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. 34-36   | 1 | 2 |

Section 1.5.4. Space groups

Gabriele Nebea*

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

1.5.4. Space groups

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In IT A (2005[link]), Section 8.1.6[link] , space groups are introduced as symmetry groups of crystal patterns.


  • (a) Let [{\bf V}_n] be the n-dimensional real vector space. A subset [{\bf L} \subseteq {\bf V}_n] is called an (n-dimensional) lattice if there is a basis [({\bf b}_1,\ldots, {\bf b}_n)] of [{\bf V}_n] such that[{\bf L}={\bb Z} {\bf b}_1 + \ldots + {\bb Z} {\bf b}_n = \{ \textstyle\sum\limits _{i=1}^n a_i {\bf b}_i \mid a_i \in {\bb Z} \}. ]

  • (b) A crystal structure is a mapping [f: {\bb E} _{n} \rightarrow {\bb R} ] of the Euclidean affine n-space into the real numbers such that [{\rm Stab}_{\tau ({\bb A} _{n}) } (f): = \{ t \in \tau({\bb A}_{n}) \mid f(P+t) = f(P) \,{\rm for \, all \,} P\in {\bb A}_{n} \}] is an n-dimensional lattice in [\tau ({\bb A}_{n})].

  • (c) The Euclidean group [{\cal E}_{n}] acts on the set of mappings [{\bb E}_{n} \rightarrow {\bb R}] via [({\sf g} \cdot f)(P): = f ({\sf g} ^{-1}P) ] for all [P\in {\bb E}_{n}] and for all [{\sf g} \in {\cal E}_{n}] and [f:{\bb E}_{n} \rightarrow {\bb R}]. A space group [{\cal R}] is the stabilizer of a crystal structure [f:{\bb E}_{n} \rightarrow {\bb R}]; [{\cal R} = {\rm Stab}_{\cal E}_{n}(f)].

  • (d) Let [{\cal R} \leq {\cal E}_{n} ] be a space group. The translation subgroup [{\cal T}({\cal R})] of [{\cal R}] is defined as [{\cal T}({\cal R}): ={\cal R}\cap {\cal T}_{n}.]

The definition introduced space groups in the way they occur in crystallography: The group of symmetries of an ideal crystal stabilizes the crystal structure. This definition is not very helpful in analysing the structure of space groups. If [{\cal R}] is a space group, then the translation subgroup [{\cal T}: = {\cal T}({\cal R})] is a normal subgroup of [{\cal R}]. It is even a characteristic subgroup of [{\cal R}], hence fixed under every automorphism of [{\cal R}]. By Definition[link], its image under the inverse [\mu '] of the mapping [\mu ] in Example[link] defined by [\mu': {{\cal T}} \rightarrow \tau ({\bb{E}}_{n}); \left(\matrix{ {\bi I}\,\vphantom{(^2{\big(_2}}&\vrule\, &{\bi v} \cr \noalign{\vskip-1pt\hrule} \cr{\bi o}^{\rm T}\,\vphantom{{\big(^2}(_2}&\vrule\, &1 } \right) \mapsto \bi{v}]in [\tau ({\bb A}_{n})] is a full lattice [{\bf L}({\cal R})]. Since [\mu '] is an isomorphism from [{\cal T} ] onto [{\bf L}({\cal R})], the translation subgroup of [{\cal R}] is isomorphic to the lattice [{\bf L}({\cal R})]. In particular, one has [\mu ' (t_1 t_2) = \mu ' (t_1) + \mu ' (t_2)] and the subgroup [{\cal T}^{p}], formed by the pth powers of elements in [{\cal T}], is mapped onto [p{\bf L}({\cal R})]. Lattices are well understood. Although they are infinite, they have a simple structure, so they can be examined algorithmically. Since they lie in a vector space, one can apply linear algebra to them.

