I reject the premise of the question. :-)
It is true, as Terry suggests, that there is a nice dynamical proof of the classification of finite abelian groups. If $A$ is finite, then for every prime $p$ has a stable kernel $A_p$ and a stable image $A_p^\perp$ in $A$, by definition the limits of the kernel and image of $p^n$ as $n \to \infty$. You can show that this yields a direct sum decomposition of $A$, and you can use linear algebra to classify the dynamics of the action of $p$ on $A_p$. A similar argument appears in Matthew Emerton's proof. As Terry says, this proof is nice because it works for finitely generated torsion modules over any PID. In particular, it establishes Jordan canonical form for finite-dimensional modules over $k[x]$, where $k$ is an algebraically closed field. My objection is that finite abelian groups look easier than finitely generated abelian groups in this question.
The slickest proof of the classification that I know is one that assimilates the ideas of Smith normal form. Ben's question is not entirely fair to Smith normal form, because you do not need finitely many relations. That is, Smith normal form exists for matrices with finitely many columns, not just for finite matrices. This is one of the tricks in the proof that I give next.
Theorem. If $A$ is an abelian group with $n$ generators, then it is a direct sum of at most $n$ cyclic groups.
Proof. By induction on $n$. If $A$ has a presentation with $n$ generators and no relations, then $A$ is free and we are done. Otherwise, define the height of any $n$-generator presentation of $A$ to be the least norm $|x|$ of any non-zero coefficient $x$ that appears in some relation. Choose a presentation with least height, and let $a \in A$ be the generator such that $R = xa + \ldots = 0$ is the pivotal relation. (Pun intended. :-) )
The coefficient $y$ of $a$ in any other relation must be a multiple of $x$, because otherwise if we set $y = qx+r$, we can make a relation with coefficient $r$. By the same argument, we can assume that $a$ does not appear in any other relation.
The coefficient $z$ of another generator $b$ in the relation $R$ must also be a multiple of $x$, because otherwise if we set $z = qx+r$ and replace $a$ with $a' = a+qb$, the coefficient $r$ would appear in $R$. By the same argument, we can assume that the relation $R$ consists only of the equation $xa = 0$, and without ruining the previous property that $a$ does not appear in other relations. Thus $A \cong \mathbb{Z}/x \oplus A'$, and $A'$ has $n-1$ generators. □
Compare the complexity of this argument to the other arguments supplied so far.
Minimizing the norm $|x|$ is a powerful step. With just a little more work, you can show that $x$ divides every coefficient in the presentation, and not just every coefficient in the same row and column. Thus, each modulus $x_k$ that you produce divides the next modulus $x_{k+1}$.
Another way to describe the argument is that Smith normal form is a matrix version of the Euclidean algorithm. If you're happy with the usual Euclidean algorithm, then you should be happy with its matrix form; it's only a bit more complicated.
The proof immediately works for any Euclidean domain; in particular, it also implies the Jordan canonical form theorem. And it only needs minor changes to apply to general PIDs.
To shed light on some questions:
The statement "$G=T\rtimes Q \Leftrightarrow Q \cong Aut(\mathcal{L}^n)$" is wrong.
The statement "if $Q \cong Aut(\mathcal{L}^n)$ and $T \cong \mathcal{L}^n$, then $G=T\rtimes Q$" is wrong.
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.
Best Answer
Edit: I couldn't resist my predilection for generalizations: Using darij grinberg's simplification, the proof below shows:
Let $k$ be a field, $q \in GL_n(k)$ a matrix of finite exponent $m$ with char$(k) \nmid m$ and $M \subseteq k^n$. Futhermore, let $E$ be the eigenspace of $q$ corresponding to the eigenvalue $1$ and let $U \le k^n$ be the space spanned by the columns of $1-q$. Then the following is true for $A := 1+q+\dots + q^{m-1}$:
(Older formulation)
Let $E \le \mathbb{C}^n$ be the eigenspace of $1$ of the matrix $q$ and let $U \le \mathbb{C}^n$ be the space spanned by the columns of $1-q$.
Set $A := 1+q+\dots + q^{m-1}$ and $X:= \lbrace x \in \mathbb{C}^n \mid A\cdot x \in \mathbb{Z}^n \rbrace$ and $L := E \cap \mathbb{Z}^n$.
Then the following holds:
Proof: Assume $\dim E = d$. Then $\dim U = \text{rank}(1-q) = n-d$.
Since each $x \in E$ satisfies $Ax = mx$, $E$ contains eigenvectors from $A$ of the eigenvalue $m$. From $A \cdot (1-q) = 0$ it follows that $U$ consists of eigenvectors of $A$ of the eigenvalue $0$. Hence $E \cap U = 0$ and for dimensional reasons $$\mathbb{C}^n = U \oplus E.$$ Since $q$ has integral entries, it's possible to chosse a basis of $E$ in $\mathbb{Q}^n$ and by multiplying with a suitable integer it's also possible to choose a basis in $\mathbb{Z}^n$. Therefore $L = E \cap \mathbb{Z}^n$ is a lattice of rank $d$. Let $\lbrace e_1, \dots, e_d \rbrace$ be a basis of $L$. Let $x \in X$ and write $$x = u + \sum_i \alpha_i e_i \text{ with } \alpha_i \in \mathbb{C}.$$ Then $Ax = \sum_i m\alpha_i e_i \in \mathbb{Z}^n$ and $q(Ax) = Ax$. It follows $Ax \in E \cap \mathbb{Z}^n = L = \oplus_i \mathbb{Z}e_i$ and therefore $m\alpha_i \in \mathbb{Z}$. This shows $X \subseteq U \oplus (1/m)L$. The converse inclusion is obvious. qed.
Edit: Also note that the image of $A$ is given by $$ Y := \lbrace Ax \mid x \in X \rbrace = L.$$