Consider the Euclidean group $E(n)$ as the semidirect product for Euclidean vector space $\mathbb{E}^n$ with its orthogonal group $O(\mathbb{E}^n)$:

$E(n)=\mathbb{E}^n\rtimes O(\mathbb{E}^n)$

Then the following short exact sequence splits

$1\rightarrow \mathbb{E}^n\rightarrow E(n)\rightarrow O(\mathbb{E}^n)\rightarrow 1$

Now consider a subgroup G of the Euclidean group which translational subgroup T (all isometries in G with trivial linear part) can be identified with a lattice $\mathcal{L}^{n}$ in Euclidean vector space, i.e. all $\mathbb Z$-linear combinations of a chosen basis. The translation subgroup T is normal in G and we can write the short exact sequence

$1\rightarrow T\rightarrow G\rightarrow Q\rightarrow 1$

where quotient group $Q=G/T$. This short exact sequence splits iff $G=T\rtimes Q$. This is the case iff Q is isomorphic to the automorphism of the lattice so that we can write for example the following split short exact sequence

$1\rightarrow \mathcal{L}^{n}\rightarrow G\rightarrow Aut(\mathcal{L}^{n}) \rightarrow 1$

The question is: why is $G=T\rtimes Q$ iff $Q$ is isomorphic to $Aut(\mathcal{L}^{n})$?

FYI: in crystallography, G are called space groups in three dimensions. Space groups which are semidirect products, are called symmorphic.

Some thoughts that might lead to the solution:

1. Since T is normal in G we can write for every isometry $(t_q,q)\in G\quad$ ($t_q$: translational component, q linear component)

$\quad\quad T=(t_q,q).T.(t_q,q)^{-1}$

$\Leftrightarrow T= (t_q,q).\lbrace (t,id)\rbrace.(-q^{-1}.t_q,q^{-1})$

$\Leftrightarrow T=\lbrace (q.t,id)\rbrace=q.T$

So q is a permutation of T. Since T is isomorphic (as a free $\mathbb Z$-module) to $\mathcal{L}^{n}$ and since q is orthogonal ($q\in O(\mathbb{E}^n)$), we find that $q\in Aut(\mathcal{L}^{n})$. This means that the set of all linear parts of G, which we'll call $Q(\mathcal{L}^{n})$ is a subgroup of $Aut(\mathcal{L}^{n})$.

2. Consider a coset $(t_q,q).T\ $ of T, then we can write

$\quad\quad (t_q,q).T=\lbrace (t_q,q).(t,id): t\in \mathcal{L}^{n}\rbrace$

$\Leftrightarrow (t_q,q).T=\lbrace (t_q+q.t,q): t\in \mathcal{L}^{n}\rbrace$

$\Leftrightarrow (t_q,q).T=\lbrace (t_q+t',q): t'\in \mathcal{L}^{n}\rbrace$

which means that each q belongs to exactly one coset of T so that $Q(\mathcal{L}^{n})$ is isomorphic to $Q$.

3. From 1. and 2. we find that in any case (i.e. also when G is not a semidirect product) $Q$ is isomorphic to a finite subgroup of $Aut(\mathcal{L}^{n})$.

4. If we have $G=T\rtimes Q$, there exists a homomorphism from $Q$ to $Aut(T)$ and since T is isomorphic to $\mathcal{L}^{n}$ (as a free $\mathbb Z$-module)

