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

International Tables for Crystallography (2011). Vol. A1, ch. 1.4, pp. 38-39   | 1 | 2 |

Section 1.4.7. Qualitative results

Gabriele Nebea*

aLehrstuhl D für Mathematik, Rheinisch-Westfälische Technische Hochschule, D-52062 Aachen, Germany
Correspondence e-mail:

1.4.7. Qualitative results

| top | pdf | General theory

| top | pdf |

In this section, we want to comment on the very subtle question of deciding whether two space groups [{\cal R}_{1}] and [{\cal R}_{2}] are isomorphic.

This problem can be treated in several stages:

Let [{\cal R}_{1}] and [{\cal R}_{2}] be space groups. Since the translation subgroups [{\cal T}({\cal R}_{i})] are characteristic subgroups of [{\cal R}_{i}] (the maximal Abelian normal subgroup of finite index), each isomorphism [\varphi: {\cal R}_{1}\rightarrow {\cal R}_{2}] induces isomorphisms of the corresponding translation subgroups [\varphi ': {\cal T}({\cal R}_{1}) \rightarrow {\cal T}({\cal R}_{2})](by restriction) as well as of the point groups [\overline{\varphi }: {\cal P}_{1}:={\cal R}_{1} / {\cal T}({\cal R}_{1}) \rightarrow {\cal R}_{2} / {\cal T}({\cal R}_{2}) =:{\cal P}_{2}. ]It is convenient to view [{\cal T}({\cal R}_{i})] as a lattice on which the point group [{\cal P}_{i}] acts as group of linear mappings (cf. the start of Section 1.4.4[link]). Then the isomorphism [\varphi '] is an isomorphism of [{\cal P}_{1}]-sets, where [{\cal P}_{1}] acts on [{\cal T}({\cal R}_{1})] via conjugation and on [{\cal T}({\cal R}_{2})] via [\ispecialfonts{\sfi g}{\cal T}({\cal R}_{1}) \cdot {\sfi t}: = \varphi ({\sfi g}) {\sfi t} \varphi ({\sfi g})^{-1} \,\, {\rm for \,\,all }\,\, {\sfi g}{\cal T}({\cal R}_{1}) \in {\cal P}_{1}, {\sfi t}\in {\cal T}({\cal R}_{2}).]Since [\varphi ({\cal T}({\cal R}_{1})) = {\cal T}({\cal R}_{2})] and [{\cal T}({\cal R}_{2}) ] centralizes itself, this action is well defined, i.e. independent of the choice of the coset representative [\ispecialfonts{\sfi g}].

The following theorem will show that the isomorphism of sufficiently large factor groups of [{\cal R}_{1}] and [{\cal R}_{2}] implies a `near' isomorphism of the space groups themselves. To give a precise formulation we need one further definition.

Definition For [d\in {\bb N}] define [O_{d}: = \{ {{a}\over{b}} \mid a,b \in {\bb Z}, b\neq 0, \gcd(b,d) = 1 \} \leq {\bb Q},]which is the set of all rational numbers for which the denominator is prime to d. For the space group [{\cal R}\leq {\cal E}_{n}] let [{\cal R}\leq {\cal R}_{(d)} \leq {\cal E}_{n}] be the group [{\cal R}_{(d)}: = \langle {\cal T}({\cal R})_{(d)}, {\cal R} \rangle ], where [{\cal T}({\cal R}) _{(d)} = \{ a t \mid a\in O_ {(d)}, t \in {\cal T}({\cal R}) \} \leq {\cal T}_{n}, ]i.e. one allows denominators that are prime to d in the translation subgroup.

One has the following:

Theorem Let [{\cal R}_1] and [{\cal R}_2] be two space groups with point groups of order [d_{i}: = | {\cal R}_i/{\cal T}({\cal R}_{i}) |]. Let [{\cal {\bf N}}({\cal R}_{i})] denote the set of normal subgroups of [{\cal R}_i] having finite index in [{\cal R}_i]. Then the following three conditions are equivalent:

  • (i) There are normal subgroups [{\cal S}_{i} \,\,{\underline{\triangleleft}}\,\, {\cal R}_i] with [{\cal R}_{1} / {\cal S}_{1} \cong {\cal R}_{2} /{\cal S}_2] and with [{\cal S}_{i} \subseteq d_{i}^{2} {\cal T}({\cal R}_{i}) ] if [d_{i} \neq 2] and [{\cal S}_{i} \subseteq 16 {\cal T} ({\cal R}_{i}) ] if [d_{i} = 2] ([i=1,2]).

