## Abstract

Point groups consist of rotations, reflections, and roto-reflections and are foundational in crystallography. Symmorphic space groups are those that can be decomposed as a semi-direct product of pure translations and pure point subgroups. In contrast, Bieberbach groups consist of pure translations, screws, and glides. These “torsion-free” space groups are rarely mentioned as being a special class outside of the mathematics literature. Every space group can be thought of as lying along a spectrum with the symmorphic case at one extreme and Bieberbach space groups at the other. The remaining nonsymmorphic space groups lie somewhere in between. Many of these can be decomposed into semi-direct products of Bieberbach subgroups and point transformations. In particular, we show that those 3D Sohncke space groups most populated by macromolecular crystals obey such decompositions. We tabulate these decompositions for those Sohncke groups that admit such decompositions. This has implications to the study of packing arrangements in macromolecular crystals. We also observe that every Sohncke group can be written as a product of Bieberbach and symmorphic subgroups, and this has implications for new nomenclature for space groups.

## Acknowledgments

The authors would like to thank Prof. Bernard Shiffman and the anonymous reviewers for their very constructive feedback.

### A Additional examples of the decomposition procedure

Here additional examples of decompositions are provided. First, we show an example where decomposing Γ/*T* does not work, but Γ/Σ does. Next we show a case when no semi-decomposition is possible. Finally, we show a case where an additional (less efficient) semi-decomposition is possible.

#### A.1 **24**, **I****2**_{1}**2**_{1}**2**_{1}

**I**

As another example, consider Γ=*I*2_{1}2_{1}2_{1}. To determine *T*, we use the Bilbao function HERMANN which gives

(where the perfunctory bottom row in the affine transformation as a 4×4 matrix has been removed to save space.)

The coset representatives in the decomposition of Γ with respect to *T* (in the basis of *T*) are given below:

Here *S*={(*x*, *y*, *z*); (−*y*, −*x*, −*z*)} is identical with *I*2_{1}2_{1}2_{1}:*C*2]=2. If ℱ_{Γ/T} can be semi-decomposed, then it should be possible to find a Bieberbach subgroup with [Γ_{B} :*P*1]=2. But the only Bieberbach group that could satisfy this is *P*2_{1}, and MAXSUB does not list *P*2_{1} as a maximal subgroup. Therefore, there is no need to investigate the case when [*I*2_{1}2_{1}2_{1}:*T*_{I} ]=4 further, and so we investigate [*I*2_{1}2_{1}2_{1}:Σ]=8, which is the next smallest candidate.

When Σ=*P*1<*T* the Bilbao COSET function (left and right are the same so Σ◁Γ) gives

Here (*x*+1/2, *y*+1/2, *z*+1/2) is a pure translation. Such will only occur when Σ<*T*. In contrast, (–*x*+1/2, –*y*, *z*+1/2); (–*x*, *y*+1/2, –*z*+1/2); (*x*+1/2, –*y*+1/2, –*z*) are all screw motions with *d*=1/2, and all of the rotation axes are mutually orthogonal with an angle of rotation of *π*. So this is a clue that it may be related to *P*2_{1}2_{1}2_{1}/*P*1. We verify this by using the Bilbao COSET function to give

This matches with four entries in _{Γ/Σ} with a candidate Bieberbach group quotient by an appropriate translation group.

In contrast, looking at the elements (–*x*, –*y*+1/2, *z*); (–*x*+1/2, *y*, –*z*); (*x*, –*y*, –*z*+1/2), we observe that they are all rotations because *d*=0. But if we seek 𝔹 and 𝕊 such that their product reproduces _{1}2_{1}2_{1}/*P*1 is of order 2 in *P*1, it is normal, and so we do not need to check that the conjugation action of 𝕊_{i} on

We see that we can define a two-element point group (which fixes a point other than the origin) from the identity and any one of (–*x*, –*y*+1/2, *z*); (–*x*+1/2, *y*, –*z*); (*x*, –*y*, –*z*+1/2). For example,

Abstractly, every two-element group is isomorphic, and so candidates for these are *P*2/*P*1 or *C*2/*T*_{C} . But concretely, we can only find affine transformations such that

There is no guarantee that this decomposition is unique.

Using the bottom-up approach employing the Bilbao MAXSUB function, we see that [*I*2_{1}2_{1}2_{1}:*P*2_{1}2_{1}2_{1}]=2. and as [*P*2_{1}2_{1}2_{1}:*P*1]=4 and [*P*2:*P*1]=2 we conclude that it is not possible to find a decomposition with smaller values than this, and so our search terminates.