$Hom:Q\rightarrow Aut(\mathcal{L}^{n})$

  • 1
    $\begingroup$ I wonder, if a counterexample couldn't be defined in the following way ? For $a \in \mathbb{R}^n$ let $t_a: \mathbb{R}^n \to \mathbb{R}^n$ be translation by $a$ and let $\sigma: \mathbb{R}^n \to \mathbb{R}^n$ be multiplication by $-1$ (involution). Then, $G := \langle \sigma, t_a \mid a \in \mathbb{Z}^n \rangle$ is a subgroup of $E(n)$ with translational subgroup $T = \langle t_a \mid a \in \mathbb{Z}^n \rangle \cong \mathbb{Z}^n$ and $T \cap \langle \sigma \rangle = \lbrace \operatorname{id}\rbrace$. Thus $G$ is the semi-direct product of $T$ and $\langle \sigma \rangle \cong \mathbb{Z}/2$. $\endgroup$
    – Ralph
    Oct 10, 2011 at 11:59
  • $\begingroup$ Why is this a counterexample? If I'm not mistaken the group you describe has cosets $(0,id)T$ and $(0,\sigma)T$ so $Q=G/T\cong\lbrace id,\sigma\rbrace$. The lattice isomorphic to T is the triclinic lattice. The automorphism group of the triclinic lattice is $\lbrace id,\sigma\rbrace$. $\endgroup$
    – Wox
    Oct 10, 2011 at 12:56
  • $\begingroup$ I think the confusion might originate from $\mathcal{L}^n\cong\mathbb{Z}^n$. The automorphism group of $\mathbb{Z}^n$ is $GL(n,\mathbb{Z})$ isn't it? However in this context, also the inner product of Euclidean vector space must be preserved. For example the automorphism group of the square lattice in $\mathbb R^2$ has order 8 (signed permutations). $\endgroup$
    – Wox
    Oct 10, 2011 at 13:03
  • $\begingroup$ Can you please give a precise mathematical definition of $Aut(\mathcal{L}^{n})$ ? $\endgroup$
    – Ralph
    Oct 10, 2011 at 13:32
  • $\begingroup$ The group of all linear maps of the lattice (module homomorphisms $Hom:\mathcal{L}^{n}\rightarrow\mathcal{L}^{n}$) that are invertible and that preserve the inner product inherited from $\mathbb R^n$. In short: all orthogonal transformations of the lattice. $\endgroup$
    – Wox
    Oct 10, 2011 at 14:10

1 Answer 1


To shed light on some questions:

  1. The statement "$G=T\rtimes Q \Leftrightarrow Q \cong Aut(\mathcal{L}^n)$" is wrong.

  2. The statement "if $Q \cong Aut(\mathcal{L}^n)$ and $T \cong \mathcal{L}^n$, then $G=T\rtimes Q$" is wrong.

  3. The following is true:

Theorem: There is a 1-1 correspondence between symmorphic space groups of dimension $n$ and the conjugacy classes of finite subgroups of $GL_n(\mathbb{Z})$.


In the following I write $L$ instead of $\mathcal{L}^n$ for a lattice and $Aut_O(L) := Aut(L) \cap O_n$ instead of $Aut(\mathcal{L}^n)$, where $O_n$ is the orthogonal group and $Aut(L)$ is the group of automorphisms of $L$ in the group theoretic sense.

1) Choose $L := \mathbb{Z}^n$. Since columns of matrices in $O_n$ have Euklid norm 1, $Aut_O(L) =GL_n(\mathbb{Z})\cap O_n$ consists of the signed permutation matrices (there are $2^n \cdot n!$). For a subgroup $Q \le Aut_O(L)$ let $G_Q := \langle t_a, r_B \mid a \in L, B \in Q \rangle$, where $t_a$ is translation by $a$ and $r_B: \mathbb{R}^n \to \mathbb{R}^n, x \mapsto Bx$. Then $G_Q$ is the semi-direct product of its translational subgroup $T = \langle t_a \mid a \in L \rangle$ and of $\langle r_B \mid B \in Q \rangle \cong Q$. This shows that $(\Rightarrow)$ is false (just take any $Q \lvertneqq Aut_O(L)$).

In order to show that $(\Leftarrow)$ is also false, it suffices to find a counterexample in dimension 2 (I guess there are counterexamples in each dimension, but that would be a - nontrivial - topic for itself). $Aut_O(\mathbb{Z}^2)$ is the dihedral group $D_8$ of order 8 and the existence of such a counterexample follows from $H^2(D_8;\mathbb{Z}^2) = \mathbb{Z}/2$.

(Under the 17 space groups of dimension 2, there are exactly 2 with point group $D_8$ and by looking into the corresponding tables you should be able to figure out the non-split one to get an explicit presentation.)

2) This was just shown by the counterexample. You misread the explication in the Havana paper: The author merely shows the existence of a space group those point group is isomorphic to $Aut_O(L)$ - namely the semi-direct product (but he does not say that, if the point group is isomorphic to $Aut_O(L)$, then the space group is symmorphic!). This is to justify the subsequent definition of Bravais groups (Def. 42).

3) In the following write $L_0 := \mathbb{Z}^n$ and $T_0 = \langle t_a \mid a \in \mathbb{Z}^n \rangle$.

Let $\mathcal{C}_n$ be the set of conjugacy classes of finite subgroups of $GL_n(\mathbb{Z})$ and let $\mathcal{I}_n$ be the set of isomorphism classes of symmorphic space groups. Define a mapping $$f: \mathcal{C}_n \to \mathcal{I}_n, [Q] \mapsto [G_Q]$$ where $[.]$ denotes the corresponding class on either side and $G_Q := L_0 \rtimes Q$, where $Q$ acts on $L_0$ through matrix multiplication.

