Group-Subgroup Relations of
Space Groups
Mois I. Aroyo
Universidad del Pais Vasco, Bilbao, Spain
Group-Subgroup Relations of
Space Groups
I. Subgroups
1.1 General considerations
Exercises
1.2 Subgroups of space groups
Exercises
II. Wyckoff-position splittings
2.1 General considerations
Exercises
apply this element first
Subgroups: General considerations
Subgroup H < G
1. H={e,h1,h2,...,hk} G
2. H satisfies the group axioms of G
Proper subgroups H < G, and
trivial subgroup: {e}, G
Index of the subgroup H in G: [i]=|G|/|H|
(order of G)/(order of H)
Maximal subgroup H of G
NO subgroup Z exists such that:
H<Z<G
Coset decomposition G:H
Group-subgroup pair H < G
left coset
decomposition
G=H+g2H+...+gmH, gi H,
m=index of H in G
right coset
decomposition
G=H+Hg2+...+Hgm, gi H
m=index of H in G
Coset decomposition-properties
(i) giH
gjH = { }, if gi
(ii) |giH| = |H|
(iii) giH = gjH, gi
gjH
gjH
Coset decomposition G:H
Normal
subgroups
Hgj= gjH, for all gj=1, ..., [i]
Theorem of Lagrange
group G of order |G|
then
subgroup H<G of order |H|
Corollary
|H| is a divisor of |G|
and [i]=|G:H|
The order k of any
element of G,
gk=e, is a divisor of |G|
Conjugate elements
Conjugate elements
gi ~ gk if g: g-1gig = gk,
where g, gi, gk, G
Classes of conjugate
elements
L(gi)={gj| g-1gig = gj, g G}
Conjugation-properties
(i) L(gi)
L(gj) = { }, if gi
(ii) |L(gi)| is a divisor of |G|
(iv) if gi, gj
L(gj)
(iii) L(e)={e}
L, then (gi)k=(gj)k= e
Conjugate subgroups
Conjugate subgroups
Let H1<G, H2<G
then, H1 ~ H2, if
g G: g-1H1g = H2
(i) Classes of conjugate subgroups: L(H)
(ii) If H1 ~ H2, then H1
H2
(iii) |L(H)| is a divisor of |G|/|H|
Normal subgroup
H G, if g-1H g = H, for
g G
Factor group
product of sets:
Kj Kk={ gjpgkq=gr | gjp
factor group G/H:
group axioms:
G={e, g2, ...,gp}
{
Kj={gj1,gj2,...,gjn}
Kk={gk1,gk2,...,gkm}
Kj, gkq Kk} Each element gr is taken
only once in the
product Kj Kk
H G
G=H+g2H+...+gmH, gi H,
G/H={H, g2H, ..., gmH}
(i) (giH)(gjH) = gijH
(ii) (giH)H =H(giH)= giH
(iii) (giH)-1 = (gi-1)H
EXERCISES
Problem 3.3.1, p.19
Demonstrate that H is always a
normal subgroup if |G:H|=2.
Problem 3.3.2, p.19
The group of the square and its
subgroups
Problem 3.3.1, p.19
(i) Classes of conjugate elements
{e}, {4,4-1}, {2}, {mx,my}, {m+,m-}
(ii) Group-subgroup diagram
SOLUTION
Problem 3.3.3, p.20
Consider the normal subgroup
{e,2} of 4mm, of index 4.
(i) Coset decomposition 4mm:{e,2}
(ii) Show that the cosets of the decomposition 4mm:{e,2}
fulfil the group axioms and form a factor group
(iii) Multiplication table of the factor group
(iv) A crystallographic point group
isomorphic to the factor group?
Problem 3.3.3, p.20
SOLUTION
(i) coset decomposition
{e,2}, {4,4-1}, {mx,my}, {m+,m-}
E
A
B
C
(ii) factor group and multiplication table
Multiplication table
of the Vierergruppe
Example: 222
Subgroups of Space groups
Coset decomposition G:TG
(I,0)
(I,t1)
(I,t2)
...
(I,tj)
...
(W2,w2) ...
(W2,w2+t1) ...
(W2,w2+t2) ...
...
...
(W2,w2+tj) ...
...
...
(Wm,wm)
...
(Wm,wm+t1) ...
