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)

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