text:
\section{Introduction}

The maximal finite subgroups of $\textnormal{Sp}_{2n}(\mathbb{Q})$ are maximal finite subgroups of $\textnormal{GL}_m(K)$ for minimal totally complex subfields $K$ with $2n=m \cdot [K:\mathbb{Q}]$. Moreover, the maximal finite subgroups of $\textnormal{Sp}_{2n}(\mathbb{Q})$ are full automorphism groups of Euclidean lattices fixing an additional nondegenerate symplectic form. So the classification yields highly symmetric
symplectic structures on interesting Euclidean lattices.

The (conjugacy classes of) maximal finite subgroups of $\textnormal{GL}_{m}(\mathbb{Q})$ have been classified up to $m=31$ in a series of papers \cite{Cryst4,Plesken_app_repr,FRMG,ERMG,FRMG25}.

The strategy for these classifications is as follows. First, it suffices to classify only the maximal finite (symplectic) matrix groups whose natural representation is irreducible over $\mathbb{Q}$. Then one reduces the problem to the so-called (symplectic) primitive matrix groups. The concept of primitivity is the key ingredient for these classifications, since it has important consequences for normal subgroups. Most notably, the restriction of the natural representation of an irreducible (symplectic) primitive matrix group $G<\textnormal{GL}_{m}(\mathbb{Q})$ onto a normal subgroup $N$ splits into copies of a single irreducible representation of $N$. In particular, each abelian subgroup of $O_p(G)$ must be cyclic. Using a theorem of Philip Hall, this restricts the Fitting subgroup $F(G)$ of $G$ to a finite number of candidates depending only on $m$.
Using the Atlas of Finite Simple Groups or the classification of Hiss and Malle \cite{HissMalle} one gets the candidates for the layer $E(G)$ of $G$. Hence there are only finitely many candidates for the generalized Fitting subgroup $F^*(G)$. Then $G$ can be constructed from $F^*(G)$ using group theory, cohomology and the fact that $G/F^*(G) \le \Out(F^*(G))$.

However, these constructions are quite cumbersome and error-prone. Moreover, they do not use the fact that $G$ is a maximal finite (symplectic) subgroup of $\textnormal{GL}_{m}(\mathbb{Q})$. In many cases, this method can be avoided using algebraic number theory and the fact that $G$ is the full automorphism group of every $G$-invariant lattice.

The main result of this classification is the following
\paragraph*{Theorem}
The number of maximal finite subgroups of $\textnormal{Sp}_{2n}(\mathbb{Q})$ (up to conjugacy in $\textnormal{GL}_{2n}(\mathbb{Q})$) are given by
$$
\begin{array}{c|ccccccccccc}
2n & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18 & 20 & 22 \\ \hline
\# \mbox{ primitive classes} & 2& 5& 2& 21& 3& 23& 3& 63& 6& 26& 4  \\
\# \mbox{ classes (total)} & 2 & 6 & 4 & 28 & 5 & 32 & 5 & 91 & 10 & 36 & 6\\
\end{array}
$$
Representatives of these conjugacy classes are available online in a {\sc MAGMA}\ \cite{Magma} readable format from\\
\url{http://www.math.rwth-aachen.de/~Markus.Kirschmer/symplectic/} . Section \ref{sectab} explains how to access the groups in the database and how to find a representative in the database that is conjugate to a given symplectic irreducible maximal finite (s.i.m.f.) matrix group.
A proof of completeness is given in my thesis \cite{MK}. The methods used are explained here as they might be of independent interest and can be applied to classify symplectic matrix groups also in higher degree.
The existence of the famous Leech lattice makes dimension 24  very interesting,
though a full classification of s.i.m.f.\ subgroups of $\textnormal{Sp}_{24}(\mathbb{Q}) $
 would be quite tedious due to the large number of possible Fitting subgroups.

This paper is organized as follows. Section \ref{secprop} contains some basic definitions for matrix groups. It is shown that it suffices to classify only the maximal finite symplectic matrix groups whose natural representation is irreducible over the rationals. These symplectic irreducible maximal finite (s.i.m.f.) matrix groups can be characterized by automorphism groups of lattices.

Section \ref{secprim} shows that it suffices to classify only the so-called symplectic primitive matrix groups and it contains the general outline of the classification. The definition and some properties of the generalized Bravais groups are also recalled.

Section \ref{secfam} introduces some infinite families of maximal finite symplectic matrix groups. In particular, all maximal finite symplectic subgroups of $\textnormal{GL}_{p-1}(\mathbb{Q})$ and $\textnormal{GL}_{p+1}(\mathbb{Q})$ whose orders are divisible by some prime $p$ are determined.

Section \ref{sec2p} shows that if $p\ge 5$ is prime, then the Fitting subgroup of a s.p.i.m.f.\ subgroup of $\textnormal{GL}_{2p}(\mathbb{Q})$ is cyclic of order $2,4$ or $6$. In particular, these groups can be constructed by the methods of Section \ref{secmet}.

Section \ref{secmet} describes some shortcuts that are used frequently in the classification. It gives two results that can be used to rule out some candidates for normal subgroups of index $2^k$. In this section, the $m$-parameter argument is recalled which can be used to find all s.i.m.f.\ matrix groups that contain a given matrix group whose commuting algebra is a field. Finally, it is explained how to find all s.i.m.f.\ matrix groups that contain a given normal subgroup whose commuting algebra is a quaternion algebra.

\section{Some definitions and basic properties}\label{secprop}

\begin{definition}
Let $G < \textnormal{GL}_m(\mathbb{Q})$.
\begin{enumerate}
\item $\mathcal{F}(G) = \{ F \in \mathbb{Q}^{m \times m} \mid gF\transpose{g} = F \mbox{ for all } g \in G\}$ is called the \emph{form space} of $G$. Further $\mathcal{F}_{sym},\; \mathcal{F}_{>0}$ and $\mathcal{F}_{skew}$ denote the subsets of symmetric, positive definite symmetric and skewsymmetric forms.
\item The group $G$ is called \emph{symplectic} if $\mathcal{F}_{skew}(G)$ contains an invertible element.
\item $\mathcal{Z}(G) = \{ L \subset \mathbb{Q}^{1\times m} \mid L \mbox{ a rank $m$ lattice with $Lg=L$ for all $g\in G$} \}$ is the set of (full) \emph{$G$-invariant lattices}.
\item $\End(G)= \{e \in \mathbb{Q}^{m \times m}\mid eg=ge \mbox{ for all } g \in G\}$ is the commuting algebra (or endomorphism ring) of $G$.
\item If $K$ is a subfield of $\End(G)$ then
$$\Aut_K(L,F) = \{g \in \mathbb{Q}^{m \times m} \mid Lg=L,\; g F \transpose{g} = F \mbox{ and } ge=eg \mbox{ for all } e \in K\}$$
denotes the group of $K$-linear automorphism of $(L, F)$. If $K \simeq \mathbb{Q}$ the subscript $K$ is usually omitted.
\end{enumerate}
\end{definition}