#### A.2 **90**, **P****42**_{1}**2**

**P**

Consider group no. 90, and use Γ=*P*42_{1}2 and *T*=*P*1. When using the identity transformation, the Bilbao COSET function gives

This is convenient because we can immediately identify

with a point group of order 4 (and index 2 in Γ). Therefore it is normal. Searching for every symmorphic space group with point group of order 4, we find that this corresponds to *C*222/*T*_{C} . The complementing subgroup can be chosen as

(In fact, there are four possible choices, but they all act in the same way as described below.) It is easy to check that this is a group.

For the nontrivial transformation in this group, *R* is a rotation around **n**=**e**_{2} by *π*. And **v**(*R*)=[1/2, 1/2, 0]^{t}. And so **v***·***n**=1/2. We conclude that 𝔹 is isomorphic with *P*2_{1}/*P*1. To find affine transformations that will make the relationship between 𝔹 and *P*2_{1}/*P*1, we solve the equation *αg*=*g*′*α* where *g* and *g*′ are respectively representatives taken from 𝔹 and *P*2_{1}/*P*1.

Clearly *α* that has been computed. We find that left and right cosets do not match modulo *P*1, and 𝔹 is not normal, and

The benefit of this top-down approach is that we need not worry about whether the two components correspond to space subgroups with a common lattice, since that is guaranteed a priori. But this does not provide a complete list of all possible decompositions. We therefore go to the Bilbao MAXSUB function, and look for other low-index symmorphic and Bieberbach subgroups. We find that [*P*42_{1}2:*P*4]=2. Hence *P*4◁*P*42_{1}2. The same subroutine provides a transformation, and we find (after moding out extraneous 1’s returned by Bilbao) that

Here we see that the first four elements correspond to 𝕊_{2}:=*P*4/*P*1, and with (–*x*+1/2, *y*+1/2, –*z*) clearly visible, we can define the complement 𝔹 as before. And it is not normal here either. Therefore, we conclude that

Though the quotient is “reverse-semi-decomposable” the Bieberbach subgroup is not normal (and no other Bieberbach subgroup with [Γ:Γ_{B} ]<[Γ:*T*] is normal either.

There is a copy of *P*2_{1}2_{1}2_{1} that has index 4, which we can identify using the HERMANN subroutine in the Bilbao server, and gradually increasing the index of candidate Bieberbach subgroups. But left and right cosets do not match, and hence it is not normal.

Another route to index-4 *P*2_{1}2_{1}2_{1} is by observing that none of the coset representatives in *z* component. Then, using the affine transformation *α* with linear part diag[1, 1, 2] and zero translation in the COSET function in Bilbao can double the size of the quotient group, giving enough room to fit both a copy of a conjugated

Γ/Σ=(*C*222/*T*_{C} ) ⋊ (*P*2_{1}2_{1}2_{1}/*P*1). But *P*2_{1}2_{1}2_{1} is not normal in *P*42_{1}2, and because this quotient group has 16 elements as opposed to the 8 of the original, we do not include this in the table.

There is an affine-conjugated copy of *P*2_{1} with *α*=diag[1, 1, 2] that is normal in *P*42_{1}2 with index 8, but this index does not match the order of the largest point subgroup, and so the compatibility conditions fail, and so it cannot be used in a semi-decomposition.

We therefore conclude that 90 cannot be efficiently semi-decomposed, but the quotient group can be “reverse-semi-decomposed” in multiple ways, each with the pure rotation part being normal rather than the Bieberbach part. As a consequence of this, and of Theorem 3, it is possible to write

where it is understood that in the above different copies of *P*2_{1} are used. Moroever, from Lemma 2.1, this is a case where even though Γ_{B} is not normal, it is nevertheless possible to perform decompositions of the form Γ=Γ_{B}*S*.

#### A.3 **93,***P***4**_{2}**22**

Using the Bilbao COSET function, and taking the top-down approach, [*P*4_{2}22:*P*1]=8 and

Of these,

is a subgroup with [*P*4_{2}22:*P*222]=2 and hence it is normal. But there is no way to construct a two-element subgroup from the remaining elements, which appear to be *P*4_{1} or *P*4_{3} transformations. And so we look for ways to conjugate so that all elements of these subgroups are present as representatives in the coset decomposition.

