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. 38-39   | 1 | 2 |

Section 1.5.6. Quantitative results

Gabriele Nebea*

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

1.5.6. Quantitative results

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

1.5.6.1. General results

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The first very easy but useful remark applies to general groups :

Remark. Let be a maximal subgroup of of finite index . Then . Hence the maximality of implies that either and is a normal subgroup of or and has i maximal subgroups that are conjugate to .

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 1.5.6.1.1. Let be a group and a subgroup of index . Then is a normal subgroup of .

Proof. Choose an element , . Then . Hence and therefore . Since this is also true if , the proposition follows. QED

Let be a subgroup of a group of index 2. Then is a normal subgroup and the factor group is a group of order 2. Since groups of order 2 are Abelian, it follows that the derived subgroup of (cf. Definition 1.5.5.2.1) (which is the smallest normal subgroup of such that the factor group is Abelian) is contained in . Hence all maximal subgroups of index 2 in contain . If one defines , then is an elementary Abelian 2-group and hence a vector space over the field with two elements. The maximal subgroups of are the maximal subspaces of this vector space, hence their number is , where .

This shows the following:

Corollary 1.5.6.1.2. The number of subgroups of of index 2 is of the form for some .

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

Proposition 1.5.6.1.3. Let be a subgroup of the group with . Then is either a normal subgroup of or and there are three subgroups of conjugate to .

Proof. is isomorphic to a subgroup of that acts primitively on . Hence either and is a normal subgroup of or , and there are three subgroups of conjugate to . QED

1.5.6.2. Three-dimensional space groups

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We now come to space groups. By Lemma l.5.5.2.3, all three-dimensional space groups are soluble. Theorem 1.5.5.3.1 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 1.5.4.2.7), we get:

Corollary 1.5.6.2.1. Let be a three-dimensional space group and a maximal subgroup. Then is a prime power.

Let be a three-dimensional space group and its point group. It is well known that the order of is of the form with or and . By Corollary 1.5.4.2.4, the t-subgroups of are in one-to-one correspondence with the subgroups of . Let us look at the t-subgroups of of index 3. It is clear that has no subgroup of index 3 if , since the index of a subgroup divides the order of the finite group by the theorem of Lagrange. If , then any subgroup of of index 3 has order and hence is a Sylow 2-subgroup of . Therefore there is such a subgroup of index 3 in by the first theorem of Sylow, Theorem 1.5.3.3.1. By the second theorem of Sylow, Theorem 1.5.3.3.2, all these Sylow 2-subgroups of are conjugate in . Therefore, by Proposition 1.5.6.1.3, the number of these groups is either 1 or 3:

Corollary 1.5.6.2.2. Let be a three-dimensional space group.

If the order of the point group of is not divisible by 3 then has no t-subgroups of index 3.

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