An averaging argument shows that $G < \textnormal{GL}_m(\mathbb{Q})$ is finite if and only if $\Forms_{>0}(G)$ and $\mathcal{Z}(G)$ are both nonempty. Thus, the finite subgroup $G < \textnormal{GL}_m(\mathbb{Q})$ is maximal finite if and only if $G = \Aut(L, F)$ for all $(L, F) \in \mathcal{Z}(G) \times \Forms_{>0}(G)$.

If $F \in \mathcal{F}(G)$ is invertible, then $\End(G) \to \mathcal{F}(G),\; e \mapsto eF$ is an isomorphism of $\mathbb{Q}$-spaces. This can be used to characterize the  maximal finite symplectic matrix groups as automorphism groups of lattices.

Before this result is stated, the next lemma shows that it suffices to consider only rationally irreducible matrix groups, i.e. matrix groups whose natural representation is irreducible over the rationals.

\begin{lemma}\label{irred_OK}
The natural representation $\triangle\colon G \to \textnormal{GL}_{2n}(\mathbb{Q}),\; g \mapsto g$ of a maximal finite symplectic subgroup $G < \textnormal{GL}_{2n}(\mathbb{Q})$ is the direct sum of rationally irreducible and pairwise nonisomorphic representations of $G$.
\end{lemma}
\begin{proof}
Suppose $\triangle$ decomposes into $\oplus_{i=1}^s n_i \triangle_i$ where $\triangle_1, \dots,\triangle_s$ are rationally irreducible and pairwise nonisomorphic.
Since the endomorphism ring and thus the form space of $G$ decomposes into direct sums, it follows that each group $n_i \triangle_i(G)$ is maximal finite symplectic. Hence it suffices to show that $n_i = 1$ for all $i$. Let $E_i = \End(\triangle_i(G))$ be the commuting algebra of $\triangle_i(G)$. If $E_i$ is not a totally real number field then $\triangle_i(G)$ is already symplectic (see Lemma \ref{char_symp}) and thus $n_i \triangle_i(G)$ is contained in the wreath product $ \triangle_i(G) \wr S_{n_i}$. If $E_i$ is a totally real number field then $n_i$ must be even since the form space of $n_i \triangle_i(G)$ contains an invertible skewsymmetric element. But then $n_i \triangle_i(G)$ is conjugate to a subgroup of the tensor product $\triangle_i(G) \otimes H$ for any maximal finite symplectic $H < \textnormal{GL}_{n_i}(\mathbb{Q})$.
\end{proof}

\begin{lemma}\label{char_symp}
Let $G < \textnormal{GL}_{2n}(\mathbb{Q})$ be finite and rationally irreducible. The following statements are equivalent:
\begin{enumerate}
\item $G$ is symplectic.
\item $\End(G)$ contains some (minimal) totally complex subfield.
\item There exists some (minimal) totally complex field $K$ of degree $d$ and some representation $\triangle_K \colon G \to \textnormal{GL}_{\frac{2n}{d}}(K)$ such that the natural character of $G$ is the trace character corresponding to $\triangle_K$.
\end{enumerate}
\end{lemma}
\begin{proof}
Let $F \in \Forms_{>0}(G)$. The space $\mathcal{F}(G)$ is closed under taking transposes, hence it decomposes into $\mathcal{F}_{sym}(G) \oplus \mathcal{F}_{skew}(G)$. Since
$E:= \End(G)$ is a skewfield it follows from $\mathcal{F}(G)=E\cdot F$ that $G$ is symplectic if and only if $\mathcal{F}(G) \ne \mathcal{F}_{sym}(G)$ which is equivalent to say that $\mathcal{F}_{skew}(G)\ne \{0\}$.

If $e\in E$ such that $eF$ is symmetric (skewsymmetric) then $e$ is a selfadjoint (antiselfadjoint) endomorphism of the Euclidean space $(\mathbb{R}^n, F)$. Hence $\mathbb{Q}[e]\le E$ is totally real (complex). Thus the first two statements are equivalent. The equivalence of the last two statements follows from ordinary representation theory.
\end{proof}

The symplectic irreducible maximal finite (s.i.m.f.) matrix groups can now be characterized as follows.
\begin{corollary}
A finite rationally irreducible symplectic matrix group $G < \textnormal{GL}_{2n}(\mathbb{Q})$ is s.i.m.f.\ if and only if $G=\Aut_K(L, F) $ for all $(L, F) \in \mathcal{Z}(G) \times \Forms_{>0}(G)$ and all minimal totally complex subfields of $\End(G)$.
\end{corollary}

It follows for example from \cite[Theorems 3.7 and 3.25]{Artin} that $\mathbb{Q}^{m \times m}$ contains an invertible skewsymmetric matrix $A$ if and only if $m$ is even. Moreover, if $m=2n$, then there exists some $T \in \textnormal{GL}_{2n}(\mathbb{Q})$  such that $T A \transpose{T}=J_n$ where $J_n:= \left( \mbox{\small $\begin{array}{rr} 0 & I_n\\ -I_n & 0  \end{array}$} \right)$. Thus each $\textnormal{GL}_{2n}(\mathbb{Q})$-conjugacy class of symplectic subgroups of $\textnormal{GL}_{2n}(\mathbb{Q})$ contains a representative in
$$\textnormal{Sp}_{2n}(\mathbb{Q}) := \{g \in \textnormal{GL}_{2n}(\mathbb{Q}) \mid g J_n \transpose{g} = J_n\}\:.$$

If $G < \textnormal{Sp}_{2n}(\mathbb{Q})$, one might ask how $\{G^x \mid x \in \textnormal{GL}_{2n}(\mathbb{Q}) \mbox{ and } G^x < \textnormal{Sp}_{2n}(\mathbb{Q})\}$ decomposes into $\textnormal{Sp}_{2n}(\mathbb{Q})$-conjugacy classes. The following example shows that there are infinitely many such classes.

