Tables for
Volume A1
Symmetry relations between space groups
Edited by Hans Wondratschek and Ulrich Müller

International Tables for Crystallography (2006). Vol. A1, ch. 1.5, pp. 36-38   | 1 | 2 |

Section 1.5.5. Maximal subgroups

Gabriele Nebea*

aAbteilung Reine Mathematik, Universität Ulm, D-89069 Ulm, Germany
Correspondence e-mail:

1.5.5. Maximal subgroups

| top | pdf | Maximal subgroups and primitive [{\cal G}]-sets

| top | pdf |

To determine the maximal t-subgroups of a space group [{\cal R}], essentially one has to calculate the maximal subgroups of the finite group [{\cal R}/{\cal T}({\cal R})]. There are fast algorithms to calculate these maximal subgroups if this finite group is soluble (see Definition[link]), which is the case for three-dimensional space groups. To explain this method and obtain theoretical consequences for the index of maximal subgroups in soluble space groups, we consider abstract groups again in this section.

For an arbitrary group [{\cal G}], one has a fast method of checking whether a given subgroup [{\cal U}\leq {\cal G}] of finite index [[{\cal G}: {\cal U}] ] is maximal by inspection of the [{\cal G}]-set [{\cal G}/{\cal U}] of left cosets of [{\cal U}] in [{\cal G}]. Assume that [{\cal U}\leq {\cal M} \leq {\cal G}] and let [{\cal M}/{\cal U}: = \{ {\sf m}_{1}{\cal U},\ldots, {\sf m}_{k}{\cal U}\}] with [{\sf m}_{i} \in {\cal M}], [{\sf m}_{1} ={\sf e}] and [{\cal G}/{\cal M}: = \{ {\sf g}_{1}{\cal M}, \ldots, {\sf g}_{l} {\cal M} \}] with [{\sf g}_{i} \in {\cal G}], [{\sf g}_{1} ={\sf e}]. Then the set [{\cal G}/{\cal U}] may be written as [ \matrix {{\cal G}/{\cal U} = &\{{\sf g}_{1} {\sf m}_{1} {\cal U}, & \ldots, & {\sf g}_{1} {\sf m}_{k} {\cal U}, \cr & \phantom{\{}{\sf g}_{2} {\sf m}_1 {\cal U},& \ldots, & {\sf g}_2 {\sf m}_k {\cal U}, \cr &\vdots, & \ldots, & \vdots, \cr & \phantom{\{}{\sf g}_{l} {\sf m}_{1} {\cal U},& \ldots, & {\sf g}_{l} {\sf m}_{k} {\cal U} \}}.]Then [{\cal G}] permutes the lines of the rectangle above: For all [{\sf g}\in {\cal G}] and all [j \in \{ 1,\ldots, l \}], the left coset [{\sf g}{\sf g}_{j} {\cal M} ] is equal to some [{\sf g}_{a} {\cal M}] for an [a\in \{ 1,\ldots, l \}]. Hence the jth line is mapped onto the set [\{ {\sf g} {\sf g}_{j}{\sf m}_{1}{\cal U}, \ldots, {\sf g} {\sf g}_{j} {\sf m}_{k}{\cal U} \} = \{ {\sf g}_{a}{\sf m}_{1}{\cal U}, \ldots, {\sf g}_{a} {\sf m}_{k}{\cal U} \}.]

Definition Let [{\cal G}] be a group and X a [{\cal G}]-set.

  • (i) A congruence [\{ S_{1}, \ldots, S_{l} \}] on X is a partition of X into non-empty subsets [X= {\dot{\cup} }_{i=1}^{\,l} S_{i}] such that for all [x_{1},x_{2} \in S_{i}], [g\in {\cal G}], [gx_{1} \in S_{j}] implies [g x_{2} \in S_{j}].

  • (ii) The congruences [\{ X \}] and [ \{ \{x \} \mid x\in X \}] are called the trivial congruences.

  • (iii) X is called a primitive [{\cal G}]-set if [{\cal G}] is transitive on X, [|X|> 1] and X has only the trivial congruences.