  • (ii) [({\cal R}_{1})_{(d_{1})} \cong ({\cal R}_{2})_{(d_{2})}].

  • (iii) There is a bijection [\mu: {\cal {\bf N}} ({\cal R}_{1}) \rightarrow {\cal {\bf N}}({\cal R}_2)] such that [{\cal R}_{1}/{\cal N}] [\cong{\cal R}_{2} /\mu ({\cal N})] for all [{\cal N}] [\in] [{\cal {\bf N}}({\cal R}_1)].

For a proof of this theorem, see Finken et al. (1980[link]).

Remark. If [{\cal R}_i] are three- or four-dimensional space groups, the isomorphism in (ii) already implies the isomorphism of [{\cal R}_1] and [{\cal R}_2], but there are counterexamples for dimension 5. Three-dimensional space groups

| top | pdf |

Corollary Let [{\cal R}] be a three-dimensional space group with translation subgroup [{\cal T}] and p be a prime not dividing the order of the point group [{\cal R}/{\cal T}]. Let [{\cal U}] be a subgroup of [{\cal R}] of index [p^{\alpha }] for some [\alpha \in {\bb Z} _{>0} ]. Then

  • (a) [{\cal U}] is a k-subgroup.

  • (b) [{\cal U}] is isomorphic to [{\cal R}].


  • (a) [{\cal U}\leq {\cal U}{\cal T} \leq {\cal R}] implies that [[{\cal R}:{\cal U}{\cal T}]] divides [[{\cal R}:{\cal U}]=p^{\alpha }]. Since [{\cal T}\leq {\cal U}{\cal T} \leq {\cal R}], one obtains [[{\cal R}:{\cal U}{\cal T}]] as a factor of [[{\cal R}:{\cal T}]]. But p is not a factor of [[{\cal R}:{\cal T}]], hence [[{\cal R}:{\cal U}{\cal T}] = 1] and [{\cal R}={\cal U}{\cal T}]. According to the remark following Definition[link], [{\cal U}] is a k-subgroup.

  • (b) Let [d_{1}: = |{\cal R}/{\cal T}| = |{\cal U}/{\cal T}({\cal U}) | ]. Let [d:=d_{1}^{2}] if [d_{1} \neq 2] and [d:=16] otherwise, and let [{\cal T}': = d {\cal T}]. Since [\gcd([{\cal R}:{\cal U}],d)] [ = 1], one has [{\cal U}{\cal T}' = {\cal R}] and [{\cal T}' \cap {\cal U} = d{\cal T}({\cal U})]. By the third iso­morphism theorem, Theorem[link], it follows that [{\cal R}/{\cal T}' = {\cal U}{\cal T}'/{\cal T}' \cong {\cal U}/ {\cal T}'\cap {\cal U} = {\cal U}/d {\cal T}({\cal U}). ]By Theorem[link] (i) [\Rightarrow ] (ii), one has [{\cal R}_{(d_{1})} \cong {\cal U}_{(d_{1})}]. By the remark above, this already implies that [{\cal R}] and [{\cal U}] are isomorphic. QED

Theorem Let [{\cal R}] be a three-dimensional space group and [{\cal U}] be a maximal subgroup of [{\cal R}] of index [>4]. Then

  • (a) [{\cal U}] is a k-subgroup.

  • (b) [{\cal U}] is isomorphic to [{\cal R}].