\begin{example}
Let $g = \mbox{\small $\left(\begin{array}{@{}r@{\;\;}r@{}} 0 & 1 \\ -1 & 0 \end{array}\right)$}$ and $G = \left\langle g \right\rangle < \textnormal{Sp}_{2}(\mathbb{Q})$. For $a \in \mathbb{Q}$ set $h_a = \Diag(1, a)$. Then $G^{h_a} < \textnormal{Sp}_2(\mathbb{Q})$ for all such $a$. Moreover, $G^{h_a}$ is conjugate to $G^{h_b}$ in $\textnormal{Sp}_2(\mathbb{Q})$ if and only if $\abs{ab} \in \Nr_{\mathbb{Q}(i)/\mathbb{Q}}(\mathbb{Q}(i)^*)$.
\end{example}
\begin{proof}
Let $t:= \mbox{\small $\left(\begin{array}{@{}r@{\;\;}r@{}} 0 & 1 \\ 1 & 0 \end{array}\right)$}$. Then $G^{h_a}$ is conjugate to $G^{h_b}$ in $\textnormal{Sp}_2(\mathbb{Q})$ if and only if there exists some $x \in \textnormal{Sp}_{2n}(\mathbb{Q})$ such that $g^{h_a} = \pm g^{h_b x} = g^{t^k h_b x}$ with $k \in \{0,1\}$. This is equivalent to say that $\alpha:= t^k h_b x h_{a^{-1}}$ is contained in the commuting algebra of $G$, which is isomorphic to $\mathbb{Q}(i)$. Taking determinants yields $(-1)^k \cdot b/a = \Nr_{\mathbb{Q}(i)/\mathbb{Q}}(\alpha) > 0$ as claimed.
\end{proof}

\section{Primitivity}\label{secprim}

Lemma \ref{irred_OK} shows that it suffices to classify only the (conjugacy classes of) rationally irreducible maximal finite symplectic matrix groups.
This set can be restricted further as follows.

\begin{definition}
Let $K$ be a field. An irreducible matrix group $G < \textnormal{GL}_m(K)$ is called primitive if it is not conjugate to some subgroup $H \wr S_k$ for some $H < \textnormal{GL}_d(K)$ with $kd=m$.

Similarly, an irreducible symplectic subgroup $G < \textnormal{GL}_{2n}(\mathbb{Q})$ is called symplectic primitive, if it is not conjugate to a subgroup of $H \wr S_k$ for some $H < \textnormal{Sp}_{2d}(\mathbb{Q})$ with $kd= n$.
\end{definition}

Clearly, the conjugacy classes of symplectic imprimitive matrix groups can be constructed from the classifications of smaller dimensions. Furthermore, an irreducible symplectic matrix group $H$ is imprimitive if and only if there exists some invariant lattice $L \in \mathcal{Z}(H)$ that admits a nontrivial decomposition $L = \perp_i L_i$ where the components $L_i$ are pairwise perpendicular with respect to the full form space $\mathcal{F}(H)$. Thus, symplectic imprimitive groups can be recognized easily.

The concept of primitivity has some important consequences for normal subgroups.

\begin{remark}
Let $G < \textnormal{GL}_m(K)$ be primitive and $N \unlhd G$. Then $G$ acts on $N$ by conjugation. Hence it also acts on the set of central primitive idempotents of $\left \langle N\right\rangle_K$. Thus $G$ permutes the homogeneous components of the natural $KN$-module $K^{1\times m}$. But since $G$ is primitive, there can only be one such component. Hence the natural $KN$-module $K^{1\times m}$ splits into a direct sum of $k$ isomorphic $KN$-modules of dimension $\frac{m}{k}$.
\end{remark}

\begin{lemma}\label{mainlemma}
Let $G < \textnormal{GL}_{2n}(\mathbb{Q})$ be irreducible and symplectic primitive.
\begin{enumerate}
\item If $N \unlhd G$ then the natural character of $N$ is a multiple of a single rationally irreducible character.
\item If $O_p(G) \ne 1$ then $p-1$ divides $2n$.
\item Every abelian characteristic subgroup of $O_p(G)$ is cyclic.
\end{enumerate}
\end{lemma}
\begin{proof}
The last two statements are immediate consequences of the first one.

Let $K< \End(G)$ be a minimal totally complex subfield of degree $d$. Let $\{f_1, \dots , f_r\}$ and $\{e_1,\dots,e_s\}$ be the central primitive idempotents of the enveloping algebras $\left\langle N \right\rangle_\mathbb{Q}$ and $\left\langle G \right \rangle_{K}$ respectively. Let $\chi_i$ denote the character corresponding to a simple $\left\langle G \right \rangle_{K}e_i$ module and let $L= \mathbb{Q}(\chi_1,\dots,\chi_s) \subseteq K$ be their character field. Then $L/\mathbb{Q}$ is Galois and $e_i = \frac{\chi_i(1)}{\abs{G}} \sum_{g \in G} \chi_i(g^{-1}) g \in \left\langle G \right \rangle_{L}$. Since $G$ is irreducible, $\{e_1,\dots, e_s\}$ is a Galois orbit under $\Gal(L/\mathbb{Q})$. For any $1 \le j \le r$ there exists some $i$ such that $e_if_j \ne 0$. Since $f_j \in \left\langle N\right\rangle_\mathbb{Q}$ is fixed under $\Gal(L/\mathbb{Q})$, it follows that $e_if_j \ne 0$ for all $i,j$.
Denote by $\triangle_K\colon G \to \textnormal{GL}_{\frac{2n}{d}}(K)$ an irreducible representation of $G$ as in Lemma \ref{char_symp}. The enveloping algebra $\left\langle \triangle_K(G) \right \rangle_{K}$ is isomorphic to $\left\langle G \right \rangle_{K} e_i$ for some $i$. Now  $\{e_if_1, \dots, e_if_r\}$ is a set of central idempotents of $\left\langle N \right \rangle_{K} e_i \simeq \left\langle \triangle_K(N) \right \rangle_{K}$. But since $G$ is symplectic primitive, $\triangle_K(G)$ is primitive as well. Therefore $\left\langle\triangle_K(N) \right\rangle_K$ is a simple algebra by the remark above. This shows $r=1$, since no $e_if_j$ vanishes.
\end{proof}

