§ 6. Extensions, solvable groups, nilpotent groups

Throughout this paragraph, the group laws are, unless expressly mentioned otherwise, written multiplicatively.

1. Extensions

Definition 1

Let $F$ and $G$ be two groups. An extension of $G$ by $F$ is a triple $\mathscr{E} = (E, i, p)$ , where $E$ is a group, $i$ an injective homomorphism of $F$ into $E$ and $p$ a surjective homomorphism of $E$ onto $G$ such that $\operatorname{Im}(i) = \operatorname{Ker}(p)$ . A homomorphism $s : G \to E$ (resp. $r : E \to F$ ) such that $p \circ s = \operatorname{Id}_G$ (resp. $r \circ i = \operatorname{Id}_F$ ) is called a section (resp. retraction) of the extension $\mathscr{E}$ .

An extension $\mathscr{E} = (E, i, p)$ of $G$ by $F$ is often denoted by the diagram $\mathscr{E} : F \xrightarrow{i} E \xrightarrow{p} G$ , in which $i$ and $p$ are sometimes omitted if no confusion can arise. It is sometimes said simply that the group $E$ is an extension of $G$ by $F$ .

For a group $E$ to be an extension of $G$ by $F$ , it is necessary and sufficient that it contain a normal subgroup $F'$ isomorphic to $F$ such that the quotient group $E/F'$ is isomorphic to $G$ .

An extension $\mathscr{E} : F \xrightarrow{i} E \xrightarrow{p} G$ is called central if the image $i(F)$ is contained in the centre of $E$ ; this is only possible if $F$ is commutative.

Let $\mathscr{E} : F \xrightarrow{i} E \xrightarrow{p} G$ and $\mathscr{E}' : F \xrightarrow{i'} E' \xrightarrow{p'} G$ be two extensions of $G$ by $F$ . A morphism of $\mathscr{E}$ into $\mathscr{E}'$ is a homomorphism $u : E \to E'$ such that $p' \circ u = p$ and $u \circ i = i'$ , or, in other words, such that the following diagram is commutative:

\begin{array}{ccc} & & E \\ & i & \\ F & \xrightarrow{i'} & E' \\ & \downarrow & \downarrow \\ & & E \\ & p & \\ & & G \end{array}

u \uparrow \quad \text{and} \quad p' \uparrow

Proposition 1

Let $\mathscr{E}: F \xrightarrow{i} E \xrightarrow{p} G$ and $\mathscr{E}': F \xrightarrow{i'} E' \xrightarrow{p'} G$ be extensions of $G$ by $F$ . If $u: E \to E'$ is a morphism of $\mathscr{E}$ into $\mathscr{E}'$ , $u$ is an isomorphism of $E$ onto $E'$ and $u^{-1}$ is a morphism of $\mathscr{E}'$ into $\mathscr{E}$ .

Let $x \in E$ be such that $u(x) = e$ . Then $p(x) = p'(u(x)) = e$ , whence $x \in i(F)$ . Let $y \in F$ be such that $x = i(y)$ ; then $i'(y) = u(i(y)) = e$ . As $i'$ is injective, $y = e$ and $x = e$ . Therefore $u$ is injective. By virtue of § 4, no. 6, Corollary 1 to Proposition 7, $u$ is surjective since $u(i(F)) = i'(F)$ . The last assertion is immediate.

In other words, the extensions $\mathscr{E}$ and $\mathscr{E}'$ are isomorphic if and only if there exists a morphism of $\mathscr{E}$ into $\mathscr{E}'$ .

Let $F$ and $G$ be two groups and write $E_0 = F \times G$ ; let $i: F \to E_0$ be the canonical injection and $p: E_0 \to G$ be the canonical projection. Every extension of $G$ by $F$ isomorphic to the extension $\mathscr{E}_0: F \xrightarrow{i} E_0 \xrightarrow{p} G$ is called a trivial extension.

Proposition 2

Let $\mathscr{E}: F \xrightarrow{i} E \xrightarrow{p} G$ be an extension of $G$ by $F$ . The following conditions are equivalent*:

(i) $\mathscr{E}$ is a trivial extension; (ii) $\mathscr{E}$ has a retraction $r$ ; (iii) $\mathscr{E}$ has a section $s$ such that $s(G)$ is contained in the centralizer of $i(F)$ .

Clearly (i) implies (ii) and (iii). If (ii) holds, the mapping $(r, p): E \to F \times G$ is a morphism of $\mathscr{E}$ into $\mathscr{E}_0$ , whence (i). If (iii) holds, the homomorphism of $F \times G$ into $E$ corresponding to $(i, s)$ ( $\S$ 4, no. 9, Proposition 12) is a morphism of $\mathscr{E}_0$ into $\mathscr{E}$ , whence (i).

It may be that an extension $\mathscr{E}: F \to E \to G$ is not trivial and yet the group $E$ is isomorphic to $F \times G$ (Exercise 6).

Definition 2

Let $F$ and $G$ be two groups and $\tau$ a homomorphism of $G$ into the automorphism group of $F$ . Write $\tau(g)(f) = gf$ for $g \in G$ and $f \in F$ . The set $F \times G$ with the law of composition

((f, g), (f', g')) \mapsto (f, g) \cdot_{\tau} (f', g') = (f \cdot gf', gg')

is called the external semi-direct product of $G$ by $F$ relative to $\tau$ .

The external semi-direct product of $G$ by $F$ relative to $\tau$ is denoted by $F \times_{\tau} G$ .

Proposition 3

The external semi-direct product $F \times_{\tau} G$ is a group. The mappings $i: F \to F \times_{\tau} G$ defined by $i(f) = (f, e)$ , $p: F \times_{\tau} G \to G$ defined by $p(f, g) = g$ , and $s: G \to F \times_{\tau} G$ defined by $s(g) = (e, g)$ are group homomorphisms. The triple $(F \times_{\tau} G, i, p)$ is an extension of $G$ by $F$ and $s$ is a section of the extension.

We have:

((f, g) \cdot_{\tau} (f', g')) \cdot_{\tau} (f'', g'') = (f.gf', gg') \cdot_{\tau} (f'', g'') = (f.gf'.gg'f'', gg'g'');

(f, g) \cdot_{\tau} ((f', g') \cdot_{\tau} (f'', g'')) = (f, g) \cdot_{\tau} (f'g'f'', g'g'') = (f.g(f'.g'f''), gg'g'').