Using HERMANN routine in Bilbao with *G*=93 and *H*=76 with index 4, we find that with

Several subgroups can be visually identified:

We find that 𝔹_{1}𝕊=𝔹_{2}𝕊=*F*_{P}222/_{P1} mod P1. And by conjugating each element of 𝔹_{1} by all elements of 𝕊 and evaluating mod P1, we see that 𝔹_{1} is closed under conjugation. The same is true for 𝔹_{2}. Hence, we find that 𝔹_{1} and 𝔹_{2} are normal in

and

#### A.4 **94,****P****4**_{2}**2**_{1}**2**

**P**

[*P*4_{2}2_{1}2, *P*1]=8. When computing

But also present are elements such as (–*x*+1/2, *y*+1/2, –*z*+1/2) and (*x*+1/2, –*y*+1/2, –*z*+1/2), each of which are conjugated versions of elements of *P*2_{1}. But, computing all right coset reps for *P*4_{2}2_{1}2/*P*2_{1} in HERMANN, and computing the corresponding left cosets in COSET, we see that the *P*2_{1} elements corresponding to this are not normal.

In contrast, (–*y*+1/2, *x*+1/2, *z*+1/2) and (*y*+1/2, –*x*+1/2, *z*+1/2) which are in *P*4_{1}, and neither of which forms a 2-element group mod *P*1 with the identity. But when using

in Bilbao, [(*P*4_{2}2_{1}2)^{α}*,**P*1]=16.

*P*4_{1}◁(*P*4_{2}2_{1}2)^{α}; [(*P*4_{2}2_{1}2)^{α}, *P*4_{1}]=4 and [(*P*4_{2}2_{1}2)^{α}, *C*222]=4.

In particular,

And

looks like

Of the remainder of elements, we can construct

𝔹 is normal in

And the same is true for *P*4_{3}:

We note also that there are normal affine-conjugated copies of *P*222 and *P*2_{1}2_{1}2_{1} inside of *P*4_{2}2_{1}2 with [(*P*4_{2}2_{1}2)^{α}:*P*222]=[(*P*4_{2}2_{1}2)^{β}:*P*2_{1}2_{1}2_{1}]=4. The difficulty with *P*222 is that it has a different lattice than the Bieberbach subgroups with compatible orders, and so it cannot be used to semi-decompose. *P*2_{1}2_{1}2_{1} does have a complement that is *C*222/*T*_{C} (about a point that is not the origin when expressed in the basis of *P*2_{1}2_{1}2_{1} in the standard setting) giving

#### A.5 **98,***I***4**_{1}**22**

*P*4_{1} with index 4 is normal. [*I*4_{1}22:*P*1]=16 in standard setting. Subgroup of pure rotations is order 4. So should be decomposable.

Of these,

This is a conjugated version of

This is of the form *α*∈*SE*(3), and 𝔹 is normal in

Similarly, by choosing different coset representatives to define 𝔹 we find

It is also possible to identify elements of *P*2_{1}2_{1}2_{1} in

#### A.6 **171****P****6**_{2}**– A case that can be alternatively decomposed with larger [Γ:Σ]**

**P**

It is obvious from examining *P*6_{2}=*P*3_{2}⋊ (*P*2/*P*1), and in this case [Γ:Γ_{B} ]=[*P*6_{2}:*P*3_{2}]=2 and [Γ_{B} :*T*]=[*P*3_{2}:*P*_{1}]=3.

Here we show that it is possible to find other decompositions where [Γ:Γ_{B′}] is the same but [Γ_{B′}:Σ] is larger. In particular, [*P*6_{2}:*P*6_{1}]=2 and [*P*6_{1}:*P*1]=6 with

The coset representatives from the two cosets are

and

From these we can construct

which is a pure rotation group.

This is a case where we could decompose Γ/Σ (which has 12 elements)

where 𝔹=*P*6_{1}/Σ and

Similar less-efficient decompositions are possible for *P*222_{1}, *C*222_{1}, *P*6_{3}, *P*6_{4}, *P*6_{2}22, *P*6_{4}22, *P*6_{3}22. These are listed in the table after the | symbol. In most cases they have the same type of complementing *S* as the Γ_{B} in the more efficient decomposition of the same group, and the answer to the question of whether 𝕊 is normal in Σ\Γ is the same as well. When the answers differ, the answer for the less efficient decomposition is written to the right of the | symbol.

#### A.7 **208 and 210**

These two groups each have large symmorphic subgroups of index 2, indicating normality of Γ_{S} . Since in a semi-decomposition all of the torsion must reside in 𝕊, if a compatible Bieberbach subgroup exists it must be the case that [Γ:Γ_{B} ]=|𝕊|. Each of the groups 208 and 210 have a symmorphic subgroup with point group of order 12 (195 in 208, and 196 in 210). Moreover, in each of these two groups are Bieberbach groups of index 12 (76 and 78 in 208, and 76, 78 and 19 in 210). Checking these (e.g. by conjugating generators of each candidate Γ_{B} by all generators of Γ) none of these Bieberbach subgroups are normal. In some cases there is a faster way to rule out Γ_{B} ’s that are not normal, based on the discussion at the end of Section 6, which leads to the following theorem.

