Maximal subgroups of space groups

Nebe, G.

doi:10.1107/97809553602060000541

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

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International Tables for Crystallography (2006). Vol. A1. ch. 1.5, pp. 35-36 | 1 | 2 |

Section 1.5.4.2. Maximal subgroups of space groups

Gabriele Nebe^a ^*

^a Abteilung Reine Mathematik, Universität Ulm, D-89069 Ulm, Germany
Correspondence e-mail: nebe@mathematik.uni-ulm.de

1.5.4.2. Maximal subgroups of space groups

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Definition 1.5.4.2.1. 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 1.5.4.2.2. 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 1.5.3.5.2, 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 1.5.3.2.11 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 1.5.3.5.2, $[{\cal R}/{\cal T} =]$ $[{\cal M}{\cal T}/{\cal T} \cong {\cal M}/({\cal M}\cap {\cal T}) =]$ $[{\cal M}/{\cal T}({\cal M})]$ , hence $[{\cal M}]$ is a k-subgroup. This is given by the following theorem:

Theorem 1.5.4.2.3. (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 1.5.4.1.1. For future reference, we note this in the following corollary:

Corollary 1.5.4.2.4. 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 1.5.3.6.1), $[{\cal C}yc_{2}]$ 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 1.5.3.6). 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.

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

Lemma 1.5.4.2.5. 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 1.5.4.2.6. 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 1.5.4.2.7. 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 1.5.4.2.8. 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 1.5.4.2.9. 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 $[{\cal M} \cap {\cal T} ={\cal S}]$ .

References

International Tables for Crystallography (2006). Vol. A1. ch. 1.5, pp. 35-36