See Kargapolov, M.I. and Merzljakov, J.I., Fundamentals of the Theory of Groups, Springer-Verlag, 1979; MacDonald, I.D., The Theory of Groups, London: Oxford University Press, 1968; Rotman, J.J., The Theory of Groups: An Introduction, 2nd ed., Boston: Allyn & Bacon, 1973.

106. This is also true if G is infinite by Exercise 31, §2.6.

107. The name honors William Burnside and Helmut Wielandt.

108. Because of (2) and the corollary to Theorem 9, the Fitting subgroup of G is also called the nilpotent radical of G.

109. The following books contain excellent introductions to the theory of nilpotent groups: Mac-Donald, I.D., The Theory of Groups, London: Oxford University Press, 1968; Rose, J.S., A Course on Group Theory, Cambridge, England: Cambridge University Press, 1978; Kargapolov, M.I. and Merzljakov, J.I., Fundamentals of the Theory of Groups, New York: Springer-Verlag, 1979; Gorenstsein, D., Finite Groups, 2nd ed., New York: Chelsea, 1980.