(Wm,wm+t2) ...
...
...
(Wm,wm+tj) ...
...
...
(Wi,wi)
(Wi,wi+t1)
(Wi,wi+t2)
...
(Wi,wi+tj)
...
Factor group G/TG
isomorphic to the point group PG of G
Point group PG = {I, W1, W2, ...,Wi}
Subgroups of space groups
Translationengleche subgroups H<G:
{
T H= T G
PH<PG
Example: P2/m
Coset decomposition
TG
TG 2
TG 1̄
(I,0)
(I,t1)
(2,0)
(2,t1)
(1̄,0) (m,0)
(1̄, t1) (m, t1)
(I,t2)
...
(I,tj)
(2,t2)
...
(2,tj)
(1̄, t2) (m,t2)
...
...
( 1̄, tj) (m, tj)
...
t-subgroups:
H1=TG TG2
P2
...
...
H2=TG TG 1̄
TG m
...
H3=TG TG m
Pm
Subgroups of space groups
Klassengleiche subgroups H<G:
non-isomorphic
Example: C2
Ti
Coset decomposition
ti=integer
tc=1/2,1/2,0
Ti t c
(I,0)
(I,t1)
Ti 2
H1=Ti Ti2
P2
Ti t c 2
(I,tc)
(2, 0) (2,tc)
(I,t1+tc) (2, t1) (2, t1+tc)
(I,t2) (I,t2+tc)
...
...
(I,tj) (I,tj+tc)
...
k-subgroups:
{
T H< T G
PH=PG
...
(2, t2) (2, t2+tc)
...
...
(2, tj) (2, tj+tc)
...
...
H2=Ti Titc 2
P21
Subgroups of space groups
Klassengleiche subgroups H<G:
isomorphic
Example: P1
t=ua+vb+wc
Coset decomposition
Te={t(u=2n,v,w)}
ta(a,0,0)
isomorphic k-subgroups:
P1(2a,b,c)
{
T H< T G
PH=PG
Te
Te t a
(I,0)
(I,t1)
(I,ta)
(I,t1+ta)
(I,t2)
...
(I,tj)
...
(I,t2+ta)
...
(I,tj+ta)
...
H1=Te
Subgroups of space groups
General subgroups H<G:
{
T H< T G
PH<PG
Theorem Hermann, 1929:
For each pair G>H, there exists a
uniquely defined intermediate subgroup
M, G M H, such that:
M is a t-subgroup of G
H is a k-subgroup of M
Corollary
A maximal subgroup
is either a
t- or k-subgroup
G
it
M
ik
H
Subgroups of space groups
Chains of maximal subgroups
Group-subgroup pair
G
G > H : G , H, [i], (P, p)
Z1!
Z1
Z1!!
Pairs: group - maximal subgroup
Zk > Zk+1, (P, p)k
Z2!
Z2
H
(P, p) =
!n
k=1(P, p)k
Group-subgroup
Group-subgrouprelations
relations
Applications
Applications
! Possible low-symmetry structures
! Domain structure
! Prediction of new structures
! Symmetry modes
Aim
AIM
G > H, [i]
chains of maximal subgroups
[i]
Hk ∼ H
[i]
classification of Hk ∼ H
Different subgroups of the same type
G = Pmm2 > H = Pm, [i] = 2
S0, G = Pmm2
S1, H = P1m1
S2, H = Pm11
H = Pm, No. 6 en ITA
[i]
Hk ∼ H ⇔ different low-symmetry structures
Different subgroups of the same type
G = P4mm > H = Pm, [i] = 4
S0, G = P4mm
S !, H = Pm11
S !!, H = P1m1
H = Pm, No. 6 en ITA
[i]
Hk ∼ H of the same class ⇔ domain structure
Subgroups calculation: SUBGROUPGRAPH
Chains of maximal subgroups
G > H, [i]
(P, p)m ⇒ Hm
↓
Direct comparison of the Hm
[i]
Hk ∼ H
↓
![i]
G = H + l=2 glH
Conjugation Hk with gl
Comparison of the subgroup elements
Classes
of
[i]
Hk ∼ H
http://www.cryst.ehu.es/subgroupgraph.html
http://www.cryst.ehu.es/subgroupgraph.html
EXERCISES
Problem 4.1.1, p.23
Construct the diagram of the
t-subgroups of P4mm using the ‘analogy’
with the subgroup diagram of 4mm
Problem 4.1.1, p.23
SOLUTION
index
P 4mm
P 4mm
4mm
[1]
mx my 2z
mx
m+ m− 2z
4z
my
2z
m+
m−
[2]
[4]
P 2z m1
Pmm2
P 1mx 1
Pm
1
Subgroup diagram of point
group 4mm
P 4z 11
P4
P 2z 1m
Cmm2
P 1my 1 P 2z 11 P 11m+ P 11m−
Pm
Cm
Cm
P2
[8]
P1
Translationengleiche subgroups of
space group P4mm
Problem 4.1.1, p.23
SOLUTION
Remark 1. Due to the convention to choose the basis vectors parallel to
the rotation axes, C -centered cells appear although the translation lattice
has not changed. If the retained twofold axes are diagonal, the conventional
basis vectors a’,b’,c’ of the subgroup are a’ = a-b, b’ = a+b, c’ = c
with respect to the basis vectors a,b,c of P 4mm. Referred to a’,b’,c’ the
cell is C -centered, see Fig.4.