Step 0: $f$ is well-defined

Let $Q,P \le GL_n(\mathbb{Z})$ be conjugated in $GL_n(\mathbb{Z})$, i.e. there is $A \in GL_n(\mathbb{Z})$ such that $P = AQA^{-1}$. Using $$\alpha: L_0 \to L_0, x \mapsto Ax,$$ $$\beta: Q \to P, B \mapsto ABA^{-1},$$ one immediately sees $\beta(B)\cdot \alpha(x) = ABA^{-1}Ax = \alpha(Bx)$. Consequently $$L_0 \rtimes Q \to L_0 \rtimes P, (x,B) \mapsto (\alpha(x),\beta(B))$$ is an isomorphism. Thus $G_Q = L_0 \rtimes Q \cong L_0 \rtimes P = G_P$.

Next I show that $G_Q$ is isomorphic to a symmorphic space group. By Maschke's theorem $Q$ is conjugate in $GL_n(\mathbb{R})$ to a subgroup $P$ of $O_n$. So there is $A \in GL_n(\mathbb{R})$ such that $P = AQA^{-1}$. Let $L := AL_0 \le \mathbb{R}^n$. Then the isomorphisms $\alpha: L_0 \to L$, $\beta: Q \to P$ (defined by the same formulars as above) yield $G_Q \cong L \rtimes P \cong \langle t_a \mid a \in L \rangle \rtimes \langle r_B \mid B \in P \rangle$ and the latter is the sought-after symmorphic space group.

1. Step: $f$ is injective

Let be $f([Q_1]) = f([Q_2])$ and write $G_i := G_{Q_i}$. Then there is an isomorphism $\phi: G_1 \to G_2$ of groups and we wan't to show that $Q_1, Q_2$ are conjugate in $GL_n(\mathbb{Z})$.

From the lemma in Step 3 it follows that $L_0$ is the unique maximal free abelian rank $n$ subgroup of each $G_Q$. Since maximality is preserved by group isomorphisms, $\phi(L_0) \cong L_0$ is a maximal free abelian subgroup of rank $n$ of $G_2$ and by uniqueness, $\phi(L_0) = L_0$. Now by a general principal in group theory, $\phi$ induces an isomorphism $\bar{\phi}: Q_1 \to Q_2$ such that there is a commutative diagramm:

$$ 1 \to L_0 \to G_1 \overset{\kappa_1}{\to} Q_1 \to 1 $$ $$ \hspace{5pt} \phi \downarrow \hspace{15pt} \phi \downarrow \hspace{17pt} \downarrow\bar{\phi} \hspace{15pt}$$ $$ 1 \to L_0 \to G_2 \underset{\kappa_2}{\to} Q_2 \to 1 $$

$\bar{\phi}$ can be described as follows: to $B \in Q_1$ choose $X \in G_1$ such that $\kappa_1(X) = B$. Then $\bar{\phi}(B) = (\kappa_2 \circ \phi)(X)$. (In the current situation we can always take $X=(0,B)$ ).

Since $\phi \in Aut(L_0) = Aut(\mathbb{Z}^n)$, there is $A \in GL_n(\mathbb{Z})$ such that $\phi(a,I) = (Aa,I)$.

For $B \in Q_1$ write $\phi(0,B) = (b,C)$ with an $C \in Q_2$. Let $e_i$ be a standard vector. From $$(CAe_i,I)= (b,C) \cdot (Ae_i,I) \cdot (b,C)^{-1} = \phi((0,B) \cdot (e_i,I) \cdot (0,B)^{-1})$$ $$ = \phi(Be_i,I)= (ABe_i,I)\hspace{80pt}$$ one finds $CA=AB$, i.e. $C=ABA^{-1}$. This yields $$\bar{\phi}(B) = (\kappa_2\circ \phi)(0,B)= \kappa_2(b,ABA^{-1}) = ABA^{-1},$$ hence $\bar{\phi}$ is conjugation with $A$ and $Q_2= \bar{\phi}(Q_1) = AQ_1A^{-1}$ is conjugated to $Q_1$ in $GL_n(\mathbb{Z})$.

Step 2: $f$ is surjective

Let $G$ be a symmorphic space group. We have to show that there exists a finite subgroup $Q \le GL_n(\mathbb{Z})$ such that $G \cong G_Q$.