Now we want to look at how this lattice [{\cal T}({\cal R})] fits into the space group [{\cal R}]. The affine group [{\cal A}_{n}] acts on [{\cal T}_{n}] by conjugation as well as on [\tau({\bb A}_{n})] via its linear part. Similarly the space group [{\cal R}] acts on [{\cal T}({\cal R})] by conjugation: For [{\sf g} \!\in \!{\cal R}] and [{\sf t} \!\in \!{\cal T}], one gets [\mu' ({\sf g} {\sf t} {\sf g} ^{-1}) = \overline{{\sf g} } \mu ' ({\sf t}) ], where [\overline{{\sf g} }] is the linear part of [{\sf g}]. Therefore the kernel of this action is on the one hand the centralizer of [{\cal T}({\cal R})] in [{\cal R}], on the other hand, since [{\bf L}({\cal R})] contains a basis of [\tau ({\bb E}_{n})], it is equal to the kernel of the mapping [\overline{} ], which is [{\cal R} \cap {\cal T}_n = {\cal T}({\cal R})], hence [{\cal C}_{\cal R}({\cal T}({\cal R})) = {\cal T}({\cal R}).]Hence only the linear part [\overline{{\cal R}} \cong {\cal R}/{\cal T}({\cal R}) ] of [{\cal R}] acts faithfully on [{\cal T}({\cal R})] by conjugation and linearly on [{\bf L}(\cal {R})]. This factor group [{\cal R} /{\cal T}({\cal R})] is a finite group. Let us summarize this:

Theorem Let [{\cal R}] be a space group. The translation subgroup [{\cal T}({\cal R})\! = \!{\cal R} \cap {\cal T}_{n} ] is an Abelian normal subgroup of [{\cal R}] which is its own centralizer, [{\cal C}_{\cal R}({\cal T}({\cal R})) = {\cal T}({\cal R})]. The finite group [{\cal R}/{\cal T}({\cal R}) ] acts faithfully on [{\cal T}({\cal R})] by conjugation. This action is similar to the action of the linear part [\overline{{\cal R}}] on the lattice [\mu ' ({\cal T} ({\cal R})) = {\bf L}({\cal R})]. Maximal subgroups of space groups

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Definition A subgroup [{\cal M} \leq {\cal G}] of a group [{\cal G}] is called maximal if [{\cal M}\neq {\cal G}] and for all subgroups [{\cal U}\leq {\cal G}] with [{\cal M}\subseteq {\cal U}] it holds that either [{\cal U}={\cal M}] or [{\cal U}={\cal G}].

The translation subgroup [{\cal T}:={\cal T}({\cal R})] of the space group [{\cal R}] plays a very important role if one wants to analyse the space group [{\cal R}]. Let [{\cal U}\neq {\cal R}] be a subgroup of [{\cal R}]. Then [{\cal U}] has either fewer translations ([{\cal T}({\cal U}) \,\lt\, {\cal T}]) or the order of the linear part of [{\cal U}], the index of [{\cal T}({\cal U})] in [{\cal U}], gets smaller ([|\overline{{\cal U}} | \,\lt\, |\overline{{\cal R}}| ]), or both happen.

Definition Let [{\cal U}] be a subgroup of the space group [{\cal R}] and [{\cal T}:= {\cal T}({\cal R})].

(t) [{\cal U}] is called a translationengleiche or a t-subgroup if [{\cal U}\cap {\cal T} = {\cal T}].

(k) [{\cal U}] is called a klassengleiche or a k-subgroup if [{\cal U}/{\cal U}\cap {\cal T} \cong {\cal R}/{\cal T} ].

Remark. The third isomorphism theorem, Theorem[link], implies that if [{\cal U}] is a k-subgroup, then [{\cal U}{\cal T}/{\cal T} \cong {\cal U}/{\cal U}\cap {\cal T} \cong {\cal R}/{\cal T}]. Hence [{\cal U}] is a k-subgroup if and only if [{\cal U}{\cal T}={\cal R}].

