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)


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



Subgroups of the Euclidean group as semidirect products
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 $GL\_n(\mathbb{Z})$.
> 
> 
> 


Explanations: 


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. 