Let $T$ be the translational subgroup of $G$ with lattice $L = \oplus_{i=1}^n \mathbb{Z}a_i$. Set $A := (a_1,...,a_n) \in GL_n(\mathbb{R})$. Then $\alpha: L_0 \to L, x \to Ax$ is an isomorphism that induces a further isomorphism $$\phi: GL_n(\mathbb{Z}) = Aut(L_0) \to Aut(L), B \mapsto ABA^{-1}.$$

Now let be $P := \mathcal{Q}(L) \le Aut(L) \cap O_n$ (notation in OP in 1.). In particular, as a discrete subgroup of a compact group, $P$ is finite. Thus $Q := \phi^{-1}(P)$ is a finite subgroup of $GL_n(\mathbb{Z})$ and $\beta: Q \to P, B \mapsto \phi(B) = ABA^{-1}$ is an isomorphism.

Since $G$ is symmorphic, $G = L \rtimes P$ where $P$ acts on $L$ via matrix multiplication. By definition, $G_Q = L_0 \rtimes Q$ where $Q$ also operates via matrix multiplication. Now $G \cong G_Q$ follows exactly as in Step 0.

Step 3: This result was used in Step 1.

Lemma: The translational subgroup is the unique maximal free abelian rank $n$ subgroup of a space group of dimension $n$.

Let $T$ be the translational subgroup of the space group $G$ (dimension $n$) and let $F \le G$ be a free abelian subgroup of rank $n$. From finiteness of $G/T$ follows that $TF/T \cong F/F \cap T$ is finite. Thus $F \cap T$ is free abelian of rank $n$, hence $F \cap T$ has finite index in $T$. Let $t_{a_i} = (a_i,I)$ $(i=1,...,n)$ be basis vectors for $T$. Then we can choose $k \in \mathbb{Z}, k \neq 0$ such that $(t_{a_i})^k = (ka_i,I) \in F$.

Let $(a,A) \in F$. Since $F$ is abelian, $(a,A)$ and $(ka_i,I)$ commute, yielding $(a+Aka_i,A)=(ka_i+a,A)$ and in particular $Aa_i = a_i$. Thus $A=I$, i.e. $(a,A)$ is a translation, which means $(a,A) \in T$. Hence $F \le T$.

Next I show that $T$ is maximal free abelian of rank $n$: Let $F$ be as above and assume $T \le F$. Since $F \le T$ as shown before, $T=F$ follows. Thus $T$ is maximal.

Finally I show that $T$ is the only maximal free abelian subgroup of rank $n$. Let $F$ be another one. Since again $F \le T$, maximality of $F$ implies $F=T.\quad\quad$ q.e.d.

  • $\begingroup$ Great stuff Ralph! The proof for the third statement would be certainly useful. My effort: consider $T=\lbrace (t,0),\forall t\in L\rbrace$ and $Q=G/T=\lbrace (t_q+L,q),q\in Q(L),t_q\in \mathbb R^n\rbrace$ where $Q(L)$ the finite group of all linear parts of isometries in G, with $Q(L)\leq Aut_{O}(L)\leq GL(\mathbb{n,Z})$. For a symmorphic space group $G=L\rtimes Q(L)$ which has $\lbrace(0,q),q\in Q(L)\rbrace$ as subgroup so that $Q=G/T=\lbrace (L,q),q\in Q(L)\rbrace$. So for each $Q(L)$ (finite subgroup of $GL(\mathbb{n,Z})$) there is exactly one symmorphic space group. $\endgroup$
    – Wox
    Oct 12, 2011 at 9:48
  • $\begingroup$ A thought on $Aut_O(L)$ vs $Aut(L)$. If a lattice is defined as a finitely generated free Z-module with positive definite symmetric bilinear form, then I can just use $Aut(L)$ which covers what you note as $Aut_O(L)$, right? $\endgroup$
    – Wox
    Oct 12, 2011 at 9:56
  • $\begingroup$ @1: To my understanding, you have only shown that for a symmorphic space group, $G/T$ is isomorphic to $Q(L)$, but that is true for all space groups. $\endgroup$
    – Ralph
    Oct 13, 2011 at 0:53
  • $\begingroup$ @2: My reason for distinguishing the two groups is to avoid inaccuracies as in the 3rd line of your first comment: In general $Aut_O(L)$ is no subgroup of $GL_n(\mathbb{Z})$ but $Aut_O(L) \le A^{-1}GL_n(\mathbb{Z})A \cap O_n$ (see Step 2). $\endgroup$
    – Ralph
    Oct 13, 2011 at 0:53

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