Let [{\cal M}] be a maximal subgroup of [{\cal R}]. Then we have the following preliminary situation: [Scheme scheme1]

Since [{\cal T} \,{\underline{\triangleleft}} \,{\cal R}] and [{\cal M}\leq {\cal R}], one has by Proposition[link] that [{\cal M}{\cal T} \leq {\cal R}]. Hence the maximality of [{\cal M}] implies that [{\cal M}{\cal T}={\cal M}] or [{\cal M}{\cal T}={\cal R}]. If [{\cal M}{\cal T}={\cal M}] then [{\cal T}\subseteq {\cal M}], hence [{\cal M}] is a t-subgroup. If [{\cal M}{\cal T}={\cal R}], then by the third isomorphism theorem, Theorem[link], [{\cal R}/{\cal T} =] [{\cal M}{\cal T}/{\cal T} \cong {\cal M}/({\cal M}\cap {\cal T}) =] [{\cal M}/{\cal T}({\cal M})], hence [{\cal M}] is a k-subgroup. This is given by the following theorem:

Theorem (Hermann) Let [{\cal M}\leq {\cal R}] be a maximal subgroup of the space group [{\cal R}]. Then [{\cal M}] is either a k-subgroup or a t-subgroup.

The above picture looks as follows in the two cases: [Scheme scheme2]

Let [{\cal M}] be a t-subgroup of [{\cal R}]. Then [{\cal T}({\cal R}) \leq {\cal M}] and [{\cal M} / {\cal T}({\cal R})] is a subgroup [{\cal S}] of [{\cal P} = {\cal R}/{\cal T}({\cal R})]. On the other hand, any subgroup [{\cal S}] of [{\cal P}] defines a unique t-subgroup [{\cal M}] of [{\cal R}] with [{\cal T}({\cal R}) \leq {\cal M}] and [{\cal M} / {\cal T}({\cal R})={\cal S}], namely [{\cal M} =] [\{ {\sf s} \in {\cal R} \mid {\sf s}{\cal T}({\cal R}) \in {\cal S} \}]. Hence the t-subgroups of [{\cal R}] are in bijection to the subgroups of [{\cal P}], which is a finite group according to the remarks below Definition[link]. For future reference, we note this in the following corollary:

Corollary The t-subgroups of the space group [{\cal R}] are in bijection with the subgroups of the finite group [{\cal R}/ {\cal T}({\cal R})].

In the case [n=3], which is the most important case in crystallography, the finite groups [{\cal R}/{\cal T}({\cal R})] are isomorphic to subgroups of either [{\cal C}yc_{2}\times {\cal S}ym_{4}] (Hermann–Mauguin symbol [m\overline{3}m]) or [{\cal C}yc_{2}\times {\cal C}yc_{2} \times {\cal S}ym_{3}] ([= 6/mmm]). Here [\times ] denotes the direct product (cf. Definition[link]), [{\cal C}yc_{2}] the cyclic group of order 2, and [{\cal S}ym_{3}] and [{\cal S}ym_{4}] the symmetric groups of degree 3 or 4, respectively (cf. Section[link]). Hence the maximal subgroups [{\cal M}] of [{\cal R}] that are t-subgroups can be read off from the subgroups of the two groups above.

An algorithm for calculating the maximal t-subgroups of [{\cal R}] which applies to all three-dimensional space groups is explained in Section 1.5.5[link].

The more difficult task is the determination of the maximal k-subgroups.

Lemma Let [{\cal M}] be a maximal k-subgroup of the space group [{\cal R}]. Then [{\cal T}({\cal M}) = {\cal T}\cap {\cal M} \,{\underline{\triangleleft}} \,\,{\cal R}] is a normal subgroup of [{\cal R}]. Hence [\mu ' ({\cal T} ({\cal M})) \leq {\bf L}({\cal R})] is an [\overline{{\cal R}}]-invariant lattice.