In particular, the candidates for $O_p(G)$ are well known by a theorem of Ph. Hall:

\begin{theorem}[Hall, \protect{\cite[Satz 13.10]{Huppert1}}]\label{Hall}
Suppose $P$ is a $p$-group such that every abelian characteristic subgroup of $P$ is cyclic. Then $P$ is the central product of two subgroups $E_1$ and $E_2$ where either
\begin{enumerate}
\item $p\ne 2$ and $E_1$ is extraspecial of exponent $p$ and $E_2$ is cyclic.
\item $E_1 \simeq 2^{1+2m}_+$ and $E_2$ is cyclic, dihedral, quasi-dihedral or a generalized quaternion $2$-group.
\item $E_1 = 2^{1+2m}_-$ and $E_2 = 1$.
\end{enumerate}
\end{theorem}

The outline of the classification of all conjugacy classes of s.i.m.f.\ subgroups of $\textnormal{Sp}_{2n}(\mathbb{Q})$ is now as follows:
\begin{enumerate}
\item The symplectic imprimitive matrix groups come from the classifications of $\textnormal{Sp}_{2d}(\mathbb{Q})$ where $d$ runs through all divisors of $n$. The wreath products $H \wr S_k$ for some s.p.i.m.f.\ $H < \textnormal{Sp}_{2d}(\mathbb{Q})$ with $dk=n$ are most often maximal finite. (The only exception up to dimension $2n=22$ is $C_4 \wr S_2 < \textnormal{Sp}_{4}(\mathbb{Q})$).

Suppose now $G < \textnormal{Sp}_{2n}(\mathbb{Q})$ is s.p.i.m.f.{}.
\item There are only finitely many candidates for the Fitting subgroup $F(G)$ according to Hall's classification.
\item There are only finitely many candidates for the layer $E(G)$ (the central product of all subnormal quasisimple subgroups of $G$). These are described in \cite{HissMalle} which is based on the ATLAS \cite{atlas}. (Note that this step depends on the completeness of the classification of all finite simple groups. For small dimensions, one can also use the results of Blichfeldt, Brauer, Lindsey, Wales and Feit (see for example \cite{Feit2}) who classified the finite (quasiprimitive) subgroups of $\textnormal{GL}_m(\mathbb{C})$ for $m \le 10$).
\item So there are only finitely many candidates for the generalized Fitting subgroup $F^*(G)$, i.e. the central product of $E(G)$ and $F(G)$. Since $G/F^*(G)$ is isomorphic to a subgroup of $\Out(F^*(G))$ it remains to construct all possible extensions of $F^*(G)$
 up to conjugacy in $\textnormal{GL}_{2n}(\mathbb{Q})$.
\end{enumerate}
The last step is the crucial one. Although it is a cohomological task to find all abstract extensions $G$, this does not yield all matrix group extension of $F^*(G)$. A first step is to replace $F^*(G)$ by its generalized Bravais group as described below.

\subsection{Generalized Bravais groups}

Let $G< \textnormal{Sp}_{2n}(\mathbb{Q})$ or $G < \textnormal{GL}_m(\mathbb{Q})$ be maximal finite. If $N \unlhd G$ then clearly, $N \unlhd \mathcal{B}^o(N) :=\{g \in G \mid g \text{ centralizes } \End(N)\} \unlhd G$.
The so-called generalized Bravais group $\mathcal{B}^o(N)$ can be constructed  independently of the group $G$ as described below. This was first introduced in \cite[page 82]{FRMG}.
This gives a general construction how to replace candidates for normal subgroups of $G$ by (hopefully) larger normal subgroups.

First, one applies the so called radical idealizer process to the $\mathbb{Z}$-order $\Lambda_0 := \left \langle N \right \rangle_\mathbb{Z}$ in the enveloping algebra $ \left \langle N \right \rangle_\mathbb{Q}$ of $N$. I.e. one iterates the following steps
\begin{itemize}
\item Let $I_i$ be the arithmetical radical of $\Lambda_i$, that is the intersection of all maximal right ideals of $\Lambda_i$ that contain the (reduced) discriminant of $\Lambda_i$.
\item Let $\Lambda_{i+1}$ be the right idealizer of $I_i$ in $ \left \langle N \right \rangle_\mathbb{Q}$.
\end{itemize}
This process stabilizes in a necessarily hereditary order $\Lambda_\infty$ see \cite[Theorems 39.11, 39.14 and 40.5]{Reiner}.

The natural $\mathbb{Q} N$-module $\mathbb{Q}^{1\times 2n}$ is isomorphic to a multiple of an irreducible $\mathbb{Q} N$-module $V$ according to Lemma \ref{mainlemma}.  Let $L_1,\dots,L_s$ represent the isomorphism classes of $\Lambda_\infty$-lattices in $V$ and fix some $F \in \Forms_{>0}(N)$. Then
$$\mathcal{B}^o(N) = \{ g \in \left \langle N \right \rangle_\mathbb{Q} \mid L_ig = L_i \mbox{ for all $1\le i \le s$ and } g F \transpose{g} = F \}\:.$$

The group $\mathcal{B}^o(N)$ is finite, since it fixes a full lattice as well as a positive definite form.
By the double centralizer property $\mathcal{B}^o(N)$ and $N$ have the same commuting algebras and thus the same form spaces. In particular, the definition of $\mathcal{B}^o(N)$ does not depend on the choice of the form $F$.
Moreover, $\mathcal{B}^o(N)$ has the following properties:

\begin{lemma}[\protect{\cite[Proposition II.10]{FRMG}}]\label{lemmabravais}
If $G$ is a s.p.i.m.f.\ subgroup of $\textnormal{Sp}_{2n}(\mathbb{Q})$ and $N \unlhd G$ then
\begin{enumerate}
\item $N \unlhd \mathcal{B}^o(N) \unlhd G$
\item If $X$ is a finite subgroup of $\left \langle N \right \rangle_\mathbb{Q}^*$ such that $N \unlhd X$ then $X \le \mathcal{B}^o(N)$.
\end{enumerate}
\end{lemma}
By the double centralizer property, it follows that if $G$ is s.p.i.m.f.\ then $\mathcal{B}^o(N) = \{g \in G \mid\text{ centralizes } \End(N) \}$ as claimed in the beginning of this section.