Now $g(f'.g'f'') = gf'.gg'f''$ , which shows that the law of composition defined by (1) is associative. The element $(e, e)$ is the identity under this law. The element $(f, g)$ admits as inverse $(g^{-1}f^{-1}, g^{-1})$ . Hence the law of composition on $F \times_{\tau} G$ is a group law. The other assertions are immediate.

Using the notation of Proposition 3, $\mathscr{E}_{\tau}$ will denote the extension

F \xrightarrow{i} F \times_{\tau} G \xrightarrow{p} G.

Let $\mathscr{E}' : F \xrightarrow{i'} E' \xrightarrow{p'} G$ be an extension of $G$ by $F$ and $s' : G \to E'$ a section of $\mathscr{E}'$ . We define an operation $\tau$ of $G$ on the group $F$ by:

(2)

i'(\tau(g, f)) = s'(g)i'(f)s'(g)^{-1} = \operatorname{Int}(s'(g))(i'(f)).

Proposition 4

With the above notation, there exists one and only one isomorphism u of $\mathscr{E}_{\tau}$ onto $\mathscr{E}'$ such that $u \circ s = s'$ .

(f, g) = (f, e) \cdot_{\tau} (e, g) = i(f) \cdot_{\tau} s(g).

Therefore, if

u

is a solution to the problem, of necessity

u(f, g) = i'(f).s'(g)

, whence the uniqueness of

u

. We prove the existence. We write

u(f, g) = i'(f).s'(g)

. Then

\begin{align*} u(f, g).u(f', g') &= i'(f)s'(g)i'(f')s'(g') \\ &= i'(f)(s'(g)i'(f')s'(g)^{-1})s'(g)s'(g') \\ &= i'(f)i'(\tau(g, f')).s'(g)s'(g') \\ &= i'(f.\tau(g, f')).s'(gg') \\ &= u((f, g) \cdot_{\tau} (f', g')). \end{align*}

Therefore, $u$ is a homomorphism of $F \times_{\tau} G$ into $E'$ . Obviously $u \circ i = i'$ , $p' \circ u = p$ and $u \circ s = s'$ .

Remark

The definition of the operation $\tau$ by formula (2) depends on the extension $\mathscr{E}'$ and the section $s'$ . When $F$ is commutative, the operation $\tau$ does not depend on $s'$ . For $\operatorname{Int}(s'(g)) | i'(F)$ depends then only on the coset of $s'(g)$ mod. $i'(F)$ .

More generally, let $\mathscr{E} : F \to E \to G$ be an extension of $G$ by a commutative group $F$ (it is not assumed that $\mathscr{E}$ admits a section). The group $E$ operates on $F$ by inner automorphisms, this image is trivial on the image of $F$ and hence defines an operation of $G$ on $F$ . If $\mathscr{E}$ admits a section, this operation is that defined by formula (2).

Corollary

Let $G$ be a group and $H$ and $K$ two subgroups of $G$ such that $H$ is normal, $H \cap K = \{e\}$ and $H.K = G$ . Let $\tau$ be the operation of $K$ on $H$ by inner automorphisms of $G$ . The mapping $(h, k) \mapsto hk$ is an isomorphism of $H \times_{\tau} K$ onto $G$ .

Under the hypotheses of this corollary, $G$ is said to be the semi-direct product of $K$ by $H$ .

Example

(1) Let $G$ be a group and $E$ a homogeneous principal $G$ -set; let $\Gamma$ denote the automorphism group of $G$ . Let $A$ be the set of permutations $f$ of $E$ with the following property: There exists $\gamma \in \Gamma$ such that $f$ is a $\gamma$ -morphism of $E$ into $E$ (that is, $f(gb) = \gamma(g)f(b)$ for $b \in E$ and $g \in G$ ). The above formula $f(gb) = \gamma(g)f(b)$ shows that if $f \in A$ there exists a unique $\gamma \in \Gamma$ such that $f$ is a $\gamma$ -morphism, we shall denote it by $p(f)$ . Let $f, f'$ be in $A$ , $\gamma = p(f), \gamma' = p(f')$ . Then, for all $b \in E$ and all $g \in G$ ,

(f' \circ f)(gb) = f'(\gamma(g)f(b)) = \gamma'(\gamma(g))f'(f(b))

which proves that

f' \circ f \in A

and that

p(f' \circ f) = p(f')p(f)

. On the other hand,

f(\gamma^{-1}(g)f^{-1}(b)) = gb

, whence

f^{-1}(gb) = \gamma^{-1}(g)f^{-1}(b)

and

f^{-1} \in A

. Thus

A

is a subgroup of

\mathscr{S}_E

and

p

is a homomorphism of

A

into

\Gamma

. The kernel of

p

is the set

\mathrm{Aut}_G(E)

of automorphisms of the

G

-set

E

. We fix

a \in E

. We have defined in § 5, no. 6 an isomorphism

\psi_a

G^0

onto

\mathrm{Aut}_G(E)

such that

\psi_a(x)(ga) = gxa

for all

g, x

G

. On the other hand, for

\gamma \in \Gamma

, let

s_a(\gamma)

be the permutation of

E

defined by

s_a(\gamma)(ga) = \gamma(g)a

for all

g \in G

; it is immediately verified that

s_a

is a homomorphism of

\Gamma

into

A

such that

p \circ s_a = \mathrm{Id}_{\tau}

. Thus

G^0 \xrightarrow{\psi_a} A \xrightarrow{p} \Gamma

is an extension of

\Gamma

G^0

and

s_a

is a section of this extension. This extension and this section define an operation of

\Gamma

G^0

s_a(\Gamma)

acting on

\psi_a(G^0)

by inner automorphisms; we write this operation exponentially. We show that this operation is the natural operation (

§ 3

, no. 1, Example 3): for

x, g

G

and

\gamma \in \Gamma

(\psi_a(\gamma x))(ga) = (s_a(\gamma) \circ \psi_a(x) \circ s_a(\gamma)^{-1})(ga) = (s_a(\gamma) \circ \psi_a(x))(\gamma^{-1}(g)a) = s_a(\gamma)(\gamma^{-1}(g)xa) = g\gamma(x)a = \psi_a(\gamma(x))ga

whence

^\gamma x = \gamma(x)

. Proposition 4 then shows that

A

is isomorphic to the semidirect product of

\Gamma = \mathrm{Aut}(G)

G^0

under the natural operation of

\mathrm{Aut}(G)

G^0

. Note that the isomorphism which we have constructed depends in general on the choice of the element

a \in E

. (2) *Let

A

be a commutative ring. The upper triangular group

T(n, A)

is the semi-direct product of the diagonal subgroup

D(n, A)

by the upper strict triangular group

T_1(n, A)

2. Commutators

Definition 3

Let $G$ be a group and $x$ and $y$ two elements of $G$ . The element $x^{-1}y^{-1}xy$ of $G$ is called the commutator of $x$ and $y$ .