b
!
!
a
b
a
Change of basis vectors: a’=a-b, b’=a+b
Problem 4.1.1, p.23
P4mm
P=(a-b, a+b, c)
SOLUTION
Problem 4.1.2, p.23
Determine the k-subgroups of Pnma, No.
53 that are obtained by doubling of the
b lattice parameter
Hint: split the cosets of Pnma relative to TG
into cosets with respect to TH
Problem 4.1.2, p.23
SOLUTION
Splitting of the translation subgroup TG
splits
TH = {t(u,v=2n,w)}
TG
TH
TH t b
tb=(0,b,0)
Splitting of the generator cosets
Problem 4.1.2, p.23
k-subgroups for b’=2b
Example: (2)’ (5)’
(2)’’ (5)’
az
nz
SOLUTION
Data on maximal subgroups of space
groups in International Tables for
Crystallography, Vol. A1 (ITA1)
ITA1 maximal subgroup data
Space-group data from International
Tables for Crystallography, Vol. A (ITA)
ITA1 maximal subgroup data
Maximal t-subgroups of P3112 (No. 151)
Remarks
{
braces for
conjugate
subgroups
(P, p): OH = OG + p
(aH,bH,cH)= (aG,bG,cG) P
ITA1 maximal subgroup data
Maximal k-subgroups of R3,
index 2,3 and 4
lattice
relations
generators
general
position
triplets
ITA1 maximal subgroup data
Series of isomorphic subgroups of R3
p
prime number
q, r
positive integers
u, v, w
integers
Relations between Wyckoff positions
General splitting rules
(Wondratschek 1993,1995)
WG
G > H, (P, p)
Ri =
WiH
[i] =
|SG (X )|
|SH(Xi)|
!q
i=1
Ri
Relations
between
Wyckoff
Symmetry
reduction
positions
G = Pmm2 > H = Pm, [i] = 2
S0, G = Pmm2
1
2h
m..