**Theorem 4**: If Σ=Γ_{B} ∩*T*◁Γ_{B} <Γ and Σ is not normal in Γ, then Γ_{B} is not normal in Γ.

*Proof:* Every element of Γ_{B} can be written as *γ*_{B} =*σb* for some *σ*∈Σ and *T* is the minimal-index translation subgroup of Γ. Conjugating by an arbitrary *γ*∈Γ gives

Conjugation does not change the nature of a rigid-body displacement, i.e. pure translations remain pure translations, pure rotations remain pure rotations, and screw displacements remain screw displacements with the same *θ* and *d*. Therefore, *γσγ*^{–1} must be a translation, but since Σ is not normal in Γ it must be that *γσγ*^{–1}∉Σ for at least one *γ*∈Γ. Therefore, *γγ*_{B}*γ*^{–1} cannot be decomposed as *σ*′*b*′. But since every element of Γ_{B} can be decomposed in this way, it must be that *γγ*_{B}*γ*^{–1}∉Γ_{B} and therefore Γ_{B} cannot be normal in Γ. □

Note that whereas the proof of this theorem is specific to space groups, it can be viewed as an example of the contrapositive of the statement

which is true in more general (abstract) settings. That is, the intersection of normal subgroups is also normal.

As a consequence of this theorem, if Σ=Γ_{B} ∩*T* is the primitive translation group for Γ_{B} , and Γ_{B} <Γ, then if Σ is not normal in Γ we need not even check if Γ_{B} is normal, because it is guaranteed not to be. Therefore, in such cases it cannot be the case that Γ=Γ_{B} ⋊𝕊.

Moreover, even if there were normal Bieberbach subgroups further down the subgroup tree, these would be irrelevant for such decompositions because there would be no way to match [Γ:Γ_{B} ] and |𝕊|. In searching for Γ_{B} and *S* such that Γ=Γ_{B}*S*, the indices and orders of coset spaces must match even if Γ_{B} is not normal in Γ because Γ_{B} ∩*S*=*e*. The above mentioned groups have compatible indices and orders for this, and these are explored in the subsections that follow.

##### A.7.1 **Details of 208**

There are no normal Bieberbach subgroups in this group of an index compatible with the condition [Γ:Γ_{B} ]=|𝕊|. The largest symmorphic subgroup has

It has index 2, and hence it is normal and the quotient is a group of order 2. The only Bieberbach group with |𝔹|=2 is Γ_{B} =*P*2_{1}. But none of the elements of *P*4_{2}32 are elements of *P*2_{1} with the same lattice. For example, {(*x*, *y*, *z*); (*y*+1/2, *x*+1/2, –*z*+1/2)} is a 2-element subgroup of *S*. Though it is isomorphic with *P*2_{1}/*P*1, it is not equal to *C*2 sharing the same lattice as *P*4_{2}32 and so we can write

But our goal was not to decompose space groups into products of symmorphic subgroups, and so we look further down the tree. To do this, we double the unit cell as before by choosing *α*=diag[1, 1, 2]. This results in 48-element coset spaces

In other words, this is a case where Σ is not normal in Γ. Nevertheless, within

and

Unfortunately,

And no other *α* was found that would enable such a decomposition. Searching further down the subgroup graph cannot result in a Γ_{B}*S* decomposition. We therefore search for products of the form Γ_{B} Γ_{S} .

Choosing *α*=diag[1, 2, 1] we can identify inside of *B*, Ξ, *S*:

and

But the product of these three does not reproduce *B*∩Ξ=Ξ∩*S*=*B*∩*S*={*e*} we find that (Ξ*S*)∩*B*≠{*e*}.