(x, y) is used to denote the commutator of x and y. Then obviously

(y, x) = (x, y)^{-1}.

For x and y to commute it is necessary and sufficient that

(x, y) = e

. More generally,

xy = yx(x, y).

On the other hand we write (3)

x^y = y^{-1}xy = x(x, y) = (y, x^{-1})x.

As the mapping

x \mapsto x^y

is the inner automorphism

\operatorname{Int}(y^{-1})

(x, y)^z = (x^z, y^z)

for all

x, y, z \in G

For $x, y, z \in G$ , we prove the following relations: (4)

(x, yz) = (x, z).(x, y)^z = (x, z).(z, (y, x)).(x, y)

(4 bis)

(xy, z) = (x, z)^y.(y, z) = (x, z).((x, z), y).(y, z)

(5)

(x^y, (y, z)).(y^z, (z, y)).(z^x, (x, y)) = e

(6)

(x, yz).(z, xy).(y, zx) = e

(6 bis)

(xy, z).(yz, x).(zx, y) = e.

Now

\begin{align*} (x, yz) &= x^{-1}z^{-1}y^{-1}xyz = (x, z)z^{-1}x^{-1}y^{-1}xyz = (x, z)(x, y)^z \\ &= (x, z)(z, (x, y)^{-1})(x, y) \end{align*}

by (3), which proves (4). Formula (4 bis) follows similarly. On the other hand,

\begin{align*} (x^y, (y, z)) &= (x^y)^{-1}(z, y)(x^y)(y, z) \\ &= y^{-1}x^{-1}yz^{-1}y^{-1}zyy^{-1}xyy^{-1}z^{-1}yz \\ &= (yzy^{-1}xy)^{-1}(zxz^{-1}yz). \end{align*}

Then writing

u = yzy^{-1}xy, v = zxz^{-1}xy

and

w = xyx^{-1}zx

, we obtain

(x^y, (y, z)) = u^{-1}v.

By cyclically permuting

x, y, z

, we deduce

(y^z, (z, x)) = v^{-1}w

and

(z^x, (x, y)) = w^{-1}u,

which immediately imply (5). Finally, (6) follows by multiplying together the two sides in the three formulae obtained by cyclically permuting

x, y, z

in the formula

(x, yz) = x^{-1}z^{-1}y^{-1}xyz = (yzx)^{-1}(xyz)

, and similarly for (6 bis).

If A and B are two subgroups of G, (A, B) denotes the subgroup generated by the commutators $(a, b)$ with $a \in A$ and $b \in B$ .† Then $(A, B) = \{e\}$ if and only if A centralizes B. $(A, B) \subset A$ if and only if B normalizes A. If A and B are normal (resp. characteristic), so is $(A, B)$ .

† We here reject the notational convention made in § 1, no 1 of extending a law of composition to subsets.

Proposition 5

Let $A, B, C$ be three subgroups of $G$ .

(i) The subgroup $A$ normalizes the subgroup $(A, B)$ .

(ii) If the subgroup $(B, C)$ normalizes $A$ , the subgroup $(A, (B, C))$ is generated by the elements $(a, (b, c))$ with $a \in A, b \in B$ and $c \in C$ .

(iii) If $A, B$ and $C$ are normal, then

(A, (B, C)) \subset (C, (B, A)).(B, (C, A)).

By (4 bis), for $a, a' \in A$ and $b \in B$ ,

(a, b)^{a'} = (aa', b).(a', b)^{-1},

whence (i). Suppose now that $(B, C)$ normalizes $A$ . For $a \in A, b \in B, c \in C$ and $x \in G$ , (4) implies

(a, (b, c).x) = (a, x).(x, ((b, c), a))(a, (b, c))

and $((b, c), a) \in A$ since $(B, C)$ normalizes $A$ , whence by induction on $p$ the fact that $\left( a, \prod_{i=1}^p (b_i, c_i) \right)$ , for $b_i \in B, c_i \in C$ , belongs to the subgroup generated by the elements of the form $(a, (b, c))$ . If finally $A, B$ and $C$ are normal, so are the subgroups $(A, (B, C))$ , $(C, (B, A))$ and $(B, (C, A))$ . It therefore suffices by (ii) to show that

(a, (b, c)) \in (C, (B, A)).(B, (C, A))

for all $a \in A, b \in B$ and $c \in C$ . Now by (5), writing $a^{b^{-1}} = u$

(a, (b, c)) = ((u^b), (b, c)) = (c^u, (u, b))^{-1}.(b^c, (c, u))^{-1},

whence (iii).

Definition 4

Let $G$ be a group. The subgroup generated by the commutators of elements of $G$ is called the derived group of $G$ .

The derived group of $G$ is thus the subgroup $(G, G)$ . It is also denoted by $D(G)$ . By an abuse of language, it is sometimes called the commutator group of $G$ although it is in general distinct from the set of commutators of elements of $G$ (Exercise 16). $D(G) = \{e\}$ if and only if $G$ is commutative.

Proposition 6

Let $f : G \to G'$ be a group homomorphism. Then $f(D(G)) \subset D(G')$ . If $f$ is surjective, the homomorphism of $D(G)$ into $D(G')$ the restriction of $f$ is surjective.

The image under $f$ of a commutator of elements of $G$ is a commutator of elements of $G'$ . If $f$ is surjective, the image under $f$ of the set of commutators of $G$ is the set of commutators of $G'$ . The proposition thus follows from § 4, no. 3, Corollary 3 to Proposition 2.