Hence the considerations above have proven the following lemma.

Lemma Let [{\cal M}\leq {\cal G}] be a subgroup of the group [{\cal G}]. Then [{\cal M}] is a maximal subgroup if and only if the [{\cal G}]-set [{\cal G}/{\cal M}] is primitive.

The advantage of this point of view is that the groups [{\cal G}] having a faithful, primitive, finite [{\cal G}]-set have a special structure. It will turn out that this structure is very similar to the structure of space groups.

If X is a [{\cal G}]-set and [{\cal N}\,{\underline{\triangleleft}}\,\,{\cal G}] is a normal subgroup of [{\cal G}], then [{\cal G}] acts on the set of [{\cal N}]-orbits on X, hence [\{ {\cal N}x \mid x\in X \}] is a congruence on X. If X is a primitive [{\cal G}]-set, then this congruence is trivial, hence [{\cal N}x=\{x\} ] or [{\cal N}x = X] for all [x\in X]. This means that [{\cal N}] either acts trivially or transitively on X.

One obtains the following:

Theorem [Theorem of Galois (ca 1830).] Let [{\cal H}] be a finite group and let X be a faithful, primitive [{\cal H}]-set. Assume that [\{ {\sf e} \} \neq {\cal N}\,{\underline{\triangleleft}}\,\,{\cal H}] is an Abelian normal subgroup. Then

  • (a) [{\cal N}] is a minimal normal subgroup of [{\cal H}] (i.e. for all [{\cal N}_1\,{\underline{\triangleleft}}\,\,{\cal H}], [{\cal N}_{1} \subseteq {\cal N} \Leftrightarrow {\cal N}_1 = {\cal N}] or [{\cal N}_{1} = \{ {\sf e} \} ]).

  • (b) [{\cal N}] is an elementary Abelian p-group for some prime p and [|X| = |{\cal N}|] is a prime power.

  • (c) [{\cal C}_{{\cal H}}({\cal N}) = {\cal N}] and [{\cal N}] is the unique minimal normal subgroup of [{\cal H}].

Proof. Let [\{ {\sf e} \} \neq {\cal N}\,{\underline{\triangleleft}}\,\,{\cal H}] be an Abelian normal subgroup. Then [{\cal N}] acts faithfully and transitively on X. To establish a bijection between the sets [{\cal N}] and X, choose [x\in X] and define [\varphi: {\cal N} \rightarrow X; {\sf n}\mapsto {\sf n}\cdot x]. Since [{\cal N}] is transitive, [\varphi ] is surjective. To show the injectivity of [\varphi], let [{\sf n}_{1}, {\sf n}_{2} \in {\cal N}] with [\varphi({\sf n}_{1}) =\varphi ({\sf n}_{2})]. Then [{\sf n}_{1} \cdot x ={\sf n}_{2} \cdot x], hence [{\sf n}_{1}^{-1} {\sf n}_{2} x = x]. But then [{\sf n}_{1}^{-1} {\sf n}_{2}] acts trivially on X, because if [y\in X] then the transitivity of [{\cal N}] implies that there is an [{\sf n}\in {\cal N}] with [{\sf n}\cdot x = y]. Then [{\sf n}_{1}^{-1} {\sf n}_{2} \cdot y = {\sf n}_{1}^{-1} {\sf n}_{2} {\sf n} \cdot x =] [{\sf n} {\sf n}_{1}^{-1} {\sf n}_{2} \cdot x ={\sf n} \cdot x = y], since [{\cal N}] is Abelian. Since X is a faithful [{\cal H}]-set, this implies [{\sf n}_{1}^{-1} {\sf n}_{2} ={\sf e}] and therefore [{\sf n}_{1} ={\sf n}_{2}]. This proves [|{\cal N}|=|X|]. Since this equality holds for all nontrivial Abelian normal subgroups of [{\cal H}], statement (a) follows. If p is some prime dividing [|{\cal N}|], then the Sylow p-subgroup of [{\cal N}] is normal in [{\cal N}], since [{\cal N}] is Abelian. Therefore it is also a characteristic subgroup of [{\cal N}] and hence a normal subgroup in [{\cal H}] (see the remarks below Definition[link]). Since [{\cal N}] is a minimal normal subgroup of [{\cal H}], this implies that [{\cal N}] is equal to its Sylow p-subgroup. Therefore, the order of [{\cal N}] is a prime power [|{\cal N}| =p^{r}] for some prime p and [r\in {\bb N}]. Similarly, the set [{\cal N}^{p}: = \{ n^{p} \mid n\in {\cal N} \}] is a normal subgroup of [{\cal H}] properly contained in [{\cal N}]. Therefore [{\cal N}^{p} =\{ {\sf e} \}] and [{\cal N}] is elementary Abelian. This establishes (b).

