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. 38-39   | 1 | 2 |

Section 1.5.6. Quantitative results

Gabriele Nebea*

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

1.5.6. Quantitative results

| top | pdf |

This section gives estimates for the number of maximal subgroups of a given index in space groups. General results

| top | pdf |

The first very easy but useful remark applies to general groups [{\cal G}]:

Remark. Let [{\cal M} \leq {\cal G}] be a maximal subgroup of [{\cal G}] of finite index [i: = [{\cal G}: {\cal M}] \,\lt\, \infty ]. Then [{\cal M} \leq {\cal N}_{{\cal G}}({\cal M}) \leq {\cal G} ]. Hence the maximality of [{\cal M}] implies that either [ {\cal N}_{{\cal G}}({\cal M}) ={\cal G} ] and [{\cal M}] is a normal subgroup of [{\cal G}] or [ {\cal N}_{{\cal G}}({\cal M}) ={\cal M} ] and [{\cal G}] has i maximal subgroups that are conjugate to [{\cal M}].

The smallest possible index of a proper subgroup is 2. It is well known and easy to see that subgroups of index 2 are normal subgroups:

Proposition Let [{\cal G}] be a group and [{\cal M}\leq {\cal G}] a subgroup of index [2=|{\cal G}/{\cal M}|]. Then [{\cal M}] is a normal subgroup of [{\cal G}].

Proof. Choose an element [{\sf g}\in {\cal G}], [{\sf g}\not \in {\cal M}]. Then [{\cal G} = {\cal M} \cup {\sf g}{\cal M} = {\cal M}\cup {\cal M}{\sf g}]. Hence [{\sf g}{\cal M} = {\cal M}{\sf g}] and therefore [{\sf g}{\cal M}{\sf g}^{-1} = {\cal M}]. Since this is also true if [{\sf g} \in {\cal M}], the proposition follows. QED

Let [{\cal M}] be a subgroup of a group [{\cal G}] of index 2. Then [{\cal M} ] [{\underline{\triangleleft}}] [{\cal G}] is a normal subgroup and the factor group [{\cal G}/{\cal M}] is a group of order 2. Since groups of order 2 are Abelian, it follows that the derived subgroup [{\cal G}_{1}] of [{\cal G}] (cf. Definition[link]) (which is the smallest normal subgroup of [{\cal G}] such that the factor group is Abelian) is contained in [{\cal M}]. Hence all maximal subgroups of index 2 in [{\cal G}] contain [{\cal G}_{1}]. If one defines [{\cal N}:= \cap \{ {\cal M} \leq {\cal G} \mid [{\cal G}:{\cal M}] = 2 \}], then [{\cal G}/{\cal N}] is an elementary Abelian 2-group and hence a vector space over the field with two elements. The maximal subgroups of [{\cal G}/{\cal N}] are the maximal subspaces of this vector space, hence their number is [2^a-1], where [a: = \dim _{{\bb Z}/2{\bb Z}}({\cal G}/{\cal N})].

This shows the following:

Corollary The number of subgroups of [{\cal G}] of index 2 is of the form [2^{a}-1] for some [a \geq 0].

Dealing with subgroups of index 3, one has the following:

Proposition Let [{\cal U}] be a subgroup of the group [{\cal G}] with [[{\cal G}:{\cal U}] = 3]. Then [{\cal U}] is either a normal subgroup of [{\cal G}] or [{\cal G}/{\rm core}({\cal U}) \cong {\cal S}_{3}] and there are three subgroups of [{\cal G}] conjugate to [{\cal U}].

Proof. [{\cal G}/\,{\rm core}\,({\cal U})] is isomorphic to a subgroup of [{\cal S}ym_{3}] that acts primitively on [\{1,2,3\}]. Hence either [{\cal G}/\,{\rm core}\,({\cal U}) \cong {\cal C}yc_{3}] and [{\cal U}= \,{\rm core}\,({\cal U})] is a normal subgroup of [{\cal G}] or [{\cal G}/\,{\rm core}\,({\cal U}) \cong {\cal S}ym_{3}], [{\cal U}/\,{\rm core}\,({\cal U}) \cong {\cal C}yc_{2}] and there are three subgroups of [{\cal G}] conjugate to [{\cal U}]. QED Three-dimensional space groups

| top | pdf |

We now come to space groups. By Lemma l., all three-dimensional space groups are soluble. Theorem[link] says that the index of a maximal subgroup of a soluble group is a prime power (or infinite). Since the index of a maximal subgroup of a space group is always finite (see Corollary[link]), we get:

Corollary Let [{\cal G}] be a three-dimensional space group and [{\cal M}\leq {\cal G}] a maximal subgroup. Then [[{\cal G}:{\cal M}]] is a prime power.

Let [{\cal R}] be a three-dimensional space group and [{\cal P} = {\cal R} / {\cal T}({\cal R})] its point group. It is well known that the order of [{\cal P}] is of the form [ 2^{a} 3^{b}] with [a=0,1,2,3] or [4] and [b=0,1]. By Corollary[link], the t-subgroups of [{\cal R}] are in one-to-one correspondence with the subgroups of [{\cal P}]. Let us look at the t-subgroups of [{\cal R}] of index 3. It is clear that [{\cal P}] has no subgroup of index 3 if [b=0], since the index of a subgroup divides the order of the finite group [{\cal P}] by the theorem of Lagrange. If [b=1], then any subgroup [{\cal S}] of [{\cal P}] of index 3 has order [|{\cal P}|/3 = 2^{a}] and hence is a Sylow 2-subgroup of [{\cal P}]. Therefore there is such a subgroup [\cal{S }] of index 3 in [{\cal P}] by the first theorem of Sylow, Theorem[link]. By the second theorem of Sylow, Theorem[link], all these Sylow 2-subgroups of [{\cal P}] are conjugate in [{\cal P}]. Therefore, by Proposition[link], the number of these groups is either 1 or 3:

Corollary Let [{\cal R}] be a three-dimensional space group.

If the order of the point group of [{\cal R}] is not divisible by 3 then [{\cal R}] has no t-subgroups of index 3.

If 3 is a factor of the order of the point group of [{\cal R}], then [{\cal R}] has either one t-subgroup of index 3 (which is then normal in [{\cal R}]) or three conjugate t-subgroups of index 3.

to end of page
to top of page