Corollary 1

The derived group of a group $G$ is a characteristic subgroup of $G$ . In particular it is a normal subgroup of $G$ .

Corollary 2

Let G be a group. The quotient group G/D(G) is commutative. Let $\pi : G \to G/D(G)$ be the canonical homomorphism. Every homomorphism f of G into a commutative group G' can be expressed uniquely in the form $f = \bar{f} \circ \pi$ , where $\bar{f} : G/D(G) \to G'$ is a homomorphism.

Now $\pi(D(G)) = \{e\}$ . As $\pi$ is surjective, it follows that $D(G/D(G)) = \{e\}$ , whence the first assertion. The second follows from § 4, no. 4, Proposition 5.

Corollary 3

Let H be a subgroup of G. The following conditions are equivalent:

(i) $H \supset D(G)$ ; (ii) H is a normal subgroup and $G/H$ is commutative. (ii) $\Rightarrow$ (i) by Corollary 2 and (i) $\Rightarrow$ (ii) by § 4, no. 7, Theorem 4, since every subgroup of a commutative group is normal.

Corollary 4

Let G be a group and X a subset of G which generates G. The group D(G) is the normal subgroup of G generated by the commutators of elements of X.

Let H be the normal subgroup of G generated by the commutators of elements of X and $\phi : G \to G/H$ the canonical homomorphism. The set $\phi(X)$ generates $G/H$ . The elements of $\phi(X)$ are pairwise permutable and hence H is commutative (§ 4, no. 3, Corollary 2 to Proposition 2). Hence (Corollary 3) H contains D(G). On the other hand, obviously $H \subset D(G)$ .

Remark

(1) Corollary 2 can also be expressed by saying that $G/D(G)$ , together with $\pi$ , is a solution of the universal mapping problem for G, relative to commutative groups and homomorphisms from G to commutative groups.

(2) Under the hypotheses of Corollary 4, the subgroup generated by the commutators of elements of X is contained in D(G) but is not in general equal to D(G) (cf. Exercise 15e).

Example

(1) If G is a non-commutative simple group, then $D(G) = G$ . Therefore every homomorphism of G into a commutative group is trivial.

(2) The derived group of the symmetric group $S_n$ is the alternating group $A_n$ . For $A_n$ is generated by the products of two transpositions; if $\tau = \tau_{x,y}$ and $\tau' = \tau_{x',y'}$ are two transpositions, let $\sigma$ be a permutation such that $\sigma(x') = x$ and $\sigma(y') = y$ . Then $\tau' = \sigma^{-1} \tau \sigma$ and $\tau \tau' = \tau^{-1} \tau' = \tau^{-1} \sigma^{-1} \tau \sigma$ is a commutator. Hence $A_n \subset D(S_n)$ . As $S_n/A_n$ is commutative, $A_n \supset D(S_n)$ (Corollary 3).

3. Lower central series, nilpotent groups

Let G be a group, H a subgroup of G and K a normal subgroup of G. The image of H in G/K is contained in the centre of G/K if and only if $(G, H) \subset K$ .

Definition 5

Let G be a group. The lower central series of G is the sequence $(C^n(G))_{n \geq 1}$ of subgroups of G defined inductively by:

C^1(G) = G, \quad C^{n+1}(G) = (G, C^n(G)).

Let $f : G \to G'$ be a group homomorphism. It is seen, by induction on $n$ , that $f(C^n(G)) \subset C^n(G')$ and that, if $f$ is surjective, $f(C^n(G)) = C^n(G')$ . In particular, for all $n \geq 1$ , $C^2(G)$ is a characteristic (and hence normal) subgroup of $G$ . For all $n \geq 1$ , $C^n(G)/C^{n+1}(G)$ is contained in the centre of $G/C^{n+1}(G)$ .

Let $(G_1, G_2, \ldots)$ be a decreasing sequence of normal subgroups of $G$ such that (1) $G_1 = G$ ; (2) for all $i$ , $G_i/G_{i+1}$ is contained in the centre of $G/G_{i+1}$ . Then $C^i(G) \subset G_i$ , as is seen by induction on $i$ .

Now

(C^m(G), C^n(G)) \subset C^{m+n}(G).

For, if this relation is denoted by

(F_{m,n})

, it follows from

(F_{m,n})

, by no. 2, Proposition 5, that

\begin{align*} (C^m(G), C^{n+1}(G)) &\subset (G, (C^m(G), C^n(G))).(C^n(G), (G, C^m(G))) \\ &\subset C^{m+n+1}(G).(C^{m+1}(G), C^n(G)). \end{align*}

Hence

((F_{m,n}) \text{ and } (F_{m+1,n})) \Rightarrow (F_{m,n+1})

. As

(F_{m,1})

and

(F_{1,n})

are obvious

(F_{m,n})

follows by induction.

Definition 6

A group $G$ is called nilpotent if there exists an integer $n$ such that $C^{n+1}(G) = \{e\}$ . The least integer $n$ such that $C^{n+1}(G) = \{e\}$ is called the nilpotency class of a nilpotent group $G$ .

If $n \in \mathbf{N}$ , a group of nilpotency class $n$ is called a nilpotent group of class $n$ . It is sometimes said that the nilpotency class of a group $G$ is finite if $G$ is nilpotent.

Example

(1) A group is nilpotent of class 0 (resp. $\leq 1$ ) if and only if it consists of the identity element (resp. is commutative).

(2) *For every commutative ring $A$ and every integer $n \geq 1$ , the upper strict triangular group $T_1(n, A)$ is nilpotent of class $\leq n - 1$ (and exactly of class $n - 1$ if $A \neq \{0\}$ ).

(3) Let $G$ be a nilpotent group of class $n$ . Every subgroup (resp. every quotient group) of $G$ is nilpotent of class $\leq n$ . For, if $H$ is a subgroup of $G$ , then $C^n(H) \subset C^n(G)$ . If $G'$ is a quotient group of $G$ and $\pi : G \to G'$ is the canonical homomorphism, then $C^n(G') = \pi(C^n(G))$ .

(4) A finite product of nilpotent groups is nilpotent.