Proof. [{\cal R} = {\cal T}{\cal M}], so every element [{\sf g}] in [{\cal R}] can be written as [{\sf g} = {\sf t}{\sf m}] where [{\sf t} \in {\cal T}] and [{\sf m}\in {\cal M}]. Therefore one obtains for [{\sf t}_{1} \in {\cal T}\cap {\cal M}] [{\sf g}^{-1} {\sf t} _{1} {\sf g} = {\sf m}^{-1} {\sf t}^{-1} {\sf t}_{1} {\sf t} {\sf m} = {\sf m}^{-1} {\sf t}_{1} {\sf m}, ]since [{\cal T}] is Abelian. Since [{\sf m} \in {\cal R}] and [{\cal T}] is normal in [{\cal R}], one has [{\sf m}^{-1} {\sf t}_{1} {\sf m} \in {\cal T}]. But [{\sf m}^{-1} {\sf t}_{1} {\sf m} ] is a product of elements in [{\cal M}] and therefore lies in the subgroup [{\cal M}], hence [{\sf m}^{-1} {\sf t}_{1} {\sf m} \in {\cal T} \cap {\cal M} ]. QED

The candidates for translation subgroups [{\cal T}({\cal M}) ] of maximal k-subgroups [{\cal M}] of [{\cal R}] can be found by linear-algebra algorithms using the philosophy explained at the beginning of this section: [{\cal R}] acts on [{\cal T}] by conjugation and this action is isomorphic to the action of the linear part [\overline{{\cal R}}\cong {\cal R}/{\cal T}] of [{\cal R}] on the lattice [{\bf L}({\cal R})] via the isomorphism [\mu ': {\cal T} \rightarrow {\bf L}({\cal R})]. Normal subgroups of [{\cal R}] contained in [{\cal T}] are mapped onto [\overline{{\cal R}} ]-invariant sublattices of [{\bf L}({\cal R})]. An example for such a normal subgroup is the group [{\cal T}^p] formed by the pth powers of elements of [{\cal T}] for any natural number [p\in {\bb N}]. One has [\mu'({\cal T} ^p) = p {\bf L}({\cal R})].

If [{\cal M}] is a maximal k-subgroup of [{\cal R}], then [{\cal T}({\cal M})] is a normal subgroup of [{\cal R}] that is maximal in [{\cal T}], which means that [\mu ' ({\cal T}({\cal M})) ={\bf L}({\cal M})] is a maximal [\overline{{\cal R}}]-invariant sublattice of [ {\bf L}({\cal R})]. Hence it contains [p {\bf L}({\cal R})] for some prime number p. One may view [{\cal T}/{\cal T}^p \cong {\bf L}({\cal R})/p{\bf L}({\cal R})] as a finite [({\bb Z}/p{\bb Z})\overline{{\cal R}} ]-module and find all candidates for such normal subgroups as full pre-images of maximal [({\bb Z} /p {\bb Z}) \overline{{\cal R}} ]-submodules of [{\bf L}({\cal R})/p{\bf L}({\cal R})]. This gives an algorithm for calculating these normal subgroups, which is implemented in the package [CARAT].

The group [{\cal G}: = {\cal T}/{\cal T}^p] is an Abelian group, with the additional property that for all [{\sf g} \in {\cal G}] one has [{\sf g}^p = {\sf e}]. Such a group is called an elementary Abelian p-group.

From the reasoning above we find the following lemma.

Lemma Let [{\cal M}] be a maximal k-subgroup of the space group [{\cal R}]. Then [{\cal T}/{\cal T}({\cal M}) ] is an elementary Abelian p-group for some prime p. The order of [{\cal T}/{\cal T}({\cal M})] is [p^r] with [r\leq n].

Corollary Maximal subgroups of space groups are again space groups and of finite index in the supergroup.

Hence the first step is the determination of subgroups of [{\cal R}] that are maximal in [{\cal T}] and normal in [{\cal R}], and is solved by linear-algebra algorithms. These subgroups are the candidates for the translation subgroups [{\cal T}({\cal M})] for maximal k-subgroups [{\cal M}]. But even if one knows the isomorphism type of [{\cal M}/{\cal T}({\cal M})], the group [{\cal T}({\cal M})] does not in general determine [{\cal M} \leq {\cal R}]. Given such a normal subgroup [{\cal S}\,{\underline{\triangleleft}}\,\,{\cal R}] that is contained in [{\cal T}], one now has to find all maximal k-subgroups [{\cal M}\leq {\cal R}] with [{\cal S}={\cal T}\cap {\cal M}] and [{\cal T}{\cal M}={\cal R}]. It might happen that there is no such group [{\cal M}]. This case does not occur if [{\cal R}] is a symmorphic space group in the sense of the following definition:

Definition A space group [{\cal R}] is called symmorphic if there is a subgroup [{\cal P} \leq {\cal R}] such that [{\cal P}\cap {\cal T}({\cal R}) = {\cal I}] and [{\cal P}{\cal T}({\cal R}) = {\cal R}]. The subgroup [{\cal P}] is called a complement of the translation subgroup [{\cal T}({\cal R})].