The candidates of $O_p(G)$ for some irreducible symplectic primitive $G< \textnormal{GL}_{2n}(\mathbb{Q})$ are well known by Theorem \ref{Hall}. Their generalized Bravais groups have been determined generically as follows.
\begin{prop}\cite[Chapter 8]{FQMG}
The candidates for $N = O_p(G)$ and their generalized Bravais groups for an irreducible symplectic primitive matrix groups $G$ are given by
$$
\begin{array}{|l|c|c|c|} \hline
N                                       & \mathcal{B}^o(N)                 & \dim_\mathbb{Q}(\left \langle N\right\rangle_\mathbb{Q}) & \End(N)    \\ \hline
C_{p^m}                                 & \pm N                 & p^{m-1}(p-1)          & \mathbb{Q}(\zeta_{p^m})       \\ \hline
p_+^{1+2n},\quad (p>2)                  & \pm N.\textnormal{Sp}_{2n}(p)     & p^n(p-1)              & \mathbb{Q}(\zeta_p)           \\ \hline
2_+^{1+2n}                              & N.O^+_{2n}(2)         & 2^n                   & \mathbb{Q}                    \\ \hline
2_-^{1+2n}                              & N.O^-_{2n}(2)         & 2^{n+1}               & Q_{\infty, 2}       \\ \hline
p_+^{1+2n}\Ydown C_{p^m}, \quad (m>1)       & \pm N.\textnormal{Sp}_{2n}(p)     & p^{m+n-1}(p-1)        & \mathbb{Q}(\zeta_p)           \\ \hline
2_+^{1+2n}\Ydown D_{2^m}, \quad (m>3)       & N.\textnormal{Sp}_{2n}(2)         & 2^{n+m-2}             & \mathbb{Q}(\theta_{2^{m-1}})  \\ \hline
2_+^{1+2n}\Ydown Q_{2^m}, \quad (m>3)       & N.\textnormal{Sp}_{2n}(2)         & 2^{n+m-1}             & Q_{\theta_{2^{m-1}}, \infty}  \\ \hline
2_+^{1+2n}\Ydown QD_{2^m}, \quad (m>3)      & N.\textnormal{Sp}_{2n}(2)         & 2^{n+m-2}             & \mathbb{Q}(\zeta_{2^{m-1}} - \zeta_{2^{m-1}}^{-1})\\ \hline
\end{array}
$$
where $\theta_k = \zeta_k + \zeta_k^{-1}$ generates the maximal totally real subfield of the cyclotomic field $\mathbb{Q}(\zeta_k)$ and $Q_{\theta_{2^{m-1}}, \infty} \simeq Q_{\infty,2} \tensordown{\mathbb{Q}} \mathbb{Q}(\theta_{2^{m-1}})$ denotes the quaternion algebra with center $\mathbb{Q}(\theta_{2^{m-1}})$ ramified only at the infinite places.

Note however that $\mathcal{B}^o(N)$ is only correct under the assumption that $N\unlhd G$.
\end{prop}

The following example shows how the generalized Bravais group can be used to eliminate possible candidates for normal subgroups.
\begin{example}
Let $q \equiv \pm 3 \mod 8$ be a prime power. Suppose $N < \textnormal{GL}_{q}(\mathbb{Q})$ is isomorphic to $\LGruppe_2(q)$ such that the natural character of $N$ is the Steinberg character. Then $\mathcal{B}^o(N)$ is conjugate to the automorphism group $\Aut(A_{q}) \simeq S_{q+1}$ of the root lattice $A_q$. In particular, there exists no s.p.i.m.f.\ $G < \textnormal{GL}_{kq}(\mathbb{Q})$ that contains a normal subgroup isomorphic to $\LGruppe_2(q)$ such that its natural character is $k$-times the Steinberg character.
\end{example}
\begin{proof}
The natural representation of $N$ is the $q$-dimensional summand of the permutation module of $\LGruppe_2(q)$ on the projectice space $\mathbb{P}(\mathbb{F}_q)$. Hence one can assume that $N$ fixes the root lattice $L:= A_q$.

Since $q\ne 7,8$ the group $\Aut(A_q)$ is a maximal finite subgroup of $\textnormal{GL}_q(\mathbb{Q})$ (see \cite{Burnside}). So it suffices to check that $N$ and $\Aut(A_q)$ have the same invariant lattices. Let $\ell$ be a prime divisor of $\abs{N} = \frac{1}{2}(q-1)q(q+1)$.
If $\ell$ divides $q(q-1)$ then the decomposition matrix of $\LGruppe_2(q)$ \cite{PSL} shows that $L/pL$ is an irreducible $\mathbb{F}_\ell \LGruppe_2(q)$-module.

Now $S_{q+1}$ and $\LGruppe_2(q)$ both act 2-transitively on $\mathbb{P}(\mathbb{F}_q)$. If $\ell\ne 2$ divides $q+1$ then the corresponding $\mathbb{F}_\ell$-modular representations have $2$ composition factors (again from \cite{PSL}). Hence it follows from \cite[Theorem 5.1]{NTA} that $N$ and $\Aut(A_q)$ have the same invariant lattices. Thus $\mathcal{B}^o(N) = \Aut(A_q)$ as claimed. The second statement follows immediately from the definition of generalized Bravais groups and Lemma \ref{lemmabravais}.
\end{proof}

\section{Some infinite families}\label{secfam}

\subsection{Some subgroups of $\textnormal{Sp}_{p-1}(\mathbb{Q})$}

Let $p\ge 5$ be a prime and write $p-1 = 2^a \cdot o$ with $o$ odd. In the spirit of \cite[Chapter V]{FRMG} this section describes all s.i.m.f.\ supergroups $G$ of $C_p$ in dimension $p-1$ where $C_p$ denotes the (up to conjugacy) unique cyclic matrix group of order $p$ in $\textnormal{GL}_{p-1}(\mathbb{Q})$.

Clearly, one possibility is that $G$ contains a normal subgroup conjugate to $C_p$. Since the commuting algebra of $C_p$ is isomorphic to $\mathbb{Q}(\zeta_p)$ it follows that $G/\pm C_p \le C_{2^a} \times C_o$. By Galois theory, $G$ is symplectic if and only if $G/\pm C_p \le C_{o}$ and therefore $G \simeq \pm C_p \! : \! C_o$ by maximality. (The group $\pm C_p \! : \! C_o$ has only one irreducible rational representation of degree $p-1$ and therefore it can be identified with this representation.)