Proposition 7

Let $G$ be a group and $n$ an integer. The following conditions are equivalent:

(a) $G$ is nilpotent of class $\leq n$ . (b) There exists a series of subgroups of $G$ :

G = G^1 \supset G^2 \supset \ldots \supset G^{n+1} = \{e\}

such that

(G, G^{k'}) \subset G^{k'+1}

for all

k \in \{1, n\}

(c) There exists a subgroup $A$ of $G$ contained in the centre of $G$ such that $G/A$ is nilpotent of class $\leq n - 1$ .

(a) $\Rightarrow$ (b): it suffices to take $G^k = C^k(G)$ . (b) $\Rightarrow$ (a): by induction on $k$ , $C^k(G) \subset G^k$ . (a) $\Rightarrow$ (c): it suffices to take $A = C^n(G)$ . (c) $\Rightarrow$ (a): let $\pi : G \to G/A$ be the canonical homomorphism; then $\pi(C^n(G)) = C^n(G/A) = \{e\}$ and hence $C^n(G) \subset A$ , whence $C^{n+1}(G) = \{e\}$ .

More briefly: a group is nilpotent of class $\leq n$ if it can be obtained from the group $\{e\}$ by $n$ successive central extensions.

Corollary

A central extension of a nilpotent group (by a necessarily commutative group) is nilpotent.

Proposition 8

Let $G$ be a nilpotent group of class $\leq n$ and let $H$ be a subgroup of $G$ . There exists a sequence of subgroups

G = H^1 \supset H^2 \supset \ldots \supset H^{n+1} = H,

such that $H^{k+1}$ is normal in $H^k$ and $H^k/H^{k+1}$ is commutative for all $k \leq n$ .

Choose a sequence $(G^k)$ of subgroups of $G$ satisfying the conditions of Proposition 7 (b) for all $k$ ; $G^k$ is normal in $G$ . Write:

H^k = H.G^k.

It is necessary to verify that $H^{k+1}$ is normalized by $H^k = H.G^k$ ; as it is normalized by $H$ , it suffices to verify that it is by $G^k$ . Now, if $s \in G^k$ and $h \in H$ ,

shs^{-1} = shs^{-1}h^{-1}.h \in (G, G^k).H

and $(G, G^k).H$ is contained in $G^{k+1}.H = H^{k+1}$ ; hence $s.H^{k+1}.s^{-1} = H^{k+1}$ , which shows that $H^{k+1}$ is normal in $H^k$ .

Finally, the canonical homomorphism $G^k/G^{k+1} \to H^k/H^{k+1}$ is obviously surjective; as the first group is commutative, so is the second.

Corollary 1

Let $G$ be a nilpotent group and $H$ a subgroup of $G$ . If $H$ is distinct from $G$ , the normalizer $N_G(H)$ of $H$ in $G$ is distinct from $H$ .

Let $k$ be the largest index such that $H^k \neq H$ . The group $H^k$ normalizes $H$ and is distinct from $H$ .

Corollary 2

Let $G$ be a nilpotent group and $H$ a subgroup of $G$ . If $H$ is distinct from $G$ , there exists a normal subgroup $N$ of $G$ , containing $H$ , distinct from $G$ and such that $G/N$ is commutative.

Let $k$ be the least index such that $H^k \neq G$ . The group $H^k$ satisfies the required conditions.

Corollary 3

Let G be a nilpotent group and H a subgroup of G. If G = H.(G, G), then G = H.

Every subgroup N of G which contains H and such that G/N is commutative contains H.(G, G). Corollary 3 thus follows from Corollary 2.

Corollary 3 can also be formulated thus: let X be a subset of G. For X to generate G, it is necessary and sufficient that the image of X in G/D(G) generate G/D(G).

Corollary 4

Let f: G' → G be a group homomorphism. Suppose that* (a) G is nilpotent. (b) The homomorphism f₁: G'/(G', G') → G/(G, G), derived from f by passing to the quotients, is surjective. *Then f is surjective.

This follows from Corollary 3 applied to the subgroup H = f(G').

Proposition 9

Let G be a nilpotent group of class ≤ n and let N be a normal subgroup of G. There exists a series of subgroups

N = N^1 \supset N^2 \supset \ldots \supset N^{n+1} = \{e\}

such that (G, N^k) ⊂ N^{k+1} for k = 1, ..., n.

If (G^k) satisfies condition (b) of Proposition 7, then take

N^k = G^k \cap N.

Corollary 1

Let G be a nilpotent group, Z the centre of G and N a normal subgroup of G. If N ≠ {e}, then N ∩ Z ≠ {e}.

Let k be the largest index such that N^k ≠ {e}. The group N^k is contained in N. On the other hand, (G, N^k) ⊂ N^{k+1} = {e}; hence N^k is contained in the centre Z of G.

Corollary 2

Let f be a homomorphism from a nilpotent group G to a group G'. If the restriction of f to the centre of G is injective, f is injective.

This is Corollary 1 applied to Ker(f).

4. Derived series, solvable groups

Definition 7

Let G be a group. The derived series of G is the series (D^n(G))_{n \in \mathbf{N}} defined inductively by:

D^0(G) = G; \quad D^{n+1}(G) = D(D^n(G)) \quad \text{for } n \in \mathbf{N}.

Then $D^0(G) = C^1(G) = G,\ D^1(G) = C^2(G) = D(G) = (G, G)$ . For all n \in \mathbf{N}, D^n(G) \subset C^{2^n}(G), as is seen by induction on n using formula (7) of no. 3.

Let $f : G \to G'$ be a group homomorphism. It is seen, by induction on n, that $f(D^n(G)) \subset D^n(G')$ and that, if $f$ is surjective, $f(D^n(G)) = D^n(G')$ . In particular, for all $n \in \mathbf{N}$ , $D^n(G)$ is a characteristic (and therefore normal) subgroup of G. For all $n \in \mathbf{N}$ , the group $D^n(G)/D^{n+1}(G)$ is a commutative normal (but not in general central) subgroup of $G/D^{n+1}(G)$ .

Let $(G_0, G_1, \ldots)$ be a decreasing sequence of subgroups of G such that: (1) $G_0 = G$ ; (2) for all $i$ , $G_{i+1}$ is normal in $G_i$ and $G_i/G_{i+1}$ is commutative. Then $D^i(G) \subset G_i$ for all $i$ , as is seen by induction on $i$ .