Therefore we search deeper down the subgroup graph (i.e. for larger [Γ:Σ]) and find that there are multiple affine transformations with linear part

or a similarity-transformed version of this corresponding to permutations of coordinate axis names. We therefore express Γ^{α} in the basis of Σ=*P*1 with *α*=(*A*, **0**) (since translations are irrelevant to identifying Bieberbach subgroups in the top-down procedure, and making the translation zero retains the origin allows us to easily identify point rotations.) This makes it easy to identify the largest *S* (corresponding to *P*23) as

We also identify

(which is a conjugated version of

and

(which is a conjugated version of

In this case^{[8]}

for *i*=1, 2. Whereas Γ_{S} =ΣΞ_{1}*S*, we find that Γ_{S} =ΣΞ_{1}*S* and so it cannot be used. But changing the order we find

##### A.7.2 **Details of 210**

Using SUBGROUPGRAPH provides a number of transformations to generate ℱ_{Σ\Γ} of order 48. We choose this order because that is what is required to match |*B*|·|*S*|. The linear parts of all of the transformations are all equivalent under relabeling of axes, and the translational part is set to zero so that it is easier to identify point transformations. Therefore, we choose

Then the resulting

has the following subsets

and

These respectively correspond to *P*2_{1}2_{1}2_{1}, *P*4_{1}, *P*4_{3}, and *F*23. It can be shown that

for *i*=1, 2, 3 and so we could write

even though none of the

#### A.8 **213,****P****4**_{1}**32 and 212,****P****4**_{3}**32**

**P**

**P**

Taking the top-down approach, we use COSET to compute representatives of the cosets in *P*4_{1}32/*P*1. The result is [*P*4_{1}32:*P*1]=24. Of these, we identify four elements to construct

We also find three representatives (*x*, *y*, *z*); (*z*, *x*, *y*); (*y*, *z*, *x*) that are clearly rotations that preserve the origin. However, these rotations all have the axis of rotation **v**=0. In particular, we find that when using

in the Bilbao COSET function,

and

We note also that

and Γ_{B′} can be identified with *P*4_{1}. Using the Bilbao Server function IDENTIFY GROUP finds that Γ_{S} =*R*32.

Alternatively, using the bottom-up approach we search for Bieberbach and symmorphic subgrouos respectively of order 4 and 6 (index 6 and 4) with SUBGROUPGRAPH. This shows that *R*32 is the only symmorphic subgroup of *P*4_{1}32 with these properties, and so we identify it with

SUBGROUPGRAPH can also be used to identify *P*4_{1} and *P*2_{1}2_{1}2_{1} as subgroups of *P*4_{1}32 with the correct index with fundamental domains *P*2_{1}2_{1}2_{1} is normal.

We note that using any *α* given by SUBGROUPGRAPH for *P*4_{1}32 and *P*4_{1} with index 6 gives

For example, when *α* is a translation by [1/4, 0, 1/2]^{t},

and we choose

which IDENTIFY GROUP identifies with *R*32/*T*_{R} . And so, even though this Γ_{B′} is not normal, we can write Γ=Γ_{B′}*S* by conjugating *S*. Even so, we do not list *P*4_{1} in the table because it is not normal, and is *P*2_{1}2_{1}2_{1} is normal, and hence the most useful for our application.

From the above it is clear that the top-down and bottom-up approaches provide the same results. And the final result is

Group 212, *P*4_{3}32 follows in a similar way, but with *α*→*α*^{–1} (corresponding to a translation by –[3/8, 3/8, 3/8]^{t} instead of [3/8, 3/8, 3/8]^{t}). *P*4_{3} and *P*2_{1}2_{1}2_{1} as subgroups of *P*4_{3}32 with the correct index with fundamental domains to decompose, but again only *P*2_{1}2_{1}2_{1} is normal, leading to

#### A.9 **214,****I****4**_{1}**32**

**I**

[Γ:*T*]=24 with minimal index symmorphic subgroup 155 (which has index 4 and point group of order 6). Bieberbach subgroups with the compatible value of |𝔹|=4 are groups 19, 76, 78. However, these all have index 12 in group 214. Therefore, Γ/*T* cannot be decomposed into the form 𝔹𝕊 because [Γ:Γ_{B} ]>|𝕊| and no _{B} ’s exists in ℱ_{Γ/T} .

The next finest lattice Σ is the lattice in the conventional setting, with [Γ:Σ]=48. Σ is normal in Γ. As with 213, translating the origin by [3/8, 3/8, 3/8]^{t} relative to that in the standard centering allows us to write

where in the case of 214,

and

and

where Γ_{B} =*P*2_{1}2_{1}2_{1} and Γ_{S} =*R*32.

As can be seen by the translation subgroup {(*x*, *y*, *z*); (*x*+1/2, *y*+1/2, *z*+1/2)}, the fundamental domain *S*. As mentioned earlier, attempting to use a finer lattice to remove this translation also prevents any normal Bieberbach groups from existing. And seeking a coarser Σ will only worsen the problem by resulting in more translational elements in

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**Received:**2015-5-21

**Accepted:**2015-10-16

**Published Online:**2015-12-1

**Published in Print:**2015-12-1

©2015 by De Gruyter