To see that (c) holds, let [{\sf g}\in {\cal C}_{{\cal H}}({\cal N})]. Choose [x\in X]. Then [{\sf g}\cdot x = y \in X]. Since [{\cal N}] acts transitively, there is an [{\sf n}\in {\cal N}] such that [{\sf n}\cdot x = y]. Hence [{\sf n}^{-1} {\sf g} \cdot x = x]. As above, let [z\in X] be any element of X. Then there is an element [{\sf n}_{1} \in {\cal N}] with [z = {\sf n}_{1} \cdot x]. Hence [{\sf n}^{-1} {\sf g} \cdot z = {\sf n}^{-1} {\sf g} {\sf n}_{1} \cdot x ={\sf n}_{1} {\sf n}^{-1} {\sf g} \cdot x ={\sf n}_{1} \cdot x =z]. Since z was arbitrary and X is faithful, this implies that [{\sf g}={\sf n} \in {\cal N}]. Therefore [{\cal C}_{{\cal H}}({\cal N}) \subseteq {\cal N}]. Since [{\cal N}] is Abelian, one has [{\cal N}\subseteq {\cal C}_{{\cal H}}({\cal N})], hence [{\cal N}={\cal C}_{{\cal H}}({\cal N})]. To see that [{\cal N}] is unique, let [{\cal P}\neq {\cal N}] be another normal subgroup of [{\cal H}]. Since [{\cal N}] is a minimal normal subgroup, one has [{\cal N}\cap {\cal P} =\{ {\sf e} \}], and therefore for [{\sf p} \in {\cal P}], [{\sf n} \in {\cal N}]: [{\sf n} ^{-1} {\sf p} ^{-1} {\sf n} {\sf p} \in {\cal N}\cap {\cal P} =\{ {\sf e} \}]. Hence [{\cal P}] centralizes [{\cal N}], [{\cal P}\subseteq {\cal C}_{{\cal H}}({\cal N}) ={\cal N}], which is a contradiction. QED

Hence the groups [{\cal H}] that satisfy the hypotheses of the theorem of Galois are certain subgroups of an affine group [{\cal A}_{n}({\bb Z}/p{\bb Z})] over a finite field [{\bb Z}/p{\bb Z}]. This affine group is defined in a way similar to the affine group [{\cal A}_{n}] over the real numbers where one has to replace the real numbers by this finite field. Then [{\cal N}] is the translation subgroup of [{\cal A}_{n}({\bb Z}/p{\bb Z})] isomorphic to the n-dimensional vector space [({\bb Z}/p{\bb Z})^{\,n} = \{ {\bi x} = \left (\matrix { x_{1} \cr \vdots \cr x_{n} } \right) \mid x_{1}, \ldots, x_{n} \in {\bb Z}/p{\bb Z} \}]over [{\bb Z}/p{\bb Z}]. The set X is the corresponding affine space [{\bb A}_n({\bb Z} /p{\bb Z})]. The factor group [\overline{{\cal H}} = {\cal H}/{\cal N}] is isomorphic to a subgroup of the linear group of [({\bb Z}/p{\bb Z})^{n} ] that does not leave invariant any non-trivial subspace of [({\bb Z}/p{\bb Z})^{n}]. Soluble groups

| top | pdf |

Definition Let [{\cal G}] be a group. The derived series of [{\cal G}] is the series [({\cal G}_{0},{\cal G}_{1}, \ldots)] defined via [{\cal G}_{0} := {\cal G}], [{\cal G}_{i} := \langle {\sf g}^{-1}{\sf h}^{-1}{\sf g} {\sf h} \mid ] [ {\sf g}, {\sf h} \in {\cal G}_{i-1} \rangle]. The group [{\cal G}_{1}] is called the derived subgroup of [{\cal G}]. The group [{\cal G}] is called soluble if [{\cal G}_{n} = \{ {\sf e} \}] for some [n\in {\bb N}].