Definition 8

A group G is called solvable if there exists an integer n such that $D^n(G) = \{e\}$ . If G is a solvable group, the least integer n such that $D^n(G) = \{e\}$ is called the solvability class of G.

A solvable group of solvability class $n$ is called a solvable group of class $n$ . A group is sometimes said to be of finite solvability class if it is solvable.

Example

(1) A group is solvable of class 0 (resp. $\leq 1$ ) if and only if it is reduced to $\{e\}$ (resp. is commutative).

(2) Every nilpotent group of class $\leq 2^n - 1$ is solvable of class $\leq n$ ; this follows from the relation $D^n(G) \subset C^{2^n}(G)$ proved above.

(3) Let G be a solvable group of class $\leq n$ . Every subgroup (resp. quotient group) of G is solvable of class $\leq n$ (proof analogous to that of no. 3, Example 3).

(4) If G is a solvable group of class $p$ and F is a solvable group of class $q$ , every extension E of G by F is a solvable group of class $\leq p + q$ . For, let $\pi : E \to G$ be the projection; then $\pi(D^p(E)) \subset D^p(G) = \{e\}$ and therefore $D^p(E) \subset F$ ; it follows that $D^{p+q}(E) = D^q(D^p(E)) \subset D^q(F) = \{e\}$ .

(5) The symmetric group $S_n$ is solvable if and only if $n < 5$ (cf. § 5, Exercises 10 and 16).

(6) *If A is a commutative ring, the upper triangular group T(n, A) is solvable but not in general nilpotent.

Proposition 10

Let G be a group and n an integer. The following conditions are equivalent:

(i) G is solvable of class $\leq n$ . (ii) There exists a series of normal subgroups of G

G = G^0 \supset G^1 \supset \ldots \supset G^n = \{e\}

such that the groups $G^k/G^{k+1}$ are commutative. (iii) There exists a series of subgroups of G

G = G^0 \supset G^1 \supset \ldots \supset G^n = e

such that, for all k, $G^{k+1}$ is a normal subgroup of $G^k$ and $G^k/G^{k+1}$ is commutative.

(iv) There exists a normal commutative subgroup $A$ of $G$ such that $G/A$ is solvable of class $\leq n - 1$ .

For (i) $\Rightarrow$ (ii) it suffices to take $G^k$ equal to $D^k(G)$ . (ii) $\Rightarrow$ (iii) trivially. (iii) $\Rightarrow$ (i) for $D^k(G)$ is necessarily contained in $G^k$ . The equivalence of (ii) and (iv) is immediate by induction on $n$ .

More briefly: a group is solvable of class $\leq n$ if it can be obtained by successive extensions of $n$ commutative groups.

Corollary

Let $G$ be a finite group and

G = G^0 \supset G^1 \supset \ldots \supset G^n = \{e\}

a Jordan-Hölder series of

G

For

G

to be solvable, it is necessary and sufficient that the quotients

G^k/G^{k+1}

be cyclic of prime order.

If the quotients of a composition series of $G$ are cyclic and hence commutative, $G$ is solvable by Proposition 10. Conversely, if $G$ is solvable, the group $G^k/G^{k+1}$ is, for all $k$ , solvable and simple ( $§ 4$ , no. 7, Proposition 9). Now, every solvable simple group $H$ is cyclic of prime order. For $D(H)$ is a normal subgroup of $H$ ; $D(H) = H$ is impossible for in that case $D^k(H) = H$ for all $k$ ; then $D(H) = \{e\}$ and $H$ is commutative. The corollary then follows from $§ 4$ , no. 10, Corollary to Proposition 20.

5. M646-groups

In this number and the following, the letter $p$ denotes a prime number ( $§ 4$ , no 10, Proposition 16).

Definition 9

A finite group whose order is a power of $p$ is called a $p$ -group.

Let $G$ be a $p$ -group of order $p^r$ . Every divisor of $p^r$ is a power of $p$ ( $§ 4$ , no. 10, Corollary to Theorem 7). Therefore every subgroup and every quotient group of $G$ is a $p$ -group ( $§ 4$ , no. 4, Corollary to Proposition 4); the cardinal of every homogeneous space of $G$ is a power of $p$ ( $§ 5$ , no. 5, Theorem 1).

An extension of a $p$ -group by a $p$ -group is a $p$ -group.

Examples. (1) A commutative $p$ -group is isomorphic to a product of cyclic groups $\mathbf{Z}/p^n\mathbf{Z}$ (cf. Exercise 19 and also VII, $§ 4$ , no. 7, Proposition 7). (2) Let $k$ be a finite field of characteristic $p$ . The strict triangular group $T_1(n, k)$ is a $p$ -group. (3) The quaternionic group $\{ \pm 1, \pm i, \pm j, \pm k \}$ is a 2-group (cf. Exercise 4).

Proposition 11

Let $E$ be a finite set and $G$ a $p$ -group operating on $E$ . Let $E^G$ denote the set of $x \in E$ such that $gx = x$ for all $g \in G$ (the fixed points). Then

\operatorname{Card}(E^G) \equiv \operatorname{Card}(E) \pmod{p}.

E — E^G is a disjoint union of orbits not reduced to a point. The cardinal of such an orbit is a power of $p$ distinct from $p^0 = 1$ and hence a multiple of $p$ .

Corollary

Let G be a p-group. If G is not reduced to e, its centre is not reduced to e.

Let G operate on itself by inner automorphisms. The set of fixed points is the centre Z of G. By Proposition 11,

\operatorname{Card}(Z) \equiv \operatorname{Card}(G) \equiv 0 \pmod{p},

whence $\operatorname{Card}(Z) \neq 1$ and $Z \neq \{e\}$ .

Theorem 1

Let G be a p-group and $p^r$ its order. There exists a sequence of subgroups of G

G = G^1 \supset G^2 \supset \ldots \supset G^{r+1} = \{e\}

such that $(G, G^k) \subset G^{k+1}, 1 \leq k \leq r,$ and $G^k/G^{k+1}, 1 \leq k \leq r,$ is cyclic of order $p$ .