(
2 , y, z)
2h m.. (1/2,y,z)
1
2f .m. (x,→
2 , z)
2f .m. (x,1/2,z)
S1, H = Pm
2c 11 (x,
y, z)
2c
(x,y,z)
1b2 m (x2, 12 , z2)
1b m (x ,1/2,z )
1b1 m (x1,212 , z1) 2
1b m (x ,1/2,z )
1
1
Splitting of Wyckoff positions
SYMMETRY REDUCTION
Wyckoff position splitting
Resrictions on the splitting schemes:
(i) G
H: H a normal subgroup of G
Ri=R in [i]=∑Ri
(ii) G ≥ Z ≥ H:
Splitting G
splitting G
H{
Splitting Z
Z
H
Example: G>H, [i]=4
(i) one orbit: R=4
(ii) two orbits: R1=R2 =2
R1=3,R2 =1
(iii) three orbits: R1=R2 =1
R3=2
(iv) four orbits:
R1=R2 =R3=R4=1
Wyckoff position splitting
General procedure:
Given G, H < G, index [i] and (P,p)
-transform (data)G
(data)H
1. Right-coset decomposition
G = H + Hg2 + ... + Hgk
2. General-position orbit splitting
OG(Xo)=OH(Xo,1)+OH(Xo,2)+ ... +OH(Xo,k)
3. Special-position orbit splitting
(i) substitution of parameters: OH(Xo,j)
OH(Xj)
(ii)assignment of OH(Xj) to the WP of H
Wyckoff position splitting
Example:
G=P42mnm
H=Cmmm
[i]=2, a’=a-b, b’=a+b, c’=c
1. General-orbit splitting
Orbit 1
Orbit 2
coset
representatives
+
t(1/2,1/2,0)
+
t(1/2,1/2,0)
Wyckoff position splitting
2. Special-orbit splitting: 2a 0,0,0
(i) Substitution of parameters
special
general
x,y,z
0,0,0
Orbit 1:
x,y,z
Orbit 2: y,x+1/2,z+1/2
(ii) Assignment
0,0,0
0,1/2,1/2
Splitting:
(2a)P42/mnm
Example:
0,0,0
0,1/2,1/2
(2a)Cmmm
(2c)Cmmm
(2a)Cmmm + (2c)Cmmm
Wyckoff position splitting
Example:
Example WYCKSPLIT: P4 /mnm>Cmmm, index 2
2
Problem 6.1.1, p.37
Consider the group
-subgroup pair P4mm>Cm
[i]=4, a’=a-b, b’=a+b, c’=c
Determine the splitting schemes for WPs 1a,1b, 2c, 4d
group P4mm
subgroup Cm
Problem 6.1.1, p.37
General-position splitting
Special-position splittings
SOLUTION
Splitting of the Wyckoff positions:
P4mm > Cm (by direct inspection)
Problem 6.1.1, p.37
Transformation of coordinates:
P=
1
1
-1
1
P-1 =
1/2
-1/2
1/2
1/2
1
1
Cm
x’
y’
P4mm
=
1/2
-1/2
x
1/2
1/2
y
z’
1
z
Splitting schemes:
1a 4mm (00z)
2a m (x0z)
2c 2mm. (1/20z)
4b 1 (xyz)
Data on Relations between Wyckoff
Positions in International Tables for
Crystallography, Vol. A1
D
14
4h
P 42 m 21 n 2 m
Axes
Coordinates
2a
I
P 42 m n m
No. 136
2b
4g
4c
8h
Wyckoff positions
4d
8i
4e
8j
4f
16k
Maximal translationengleiche subgroups
[2] P4̄n2 (118)
x
1
2
yz
1
4
2d
2c
4f
4e
2 4e
2a; 2b
8i
4h
8i
4g
2 8i
[2] P4̄21 m 113
x
1
2
yz
1
4
2c
2c
4e
4d
2 4d
2a; 2b
8f
2 2c
2 4e
4e
2 8f
[2] P42 nm 102
2a
2a
4c
4b
2 4b
4b
8d
2 2a
2 4c
4c
2 8d
[2] P42 21 2 (94)
2a
2b
4f
4d
2 4d
4d
8g
4c
8g
4e
2 8g
2d
2c
4j
2a; 2b
4g; 4h
2e; 2 f
2 4j
4i
8k
4j
2 8k
2a
2b
4g
2c; 2d
2 4f
4f
2 4g
4e
8h
4g
2 8h
2a; 2c
2b; 2d
4g; 4 j
4e; 4 f
2 8m
8m
8p; 8q
4k; 4l
8n; 8o
4h; 4i
2 16r
[2] P42 m (84)
x
1
2
yz
[2] Pnnm (58)
[2] Cmmm 65
II
a b
a b c
1
2
1
2
x–y
x y z;
1 1
2 2 0
Maximal klassengleiche subgroups
Wyckoff position relations for
maximal t-subgroups
conventional HM
non-conventional HM
ITA1 data
transformed
coordinates
(a,b,c)H in terms
of (a,b,c)G
ITA1 data
Wyckoff position relations for
maximal non-isomorphic k-subgroups
two different subgroups
differ by an origin shift
ITA1 data
Wyckoff position relations for
maximal isomorphic k-subgroups
series of isomorphic
subgroups
parametric description of
the splitting
ADDITIONAL
my
m+
m−
mx
mx
y
x
m+
m−
my
Symmetry group 4mm=C4v of the square