Note that the group [{\cal P}] in the definition is isomorphic to [{\cal R}/{\cal T}({\cal R})] and hence a finite group.

If [{\cal R}] is symmorphic and [{\cal P} \leq {\cal R}] is a complement of [{\cal T}], then one may take [{\cal M}:={\cal S}{\cal P}]. [Scheme scheme3]

This shows the following:

Lemma Let [{\cal R}] be a symmorphic space group with translation subgroup [{\cal T}] and [{\cal T}_{1} \leq {\cal T}] an [{\cal R}]-invari­ant subgroup of [{\cal T}] (i.e. [{\cal T}_{1} \,{\underline{\triangleleft}}\,\,{\cal R}]). Then there is at least one k-subgroup [{\cal U}\leq {\cal R}] with translation subgroup [{\cal T}_{1}].

In any case, the maximal k-subgroups, [{\cal M}], of [{\cal R}] satisfy [\eqalign{&{\cal M}{\cal T} = {\cal R} \, {\rm \,and }\, \cr &{\cal M}\cap {\cal T} = {\cal S} \, {\rm is\, a\, maximal}\, {\cal R}\hbox{-invariant subgroup of} \, {\cal T}. }]

To find these maximal subgroups, [{\cal M}], one first chooses such a subgroup [{\cal S}]. It then suffices to compute in the finite group [{\cal R}/{\cal S} =: \overline{{\cal R}}]. If there is a complement [\overline{{\cal M}}] of [\overline{{\cal T}} ={\cal T}/{\cal S}] in [\overline{{\cal R}}], then every element [{\sf x}\in \overline{{\cal R}}] may be written uniquely as [{\sf x}={\sf m}{\sf t}] with [{\sf m}\in \overline{{\cal M}}], [{\sf t}\in \overline{{\cal T}}]. In particular, any other complement [\overline{{\cal M}'}] of [\overline{{\cal T}}] in [\overline{{\cal R}}] is of the form [\overline{{\cal M}'} =\{ {\sf m}{\sf t}_{m} \mid {\sf m}\in \overline{{\cal M}}, {\sf t}_{m} \in \overline{{\cal T}} \}]. One computes [{\sf m}_{1}{\sf t}_{m_{1}} {\sf m}_{2} {\sf t}_{m_{2}} ={\sf m}_{1}{\sf m}_{2} ({\sf m}_{2}^{-1} {\sf t}_{m_{1}} {\sf m}_{2}) {\sf t}_{m_{2}}]. Since [\overline{{\cal M}'} ] is a subgroup of [\overline{{\cal R}}], it holds that [{\sf t}_{ m_{1}m_{2}} =({\sf m}_{2}^{-1} {\sf t}_{m_{1}} {\sf m}_{2}) {\sf t}_{m_{2}}]. Moreover, every mapping [\overline{{\cal M}} \rightarrow \overline{{\cal T}}; {\sf m}\mapsto {\sf t}_{{\sf m}}] with this property defines some maximal subgroup [{\cal M}'] as above. Since [\overline{{\cal M}}] and [\overline{{\cal T}}] are finite, it is a finite problem to find all such mappings.

If there is no such complement [\overline{{\cal M}}], this means that there is no (maximal) k-subgroup [{\cal M}] of [{\cal R}] with [{\cal M} \cap {\cal T} ={\cal S}].


International Tables for Crystallography (2005). Vol. A, Space-group symmetry, edited by Th. Hahn, 5th ed. Heidelberg: Springer. (Abbreviated IT A.)

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