The theorem is true for $G = \{e\}$ . We prove it by induction on $\operatorname{Card}(G)$ . Let Z be the centre of G, $x \neq e$ an element of Z (Corollary to Proposition 11) and $p^s, s \neq 0,$ the order of x. Then $x^{p^s - 1}$ is an element of order $p$ and therefore Z contains a subgroup $G'$ which is cyclic of order $p$ . By the induction hypothesis, the group $G' = G/G'$ has a series of subgroups $({G'}^k)_{1 \leq k \leq r}$ with the required properties. Let $\pi : G \to G'$ be the canonical homomorphism. The sequence of subgroups of G defined by $G^k = \pi^{-1}({G'}^k), 1 \leq k \leq r, G^{r+1} = \{e\}$ is a solution for $G^k/G^{k+1}$ is isomorphic to ${G'}^k/{G'}^{k+1}$ for $1 \leq k \leq r$ (§ 4, no. 7, Theorem 4).

Corollary

Every p-group is nilpotent.

This follows from no. 3, Proposition 7.

Proposition 12

Let G be a p-group and H a subgroup of G distinct from G. Then:

(a) The normalizer $N_G(H)$ of H in G is distinct from G. (b) There exists a normal subgroup N of G of index p in G, which contains H.

Assertion (a) follows from no. 3, Corollary 1 to Proposition 8. We prove (b). By no. 3, Corollary 2 to Proposition 8, there exists a normal subgroup $N'$ of G containing H, distinct from G and such that $G/N'$ is commutative. Let N be a maximal subgroup distinct from G containing $N'$ . Then N is normal (no. 2, Corollary 3 to Proposition 6) and $G/N$ is a simple commutative $p$ -group and hence cyclic of order $p$ (§ 4, no. 10, Corollary to Proposition 20).

Corollary

Let G be a p-group. Every subgroup of G of index p is normal.

6. Sylow subgroups

Definition 10

Let G be a finite group. A Sylow p-subgroup of G is any subgroup P of G satisfying the two following conditions:

(a) P is a p-group. (b) (G : P) is not a multiple of p.

If the order of G is written in the form $p^r m$ , where m is not a multiple of $p$ , conditions (a) and (b) are equivalent to $\mathrm{Card}(P) = p^r$ .

Example

(1) In the group $\mathfrak{S}_p$ let $\zeta$ be a cycle of order $p$ . The subgroup generated by $\zeta$ is a Sylow $p$ -subgroup of $\mathfrak{S}_p$ for $p$ does not divide $(p - 1)!$ . (2) *Let k be a finite field of characteristic $p$ and let n be a positive integer. The strict triangular group $T_1(n, k)$ is a Sylow $p$ -subgroup of the group $\mathbf{GL}(n, k)$ .

Theorem 2

Every finite group contains a Sylow $p$ -subgroup.

The proof depends on the following lemma.

Lemma

Let $n = p^r m$ , where m is an integer which is not a multiple of $p$ . Then

\binom{n}{p^r} \not\equiv 0 \pmod{p}.

Let S be a group of order $p^r$ (for example $\mathbf{Z}/p^r \mathbf{Z}$ ) and T a set with m elements. Write $X = S \times T$ and let E be the set of subsets of X with $p^r$ elements. Then $\mathrm{Card}(X) = n$ , whence $\mathrm{Card}(E) = \binom{n}{p^r}$ (Set Theory, III, § 5, no. 8, Corollary 1 to Proposition 11). Let S operate on X by $s.(x, y) = (sx, y) (s, x \in S, y \in T)$ and consider the canonical extension of this operation to E. In the notation of no. 5, Proposition 11, the set $E^S$ is the set of orbits of X, that is the set of subsets $Y \subset X$ of the form $S \times \{ t \}, t \in T$ , whence $\mathrm{Card}(E^S) = m$ . By no. 5, Proposition 11,

\binom{n}{p^r} = \mathrm{Card}(E) \equiv \mathrm{Card}(E^S) = m \not\equiv 0 \pmod{p},

which proves the lemma.

We now prove the theorem. Let G be a finite group and n its order; we write $n = p^r m$ , where m is not a multiple of $p$ . Let E be the set of subsets of G with $p^r$ elements. Then

\mathrm{Card}(E) = \binom{n}{p^r};

whence, by virtue of the lemma,

\mathrm{Card}(E) \not\equiv 0 \pmod{p}

. Consider the extension to E of the operation of G on itself by left translation. There exists

X \in E

whose orbit has non-zero cardinal mod.

p

. If

H_X

denotes the stabilizer of X, then

$(G:H_x) \not\equiv 0 \pmod{p}$ , which means that $p^r$ divides $\mathrm{Card}(H_x)$ . But $H_x$ consists of the $s \in G$ such that $sX = X$ ; if $x \in X$ , then $H_x \subset X.x^{-1}$ , whence $\mathrm{Card}(H_x) \leqslant \mathrm{Card}(X) = p^r$ . Hence $\mathrm{Card}(H_x) = p^r$ .

Corollary

If the order of G is divisible by p, the group G contains an element of order p.

By virtue of Theorem 2, this is reduced to the case where G is a $p$ -group $\neq \{e\}$ ; if $x \in G$ is different from e, the cyclic group generated by x is then of order $p^n$ with $n \geqslant 1$ and it therefore contains a subgroup of order $p$ .

Remark

For every prime number q dividing Card(G), let P_q be a Sylow q-subgroup of G. Then the subgroup H of G generated by the P_q is of order a multiple of Card(P_q) for each q and of order a divisor of Card(G), hence it is equal to G.

Theorem 3

Let G be a finite group.* (a) The Sylow p-subgroups of G are conjugate to one another. Their number is congruent to 1 mod. p. (b) *Every subgroup of G which is a p-group is contained in a Sylow p-subgroup.

Let P be a Sylow $p$ -subgroup of G (Theorem 2) and let H be a $p$ -subgroup of G. Let E = G/P and consider the operation of H on G/P. As Card(E) $\not\equiv 0$ mod. $p$ , Proposition 11 of no. 5, shows that there exists $x \in G/P$ such that $hx = x$ for all $h \in H$ . If g is a representative of x in G, this means that $H \subset gPg^{-1}$ , whence assertion (b).

If H is a Sylow $p$ -subgroup, then Card(H) = Card(P) = Card( $gPg^{-1}$ ), whence $H = gPg^{-1}$ , which proves the first assertion of (a).