Another class of candidates are extensions of $\LGruppe_2(p)$. The smallest faithful irreducible complex representations of $\LGruppe_2(p)$ are of degree $\frac{p-1}{2}$ and algebraically conjugate. The corresponding character field is $\mathbb{Q}(\sqrt{\pm p})$ with the $-$~sign if and only if $p \equiv -1\mod 4$ (see \cite{SchurPSL}).

If $p \equiv -1 \mod 4$ then $\LGruppe_2(p)$ contains a subgroup $U$ isomorphic to $C_p \! : \! C_{\frac{p-1}{2}}$. The restriction of the natural representation of $\LGruppe_2(p)$ on $U$ is irreducible and has the same character field $\mathbb{Q}(\sqrt{-p})$ (\cite{SchurPSL}). By \cite[Satz 1.2.1]{Lorenz}, the Schur index of $\LGruppe_2(p)$ is equal to the Schur index of $U$ which is $1$. Thus $C_2 \times \LGruppe_2(p)$ has a unique $p-1$~dimensional rationally irreducible representation (denoted by ${}_{\sqrt{-p}}[\pm \LGruppe_2(p)]_{\frac{p-1}{2}}$ in the sequel) with commuting algebra $\mathbb{Q}(\sqrt{-p})$.

The next result shows that there are no further possibilities. More precisely:

\begin{theorem}\label{pminus1}
Let $p\ge 5$ be prime and $G < \textnormal{Sp}_{p-1}(\mathbb{Q})$ such that $p$ divides the order $\abs{G}$.  Write $p-1=2^a\cdot o$ with $o$ odd.\\
Then $G$ is s.i.m.f.\ if and only if $G$ is conjugate to
$
\begin{cases}
\pm C_p \! : \! C_{o} & \text{ if } p \equiv +1\mod 4\\
{}_{\sqrt{-p}}[\pm \LGruppe_2(p)]_{\frac{p-1}{2}} & \text{ if } p \equiv -1 \mod 4
\end{cases}
$.
\end{theorem}

To state the proof, an upper bound on $\abs{G}$ is needed.
\begin{lemma}[Minkowski's bound, \cite{Minkowski}]\label{Minkowski}
The least common multiple of the orders of all finite subgroups of $\textnormal{GL}_n(\mathbb{Q})$ is given by
$$\prod_{p} p^{ \left\lfloor \frac{n}{(p-1)}\right\rfloor + \left\lfloor \frac{n}{p(p-1)}\right\rfloor + \left\lfloor \frac{n}{p^2 (p-1)}\right\rfloor + \dots}$$
where the product is taken over all primes $p \le n+1$.
\end{lemma}

\begin{proof}[Proof of Theorem \ref{pminus1}]
The cases $p=5$ and $7$ can be checked explicitly using the 2-parameter argument (see Corollary \ref{mparam}). Let $p \ge 11$ be prime.
Let $\pm I_{p-1} < G < \textnormal{Sp}_{p-1}(\mathbb{Q})$ such that $p$ divides $\abs{G}$. Further let $P \in \Syl_p(G)$. Then by Minkowski's bound, $\abs{P} = p$.
So the commuting algebra of $P$ is isomorphic to $\mathbb{Q}(\zeta_p)$. Thus $\pm I_{p-1} \le Z(G) \le C_G(P) = \pm P$. Since $G$ must have a faithful irreducible complex character of degree $\le \frac{p-1}{2}$ it follows from a theorem of H. Blau \cite[VIII Theorem 7.2]{Feit} that either $P\unlhd G$ or $G/Z(G) \simeq \LGruppe_2(p)$. In the first case, $G$ is conjugate to a subgroup of $\pm C_{p} \! : \! C_o$. In the second case, $Z(G) = \pm P$ would imply that $G=\pm P$ since $\pm P$ is self-centralizing in $G$. So $Z(G) = \pm I_{p-1}$. Since $\LGruppe_2(p)$ is perfect, $G$ is either isomorphic to $\pm \LGruppe_2(p)$ or $\SL_2(p)$. By the generic character table of $\LGruppe_2(p)$ (see \cite{SchurPSL}) this implies $p \equiv -1 \mod 4$. But then the real Schur indices of the $\frac{p-1}{2}$ dimensional complex characters of $\SL_2(p)$ are $2$. Thus $G$ is conjugate to ${}_{\sqrt{-p}}[\pm \LGruppe_2(p)]_{\frac{p-1}{2}}$. Since ${}_{\sqrt{-p}}[\pm \LGruppe_2(p)]_{\frac{p-1}{2}}$ contains a subgroup conjugate to $\pm C_p \! : \! C_o$ the result follows.
\end{proof}

\subsection{Some subgroups of $\textnormal{Sp}_{p+1}(\mathbb{Q})$}

Let $p \ge 5$ be a prime. If $p \equiv -1\mod 4$ then $G:= \SL_2(p)$ has only two algebraically conjugate complex representations of degree $\frac{p+1}{2}$ as the generic character table \cite{SchurPSL} shows. Let $\chi$ denote one of the corresponding characters and let $P \in \Syl_p(G)$. An explicit calculation shows $(1_P^G, \chi)_G = (1_P, \chi\vert_P)_P =1$. Thus (by \cite[Corollary 10.2(c)]{Isaacs}) $\chi$ is realizable over its character field, which is $\mathbb{Q}(\sqrt{-p})$. So $\chi$ gives rise to a subgroup of $\textnormal{Sp}_{p+1}(\mathbb{Q})$ denoted by ${}_{\sqrt{-p}}[\SL_2(p)]_\frac{p+1}{2}$.