  • (i) The [{\cal G}_{i}] are characteristic subgroups of [{\cal G}].

  • (ii) [{\cal G}] is Abelian if and only if [{\cal G}_{1} = \{ {\sf e} \}].

  • (iii) [{\cal G}_{1}] is characterized as the smallest normal subgroup of [{\cal G}], such that [{\cal G}/{\cal G}_{1}] is Abelian, in the sense that every normal subgroup of [{\cal G}] with an Abelian factor group contains [{\cal G}_{1}].

  • (iv) Subgroups and factor groups of soluble groups are soluble.

  • (v) If [{\cal N}\,{\underline{\triangleleft}}\,\,{\cal G}] is a normal subgroup, then [{\cal G}] is soluble if and only if [{\cal G}/{\cal N}] and [{\cal N}] are both soluble.


The derived series of [{\cal C}yc_{2} \times {\cal S}ym_{4}] is: [{\cal C}yc_{2}\times {\cal S}ym_{4} \, {\underline{\triangleright}} \, {\cal A}lt_{4} \, {\underline{\triangleright}}\, {\cal C}yc_{2} \times {\cal C}yc_{2} \, {\underline{\triangleright}}\, {\cal I} ](or in Hermann–Mauguin notation [m\overline{3}m\, \,{\underline{\triangleright}}\,\,23 \,\,{\underline{\triangleright}}\,\,222\,\,{\underline{\triangleright}}\,\,1]) and that of [{\cal C}yc_{2}\times {\cal C}yc_{2}\times {\cal S}ym_{3}] is [{\cal C}yc_{2}\times {\cal C}yc_{2} \times {\cal S}ym_{3} \, {\underline{\triangleright}} \, {\cal C}yc_{3} \, {\underline{\triangleright}} \, {\cal I} ](Hermann–Mauguin notation: [6/mmm\,\,{\underline{\triangleright}}\,\,3\,\,{\underline{\triangleright}}\,\,1]).

Hence these two groups are soluble. (For an explanation of the groups that occur here and later, see Section[link].)

Now let [{\cal R} \leq {\cal E}_{3}] be a three-dimensional space group. Then [{\cal T}({\cal R})] is an Abelian normal subgroup, hence [{\cal T}({\cal R}) ] is soluble. The factor group [{\cal R}/{\cal T}({\cal R})] is isomorphic to a subgroup of either [{\cal C}yc_{2}\times {\cal S}ym_{4} ] or [{\cal C}yc_{2}\times {\cal C}yc_{2} \times {\cal S}ym_{3}] and therefore also soluble. Using the remark above, one deduces that all three-dimensional space groups are soluble.

Lemma Let [{\cal R}] be a three-dimensional space group. Then [{\cal R}] is soluble. Maximal subgroups of soluble groups

