International
Tables for
Crystallography
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. 39-40   | 1 | 2 |

## Section 1.5.7. Qualitative results

Gabriele Nebea*

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

### 1.5.7. Qualitative results

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#### 1.5.7.1. General theory

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In this section, we want to comment on the very subtle question of deciding whether two space groups and are isomorphic.

This problem can be treated in several stages:

Let and be space groups. Since the translation subgroups are characteristic subgroups of (the maximal Abelian normal subgroup of finite index), each isomorphism induces isomorphisms of the corresponding translation subgroups (by restriction) as well as of the point groups It is convenient to view as a lattice on which the point group acts as group of linear mappings (cf. the start of Section 1.5.4). Then the isomorphism is an isomorphism of -sets, where acts on via conjugation and on via Since and centralizes itself, this action is well defined, i.e. independent of the choice of the coset representative .

The following theorem will show that the isomorphism of sufficiently large factor groups of and implies a `near' isomorphism of the space groups themselves. To give a precise formulation we need one further definition.

Definition 1.5.7.1.1. For define which is the set of all rational numbers for which the denominator is prime to d. For the space group let be the group , where i.e. one allows denominators that are prime to d in the translation subgroup.

One has the following:

Theorem 1.5.7.1.2. Let and be two space groups with point groups of order . Let denote the set of normal subgroups of having finite index in . Then the following three conditions are equivalent:

 (i) There are normal subgroups with and with if and if (). (ii) . (iii) There is a bijection such that for all .

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

Remark. If are three- or four-dimensional space groups, the isomorphism in (ii) already implies the isomorphism of and , but there are counterexamples for dimension 5.

#### 1.5.7.2. Three-dimensional space groups

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Corollary 1.5.7.2.1. Let be a three-dimensional space group with translation subgroup and p be a prime not dividing the order of the point group . Let be a subgroup of of index for some . Then

 (a) is a k-subgroup. (b) is isomorphic to .

Proof:

 (a) implies that divides . Since , one obtains as a factor of . But p is not a factor of , hence and . According to the remark following Definition 1.5.4.2.2, is a k-subgroup. (b) Let . Let if and otherwise, and let . Since , one has and . By the third isomorphism theorem, Theorem 1.5.3.5.2, it follows that By Theorem 1.5.7.1.2 (i) (ii), one has . By the remark above, this already implies that and are isomorphic. QED

Theorem 1.5.7.2.2. Let be a three-dimensional space group and be a maximal subgroup of of index . Then

 (a) is a k-subgroup. (b) is isomorphic to .

Proof. Since is soluble, the index is a prime power (see Theorem 1.5.5.3.1). If p is not a factor of , the statement follows from Corollary 1.5.7.2.1. Hence we only have to consider the cases , and , . Since 9 is not a factor of the order of any crystallographic point group in dimension 3, assertion (a) follows if the index of is divisible by 9. If is a maximal t-subgroup, then is a primitive -set for the point group of . Since the point groups of dimension 3 have no primitive -sets of order divisible by 8, assertion (a) also follows if the index of is divisible by 8.

For all three-dimensional space groups , the module [where is identified with the corresponding lattice in as in Section 1.5.4] is not simple as a module for the point group . [It suffices to check this property for the two maximal point groups () and .] This means that is not a maximal -invariant sublattice of . Since the translation subgroup of a maximal k-subgroup of index equal to a power of 2 in is a maximal -invariant subgroup of that contains , one now finds that has no maximal k-subgroup of index 8.

Now assume that . By Corollary 1.5.7.2.1, one only needs to deal with groups such that the order of the point group is divisible by 3. is isomorphic to a subgroup of or . If is a subgroup of , then is simple and is of index 27 in [with ]. It turns out that is isomorphic to in these cases. If does not contain a subgroup isomorphic to , then the maximality of implies that is of index 3 in . Hence in this case. QED

Corollary 1.5.7.2.3. Let be a maximal subgroup of the three-dimensional space group .

 (a) If the index of is a power of 2, then or 4. (b) If 3 is a factor of the order of the point group and the index of is a power of 3, then or 27. For , is necessarily isomorphic to (by Theorem 1.5.7.2.2).

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 of a three-dimensional space group of index 9, then the order of the point group of is not divisible by three and the subgroup is a k-subgroup and isomorphic to .

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.

### References

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