\begin{theorem}\label{pplus1}
Let $p\ge 11$ be prime and $G < \textnormal{Sp}_{p+1}(\mathbb{Q})$ such that $p$ divides $\abs{G}$.
Then $G$ is s.i.m.f.\ if and only if $p \equiv -1\mod 4$ and $G$ is conjugate to ${}_{\sqrt{-p}}[\SL_2(p)]_\frac{p+1}{2}$.
\end{theorem}
\begin{proof}
Let $P \in \Syl_p(G)$. Then again, by Minkowski's bound, $\abs{P}=p$. Thus the natural representation of $P$ splits into twice the trivial one and the $\mathbb{Q}$-irreducible of degree $p-1$. Since there exists an embedding $\delta \colon G \to \textnormal{GL}_m(K)$ for some totally complex number field $K$ with $m\cdot [K:\mathbb{Q}] = p+1$ there is only the possibility $K=\mathbb{Q}(\sqrt{-p})$ and $m=\frac{p+1}{2}$. So the commuting algebra of $\delta(P)$ in $K^{m\times m}$ is isomorphic to $\mathbb{Q}(\zeta_p) \times \mathbb{Q}(\sqrt{-p})$. In particular, $Z(G)$ is isomorphic to a subgroup of $C_{2p} \times C_2$. By Lemma \ref{mainlemma} it follows that $P \ntrianglelefteq G$ and therefore $Z(G) \le C_2 \times C_2$. Since $G$ cannot be imprimitive by Minkowski's bound, $Z(G)$ is not isomorphic to $C_2\times C_2$. This shows $Z(G) = \pm I_{p+1}$. From Blau's theorem \cite[VIII Theorem 7.2]{Feit} it follows that $G$ must be isomorphic to $\pm \LGruppe_2(p)$ or $\SL_2(p)$. Since $K \simeq \mathbb{Q}(\sqrt{-p})$ it follows from the character table of $\SL_2(p)$ \cite{SchurPSL} that $p \equiv -1\mod 4$ and that $\pm \LGruppe_2(p)$ has no faithful complex character of degree $\frac{p+1}{2}$. So $G$ must be conjugate to ${}_{\sqrt{-p}}[\SL_2(p)]_\frac{p+1}{2}$ and the result follows.
\end{proof}

The classification of all conjugacy classes of s.i.m.f.\ subgroups of $\textnormal{Sp}_k(\mathbb{Q})$ for $k \in \{6, 8\}$ show that the above result also holds for $p=5$. But  the unique s.i.m.f.\ subgroup of $\textnormal{Sp}_8(\mathbb{Q})$ whose order is divisible by $7$ is ${}_{\sqrt{-7}}[2.\Alt_7]_4$ (which contains a subgroup conjugate to ${}_{\sqrt{-7}}[\SL_2(7)]_4$).

\subsection{The group $QD_{2^n}$}

Let $n\ge 4$. The group $QD_{2^n} = \left\langle x, y \mid x^{2^{n-1}}, y^2, x^y = x^{2^{n-2}-1}\right\rangle $ has one rationally irreducible representation of degree $2^{n-2}$. This representation has $\mathbb{Q}(\zeta_{2^{n-1}} - \zeta_{2^{n-1}}^{-1})$ as commuting algebra and is denoted by ${}_{\zeta_{2^{n-1}} - \zeta_{2^{n-1}}^{-1}} [QD_{2^n}]_2$.

\begin{prop}\label{QuasiDihedral}
If $n\ge 5$ then ${}_{\zeta_{2^{n-1}} - \zeta_{2^{n-1}}^{-1}} [QD_{2^n}]_2$ is a s.i.m.f.\ subgroup of $\textnormal{Sp}_{2^{n-2}}(\mathbb{Q})$.
\end{prop}
\begin{proof}
The commuting algebra $K$ of $H:= {}_{\zeta_{2^{n-1}} - \zeta_{2^{n-1}}^{-1}} [QD_{2^n}]_2$ is isomorphic to $\mathbb{Q}(\zeta_{2^{n-1}} - \zeta_{2^{n-1}}^{-1})$. This is the fixed field of the automorphism of $\mathbb{Q}(\zeta_{2^{n-1}})$ induced by $\zeta_{2^{n-1}}\mapsto -\zeta_{2^{n-1}}^{-1} = \zeta_{2^{n-1}}^{2^{n-2}-1}$. The subfields of $K$ are linearly ordered by Galois theory and the maximal subfield is the fixed-field of complex conjugation i.e. totally real. Thus each s.i.m.f.\ supergroup $G$ of $H$ embeds into $\textnormal{GL}_2(K)$. Therefore $\bar{G}:= G / Z(G) = G/\left\langle \pm I_{2^{n-2}}\right\rangle $ embeds into $\PGL_2(K)$. Since $n\ge 5$ is assumed one finds that $\bar{G}$ is a dihedral group (of order $2^{n-1}$) according to Blichfeldt's classification \cite{Blichfeldt}. This shows $G=H$ as claimed.
\end{proof}

\subsection{The group $2_+^{1+2n}$}

Let $T_n < \textnormal{GL}_{2^n}(\mathbb{Q}) $ be the $n$-fold tensor product of $\left\langle \left( \begin{smallmatrix} 0&1\\1&0 \end{smallmatrix}\right),\; \left( \begin{smallmatrix} 1&0\\0&-1 \end{smallmatrix}\right) \right\rangle \simeq D_8$. So $T_n$ is isomorphic to the extraspecial group $2_+^{1+2n}$.

This section, which is heavily based on Section 5 of \cite{cliff}, describes the construction of $\mathcal{B}^o(T_n)$ and defines a maximal finite subgroup of $\textnormal{GL}_{2^n}(\mathbb{Q}(\sqrt{-2}))$ which gives rise to a s.p.i.m.f.\ subgroup of $\textnormal{Sp}_{2^{n+1}}(\mathbb{Q})$.

Let $(b_0, \dots, b_{2^n-1})$ be the standard basis of $\mathbb{Q}^{1 \times 2^n}$. If one identifies $v\in \mathbb{F}_2^n$ with $j= \sum_i v_i 2^{i-1}$ then the basis vectors $b_j$ can be indexed by elements of $\mathbb{F}_2^n$. For an affine subspace $U$ of $\mathbb{F}_2^n$ let $\chi_U = \sum_{u \in U} b_u$. Then $L_n$ and $L'_n$ are the $\mathbb{Z}$-lattices in $\mathbb{Q}^{1 \times 2^n}$ spanned by
$$ \left \lbrace 2^{\lfloor(n-\dim(U)+\delta)/2\rfloor} \chi_U \mid  U \text{ an affine subspace of } \mathbb{F}_2^n \right\rbrace $$
where $\delta=0$ for $L_n$ and $\delta=1$ for $L'_n$.

In \cite[Theorem 3.2]{Wall} it is shown that $H_n:= \Aut(L_n, I_{2^n}) \cap \Aut(L'_n, I_{2^n})$ is isomorphic to $2_+^{1+2n}.O^+_{2n}(2)$ and further $O_2(H_n)$ is conjugate to $T_n$.