Proof. Since [{\cal R}] is soluble, the index [[{\cal R}:{\cal U}]=p^{\alpha }] is a prime power (see Theorem[link]). If p is not a factor of [|{\cal R}/{\cal T}({\cal R})|], the statement follows from Corollary[link]. Hence we only have to consider the cases [p=2], [\alpha> 2] and [p=3], [\alpha>1]. Since 9 is not a factor of the order of any crystallographic point group in dimension 3, assertion (a) follows if the index of [{\cal U}] is divisible by 9. If [{\cal U}] is a maximal t-subgroup, then [{\cal R}/{\cal U}] is a primitive [{\cal P}]-set for the point group [{\cal P}] of [{\cal R}]. Since the point groups [{\cal P}] of dimension 3 have no primitive [{\cal P}]-sets of order divisible by 8, assertion (a) also follows if the index of [{\cal U}] is divisible by 8.

For all three-dimensional space groups [{\cal R}], the module [{\bf L}({\cal R})/2{\bf L}({\cal R})] [where [{\cal T}({\cal R})] is identified with the corresponding lattice [{\bf L}({\cal R})] in [\tau ({\bb E}_3)] as in Section 1.4.4[link]] is not simple as a module for the point group [{\cal P} = {\cal R}/{\cal T}({\cal R})]. [It suffices to check this property for the two maximal point groups [{\cal C}yc_{2}\times {\cal S}ym_4] ([=m\overline{3}m]) and [{\cal C}yc_{2}\times {\cal C}yc_{2} \times {\cal S}ym_3] [(=6/mmm)].] This means that [2{\bf L}({\cal R})] is not a maximal [{\cal R}]-invariant sublattice of [{\bf L}({\cal R})]. Since the translation subgroup [{\cal T}({\cal U})] of a maximal k-subgroup [{\cal U}] of index equal to a power of 2 in [{\cal R}] is a maximal [{\cal R}]-invariant subgroup of [{\cal T}({\cal R})] that contains [2{\cal T}({\cal R})], one now finds that [{\cal R}] has no maximal k-subgroup of index 8.

Now assume that [[{\cal R}: {\cal U}] = 9]. By Corollary[link], one only needs to deal with groups [{\cal R}] such that the order of the point group [{\cal P}:={\cal R}/{\cal T}({\cal R})] is divisible by 3. [{\cal P}] is isomorphic to a subgroup of [{\cal C}yc_{2} \times {\cal S}ym_4] or [{\cal C}yc_{2}\times {\cal C}yc_{2} \times {\cal S}ym_3]. If [{\cal A}lt_{4} \leq {\cal P} ] is a subgroup of [{\cal P}], then [{\bf L}({\cal R})/3{\bf L}({\cal R})] is simple and [{\cal U}] is of index 27 in [{\cal R}] [with [{\bf L}({\cal U}) = 3{\bf L}({\cal R})]]. It turns out that [{\cal U}] is isomorphic to [{\cal R}] in these cases. If [{\cal P}] does not contain a subgroup isomorphic to [{\cal A}lt_4], then the maximality of [{\cal U}] implies that [{\cal T}({\cal U}) \leq {\cal T}({\cal R})] is of index 3 in [{\cal T}({\cal R})]. Hence [[{\cal R}:{\cal U}] =3] in this case. QED

Corollary Let [{\cal M}] be a maximal subgroup of the three-dimensional space group [{\cal R}].

  • (a) If the index of [{\cal M}] is a power of 2, then [[{\cal R}:{\cal M}] =2] or 4.

  • (b) If 3 is a factor of the order of the point group [[{\cal R}: {\cal T}({\cal R})] ] and the index of [{\cal M}] is a power of 3, then [[{\cal R}:{\cal M}] =3] or 27. For [[{\cal R}: {\cal M}] = 27], [{\cal M}] is necessarily isomorphic to [{\cal R}] (by Theorem[link]).

This interesting fact explains why there are no maximal subgroups of index 8 in a three-dimensional space group. If there is a maximal subgroup [{\cal M}] of a three-dimensional space group [{\cal R}] of index 9, then the order of the point group of [{\cal R}] is not divisible by three and the subgroup [{\cal M}] is a k-subgroup and isomorphic to [{\cal R}].

In particular, there are no maximal subgroups of index 9 for trigonal, hexagonal or cubic space groups, whereas there are such subgroups of tetragonal space groups.


Finken, H., Neubüser, J. & Plesken, W. (1980). Space groups and groups of prime power order II. Arch. Math. 35, 203–209.

to end of page
to top of page