| top | pdf |

Now let [{\cal G}] be a soluble group and [{\cal M} \leq {\cal G}] a maximal subgroup of finite index in [{\cal G}]. Then the set of left cosets [X:={\cal G}/ {\cal M}] is a primitive finite [{\cal G}]-set. Let [{\cal K} = \, {\rm core}\, ({\cal M})] be the kernel of the action of [{\cal G}] on X. Then the factor group [{\cal H}:={\cal G}/{\cal K}] acts faithfully on X. In particular, [{\cal H}] is a finite group and X is a primitive, faithful [{\cal H}]-set. Since [{\cal G}] is soluble, the factor group [{\cal H}] is also a soluble group. Let [{\cal H}\,\,{\underline{\triangleright}}\,\,{\cal H} _1\,\,{\underline{\triangleright}}\,\ldots \,{\underline{\triangleright}}\,\,{\cal H}_{n-1}\,\,{\underline{\triangleright}}\,\,\{ {\sf e} \} ] be the derived series of [{\cal H}] with [{\cal N}:= {\cal H}_{n-1} \neq \{ {\sf e} \}]. Then [{\cal N}] is an Abelian normal subgroup of [{\cal H}]. The theorem of Galois (Theorem[link]) states that [{\cal N}] is an elementary Abelian p-group for some prime p and [|X| = | {\cal N} | = p^{r}] for some [r \in {\bb N}]. Since [X = {\cal G} / {\cal M}], the order of X is the index [[{\cal G}: {\cal M}] ] of [{\cal M}] in [{\cal G}]. Therefore one gets the following theorem:

Theorem If [{\cal M}\leq {\cal G}] is a maximal subgroup of finite index in the soluble group [{\cal G}], then its index [|{\cal G}/{\cal M}|] is a prime power.

In the proof of Theorem[link], we have established a bijection between [{\cal N}] and the [{\cal H}]-set X, which is now [X: = {\cal G}/{\cal M}]. Taking the full pre-image [{\cal N}': = {\cal N}\, {\rm core}\, ({\cal M})]of [{\cal N}] in [{\cal G}], then one has [{\cal G} ={\cal N}' {\cal M}] and [{\cal M} \cap {\cal N}' =\,{\rm core}\, ({\cal M})]. Hence we have seen the first part of the following theorem:

Theorem Let [{\cal M}\leq {\cal G}] be a maximal subgroup of the soluble group [{\cal G}]. Then the factor group [{\cal H}:={\cal G}/{\rm core}({\cal M})] acts primitively and faithfully on [X :={\cal G}/{\cal M}], and there is a normal subgroup [{\cal N}'\,\,{\underline{\triangleleft}}\,\,{\cal G}] with [{\cal M}{\cal N}'={\cal G}] and [{\cal M}\cap {\cal N}' =\,{\rm core}({\cal M})]. Moreover, if [{\cal M}'] is another subgroup of [{\cal G}], with [{\cal M}'{\cal N}'={\cal G}] and [{\cal M}'\cap {\cal N}' =\,{\rm core}({\cal M})], then [{\cal M}'] is conjugate to [{\cal M}]. [Scheme scheme4]


[{\cal G}={\cal S}ym_{4} \cong T_{d}] is the tetrahedral group from Section[link] and [{\cal S}ym_{3} \cong {\cal M} = C_{3v} \leq {\cal G}] is the stabilizer of one of the four apices in the tetrahedron. Then [{\rm core}({\cal M}) = \{ {\sf e} \} ] and [{\cal G}/{\cal M}] is a faithful [{\cal G}]-set which can be identified with the set of apices of the tetrahedron. The normal subgroup [{\cal N} = {\cal N}'] is the normal subgroup [{\cal U}] of Section[link]. [Scheme scheme5]

Now let [{\cal G}={\cal S}ym_{4} \cong T_{d}] be as above, and take [D_{2d} \cong {\cal M}\leq {\cal G}] a Sylow 2-subgroup of [{\cal G}]. Then [{\rm core}({\cal M}) = D_{2} \cong {\cal C}yc_{2}\times {\cal C}yc_{2}] is the normal subgroup [{\cal U}] from Section[link] and [{\cal H}={\cal G}/{\rm core}({\cal M}) \cong {\cal S}ym_{3}]. [Scheme scheme6]

These observations result in an algorithm for computing maximal subgroups of soluble groups [{\cal G}]:

  • (1) compute normal subgroups [{\cal C}] [candidates for [{\rm core}({\cal M})]];

  • (2) compute a minimal normal subgroup [{\cal N}/{\cal C}] of [{\cal G}/{\cal C}];

  • (3) find [{\cal M}/{\cal C}] as a complement of [{\cal N}/{\cal C}] in [{\cal G}/{\cal C}].

to end of page
to top of page