Translations are normal subgroup of Space Group: Dresselhaus's proof not convincing In Group Theory: Applications to the physics of condensed matter, eq. 9.15, Dresselhaus gives the following proof that the translation group is a normal subgroup of the space group:
\begin{align*}
\{R_{\alpha}|\tau\}\{I|t\}\{R_{\alpha}|\tau\}^{-1} = \{I|R_{\alpha}t\}
\end{align*}
However, if the space group is non-symmorphic, then it could be the case that $\{R_{\alpha}|\tau\}$ cannot be factored as $\{I|\tau\}\{R_{\alpha}|0\}$, where $\{I|\tau\}$ is an element of the translation group and $\{R_{\alpha}|0\}$ is an element of the space group.
If $\{I|\tau\}$ is not an element of the translation group (i.e. is a non-symmorphic translation vector), there is no problem, but what if $\{R_{\alpha}|0\}$ is not an element of the space group? Then it is not necessarily the case that $\{I|R_{\alpha}t\}$ is an element of the translation group.
So Dresselhaus's proof requires that for all non-symmorphic space group elements $\{R_{\alpha}|\tau\}$, $\{\mathbf{I}|R_{\alpha}\mathfrak{t}\}$ is an element of the translation group for all $\mathfrak{t}$ in the translation group.
Is this true and what is the proof?
 A: I have also put some thought into the question how to show that all primitive
translations are a normal subgroup of a space group $\mathcal{G}$, in particular, of a
non-symmorphic space group which excludes the route via semidirect products.
First, the OP says correctly that the factoring into translational and
rotational ops is not possible for non-symmorphic space groups. However,
Dresselhaus never stated or implicitly used this.
Also Dresselhaus did not say that $\{I|\tau\}$ for $\tau$ being non-primitive is an element of the translation group.
I struggled as the OP with the fact that $\{I|R_{\alpha}t\}$ is in general an
element of the translation group for all possible occuring $R_\alpha$ of non-symmorphic space groups.
However, I could resolve this as following.
One should remind her-/himself in this context about
following two important points which Dresselhaus did not mention explicitly in
this paragraph but which IMHO should have been done:


*

*All space groups are subgroups of the Euclidean group $E$, i.e. group of
isometries (distance-preserving transformations): $|g_i x| = |x| \quad \forall
g_i\in E, x\in \mathbb{R}^n$ where $|\cdot|$ denotes the Euclidean length.
Since any translational part $\tau$ (primitive or non-primitive) does not
change lengths, all matrix parts $R_\alpha$ can also not change lengths. This has the consequence that
$$ |R_\alpha t| = |t|.$$

*Closure and existence of inverses for any mathematical group means the RHS
of eq. (9.15) is necessarily an element of the space group $\mathcal{G}$ under
consideration again:
$$ gh \in \mathcal{G} \quad \forall g,h\in  \mathcal{G}$$
$$ g^{-1}  \in \mathcal{G} \quad \forall g\in  \mathcal{G}$$
$$ \Rightarrow ghg^{-1} = h' \in \mathcal{G}$$
Together with point 1 we should be assured that $\{I|R_{\alpha}t\}$ is an element of  $T$: 
1) rotational part is the identity (see below for full eq. (9.15)),
2) the translational part has the length of a primitive translational vector (distance-preserving ops), and
3) it is an element of the space group (closure).


Therefore, the answer is yes: also for all non-symmorphic space groups
$\{\mathbf{I}|R_{\alpha}t\}$ is an element of
the translation group for all $t$ in the translation group and all $R_\alpha$ of the space group.
I think the recent book by Hergert and Geilhufe (2018) is a little bit better at explaining this (Section 4.2.2, p. 63).
However, I am not happy about other parts of this book.
For others who might not have all the details, I want to add the complete eq. (9.15) which shows that the translational part, denoted by $\tau$, of a general space group
element $\{R_{\alpha}|\tau\}$ cancels out when calculating the
conjugate element for any $\{\mathbf{I}|t\}$. Previous formulas for composition and inverse of Seitz symbols, which are used in eq. (9.15), are
$$ \{\beta|\tau'^{-1}\}\{\alpha|\tau^{-1}\} = \{\beta\alpha|-\beta\tau+\tau'\} \qquad (9.4)$$
$$ \{\alpha|\tau\}^{-1} = \{\alpha^{-1}|-\alpha^{-1}\tau\} \qquad (9.7)$$
The complete eq (9.15) is:
\begin{eqnarray}
 \{R_{\alpha}|\tau\}\{I|t\}\{R_{\alpha}|\tau\}^{-1} &=& \{R_{\alpha}|\tau\}\{I|t\}\{R_{\alpha}^{-1}|-R_\alpha^{-1}\tau\}\\
&=&\{R_{\alpha}|\tau\} \{R_{\alpha}^{-1}|-R_\alpha^{-1}\tau+t\}\\
&=& \{I|-R_\alpha R_\alpha^{-1}\tau+R_\alpha t + \tau\}\\
&=& \{I|R_{\alpha}t\}
\end{eqnarray}
The cancellation occurs from the third to the fourth and final line.
A: One way of proving a group $T $  is normal in group $G$ is by providing a homomorphism $\phi$ on $G$ whose kernel is T i.e $\text{ker}(\phi) = T$. So let $H$ be the point group of certain space group $G$ and let $\phi$ be the homomorphism:
$G \rightarrow H $. 
A general space group element,$g$, has the general action of $ g(x) = E x + v$ on a point x . where $E$ is a matrix from the point group So let $ \phi : g \longmapsto E $
The kernel of this map are the translations. So the translations form a normal subgroup.
A way of seeing that the space group $ G $ is a semi-direct product of translations and the point group is to note the following: we have already proved that the translation group is a normal subgroup. Now all we need to note is that for the translation group $T$ and the point group $H$,  $T\cap H = \{ e\} $. It is a theorem that that is equivalent to saying $ G = T \rtimes H $