We now prove the second assertion of (a). Let $\mathscr{S}$ be the set of Sylow $p$ -subgroups of G and let P operate on $\mathscr{S}$ by inner automorphisms. The element $P \in \mathscr{S}$ is a fixed point under this operation, we show that it is the only one. Let $Q \in \mathscr{S}$ be a fixed point; Q is a Sylow subgroup of G normalized by P and hence P is contained in the normalizer N of Q. The groups P and Q are Sylow $p$ -subgroups of N; hence there exists $n \in N$ such that $P = nQn^{-1} = Q$ . By no. 5, Proposition 11, Card( $\mathscr{S}^P$ ) $\equiv$ Card( $\mathscr{S}^{P^P}$ ) = 1 (mod. $p$ ).

Corollary 1

Let P be a Sylow p-subgroup of G, let N be its normalizer in G and let M be a subgroup of G containing N. The normalizer of M in G is equal to M.

Let $s \in G$ be such that $sMs^{-1} = M$ . The subgroup $sPs^{-1}$ of M is a Sylow $p$ -subgroup of M. There thus exists $t \in M$ such that $sPs^{-1} = tPt^{-1}$ ; then $t^{-1}s \in N$ , whence $s \in tN \subset M$ .

Corollary 2

Let $f : G_1 \to G_2$ be a homomorphism of finite groups. For every Sylow p-subgroup $P_1$ of $G_1$ there exists a Sylow p-subgroup $P_2$ of $G_2$ such that $f(P_1) \subset P_2$ .

This follows from Theorem 3 (b) applied to the subgroup $f(P_1)$ of $G_2$ .

Corollary 3

(a) Let $H$ be a subgroup of $G$ . For every Sylow $p$ -subgroup $P$ of $H$ there exists a Sylow $p$ -subgroup $Q$ of $G$ such that $P = Q \cap H$ .

(b) Conversely, if $Q$ is a Sylow $p$ -subgroup of $G$ and $H$ is normal in $G$ , the group $Q \cap H$ is a Sylow $p$ -subgroup of $H$ .

(a) The $p$ -group $P$ is contained in a Sylow $p$ -subgroup $Q$ of $G$ and $Q \cap H$ is a $p$ -subgroup of $H$ containing $P$ and is hence equal to $P$ .

(b) Let $P'$ be a Sylow $p$ -subgroup of $H$ . There exists an element $g \in G$ such that $gP'g^{-1} \subset Q$ . As $H$ is normal, $P = gP'g^{-1}$ is contained in $H$ and hence in $Q \cap H$ . As $Q \cap H$ is a $p$ -subgroup of $H$ and $P$ is a Sylow $p$ -subgroup of $H$ , $P = Q \cap H$ .

Corollary 4

Let $N$ be a normal subgroup of $G$ . The image in $G/N$ of a Sylow $p$ -subgroup of $G$ is a Sylow $p$ -subgroup of $G/N$ and every Sylow $p$ -subgroup of $G/N$ is obtained in this way.

Let $G' = G/N$ and $P'$ be the image in $G'$ of a Sylow $p$ -subgroup $P$ of $G$ . The group $G$ operates transitively on $G'/P'$ and hence $G'/P'$ is equipotent to $G/S$ , where $S$ is a subgroup of $G$ containing $P$ . Therefore $(G':P')$ divides $(G:P)$ , is thus not a multiple of $p$ and the $p$ -group $P'$ is a Sylow $p$ -subgroup of $G'$ . Let $Q'$ be another Sylow $p$ -subgroup of $G'$ ; then $Q' = g'P'{g'}^{-1}$ for some $g' \in G'$ ; if $g \in G$ is a representative of $g'$ , the group $Q'$ is the image of $Q = gPg^{-1}$ .

7. Finite nilpotent groups

Theorem 4

Let $G$ be a finite group. The following conditions are equivalent:

(a) $G$ is nilpotent. (b) $G$ is a product of $p$ -groups. (c) For every prime number $p$ there exists a normal Sylow $p$ -subgroup of $G$ . (b) $\Rightarrow$ (a) (no. 5, Corollary to Theorem 1).

Suppose (a) holds and let $P$ be a Sylow $p$ -subgroup of $G$ . If $N$ is the normalizer of $P$ in $G$ , Corollary 1 to Theorem 3 shows that $N$ is its own normalizer. By § 6, no. 3, Corollary to Proposition 8, this shows that $N = G$ . Hence (a) $\Rightarrow$ (c).

Suppose (c) holds and let $I$ be the set of prime numbers dividing $\mathrm{Card}(G)$ . For all $p \in I$ , let $P_p$ be a normal Sylow $p$ -subgroup of $G$ . For all $p \neq q$ , $P_p \cap P_q$ is reduced to $e$ for it is both a $p$ -group and a $q$ -group, hence $P_p$ and $P_q$ centralize one another ( $§ 4$ , no. 9, Proposition 15). Let $\phi$ be the canonical homomorphism ( $§ 4$ , no. 9, Proposition 12) of $\prod_{p \in I} P_p$ into $G$ . The homomorphism $\phi$ is surjective by the Remark of no. 6. As $\mathrm{Card}\left(\prod_{p \in I} P_p\right) = \mathrm{Card}(G)$ , it follows that $\phi$ is bijective.

Remark

(1) Let G be a finite group and $p$ a prime number. By no. 6, Theorem 3 (a) and no. 6, Theorem 2, the following conditions are equivalent: (i) there exists a normal Sylow $p$ -subgroup of G; (ii) every Sylow $p$ -subgroup of G is normal; (iii) there exists only one Sylow $p$ -subgroup of G.

(2) Let G be a nilpotent finite group. Let I be the set of prime divisors of Card(G). By Theorem 4 and Remark 1, $G = \prod_{p \in I} G_p$ , where $G_p$ is the unique Sylow $p$ -group of G.

(3) Applied to commutative groups, Theorem 4 gives the decomposition, of commutative finite groups as a product of primary components, which will be studied from another point of view in Chapter VII.

Example

The group $\mathfrak{S}_3$ is of order 6. It contains a normal Sylow 3-subgroup of order 3: the group $A_3$ . It contains three Sylow 2-subgroups of order 2: the groups $\{e, \tau\}$ , where $\tau$ is a transposition. The group $\mathfrak{S}_3$ is thus not nilpotent.

Exercises

See the exercises for § 6.