It is shown in \cite{Winter} that $\Out(2_+^{1+2n}) \simeq O^+_{2n}(2) \! : \! 2 \simeq GO^+_{2n}(2)$ is the full orthogonal group of a quadratic form of Witt defect $0$. Conjugation by $h_n:= \left(\begin{smallmatrix} 1&1\\1&-1 \end{smallmatrix}\right) \otimes I_{2^{n-1}}$ induces an outer automorphism on $T_n$ which is not realized by $H_n$. Since $\End({T}_n) \simeq \mathbb{Q}$ and $h_n^2 = 2 I_{2^n}$, there exists no element in $\textnormal{GL}_{2^n}(\mathbb{Q})$ of finite order which induces the same automorphism on $T_n$. Thus $H_n$ is conjugate to $\mathcal{B}^o(T_n)$ (once it is shown that $T_n \unlhd \mathcal{B}^o(T_n)$). Moreover $\frac{1}{\sqrt{-2}} h_n$ normalizes $T_n$ and therefore $\mathcal{B}^o(T_n)$.
A comparison of the orders of $H_n$ and $\Out(T_n)$ shows that $\mathcal{H}_n:= \left\langle H_n,\; \frac{1}{\sqrt{-2}} h_n\right\rangle$ is the unique extension of $H_n$ by $C_2$ in $\textnormal{GL}_{2^n}(\mathbb{Q}(\sqrt{-2}))$. The group $\mathcal{H}_n$ gives rise to a finite symplectic matrix group in $\textnormal{Sp}_{2^{n+1}}(\mathbb{Q})$ which will be denoted by ${}_{\sqrt{-2}} [2_+^{1+2n}.(O^+_{2n}(2) \! : \! 2)]_{2^n}$ in the sequel.

Finally, by extending scalars $M_n := \sqrt{-2} L'_n + L_n$ is a $\mathbb{Z}[\sqrt{-2}]$-lattice generated by
$$\left\lbrace \sqrt{-2}^{n-\dim(U)} \chi_U \mid U \text{ an affine subspace of } \mathbb{F}_2^n \right\rbrace\:.$$

\begin{lemma}
The lattice $M_n$ is the $n$-fold tensor product $M_1 \tensordown{\mathbb{Z}[\sqrt{-2}]} \dots  \tensordown{\mathbb{Z}[\sqrt{-2}]} M_1$.
\end{lemma}
\begin{proof}
Let $V_{n-1} = \left\langle e_1, \dots, e_{n-1} \right\rangle$ and $V_1 = \left\langle e_n \right\rangle$. One checks that $b_x \otimes b_y = b_{x+y}$ for all $x\in V_{n-1}, y \in V_1$.

Let $U$ be a $d$-dimensional affine subspace of $V_{n-1}$. For $y$ in $V_1$ it follows that $\sqrt{-2}^{n-1-d}\chi_U \otimes \sqrt{-2} b_y = \sqrt{-2}^{n-d} \chi_{y+U} \in M_n$. Similarly, since $U + V_1$ has dimension $d+1$, it follows that $\sqrt{-2}^{n-1-d} \chi_U \otimes \chi_{V_1} = \sqrt{-2}^{n-(d+1)} \chi_{U+V_1} \in M_n$. Thus $M_{n-1} \otimes M_1 \subseteq M_n$.

Conversely, suppose $U$ is a $d$-dimensional affine subspace of $\mathbb{F}_2^n$. Write $U = x+y + U_0$ where $U_0$ is a subspace of $\mathbb{F}_2^n$ and $x \in V_{n-1}$, $y \in V_1$.

If $U_0 \le V_{n-1}$ then $\sqrt{-2}^{n-d} \chi_U = \sqrt{-2}^{n-1-d} \chi_{x+U} \otimes \sqrt{-2} b_y \in M_{n-1} \otimes M_1$. Otherwise $U_{n-1} := U_0 \cap V_{n-1}$ is a $(d-1)$-dimensional subspace and $U_0 = U_{n-1} \cup (z+e_n + U_{n-1})$ for some $z \in V_{n-1}$.

If $z \in U_{n-1}$, then $\sqrt{-2}^{n-d} \chi_U = \sqrt{-2}^{n-1 - (d-1)} \chi_{x+U_{n-1}} \otimes \chi_{V_1}$ otherwise
\begin{align*}
\sqrt{-2}^{n-d} \chi_U &=
\sqrt{-2}^{n-1-d} \chi_{x+U_{n-1}+\left\langle z \right\rangle} \otimes \sqrt{-2}b_y\\
&+ \sqrt{-2}^{n-1-(d-1)} \chi_{x+z+U_{n-1}}\otimes \chi_{V_1} \\
&- \sqrt{-2}(\sqrt{-2}^{n-1-(d-1)} \chi_{x+z+U_{n-1}}\otimes \sqrt{-2}b_y )
\end{align*}
This shows $M_n \subseteq M_{n-1} \otimes M_1$.
\end{proof}

\begin{lemma} The group $\mathcal{H}_n$ is conjugate to
the Hermitian automorphism group $\Aut_{\mathbb{Q}(\sqrt{-2})}(M_n) := \Stab_{U_{2^n}(\mathbb{Q}(\sqrt{-2}))}(M_n)$ in $\textnormal{GL}_{2^n}(\mathbb{Q}(\sqrt{-2}))$.
\end{lemma}
\begin{proof}
For $n=3$ the result can be checked explicitly. So let $n \ne 3$ denote by $(v_1,\dots,v_{2^n})$ a $\mathbb{Z}$-basis of $L'_n$ such that $(2v_1,\dots,2v_{2^{n-1}}, v_{2^{n-1}+1},\dots,v_{2^n})$ is a $\mathbb{Z}$-basis of $L_n$. Then $(\sqrt{-2}v_1,\dots,\sqrt{-2}v_{2^n},2v_1,\dots,2v_{2^{n-1}}, v_{2^{n-1}+1},\dots,v_{2^n})$ is a $\mathbb{Z}$-basis of $M_n = \sqrt{-2} L'_n \oplus L_n$. In particular, the $\mathbb{Z}$-lattices $L_n$ and $\sqrt{-2}L'_n$ are perpendicular with respect to the scalar product $(x,y) \mapsto \frac{1}{2} \Tr_{\mathbb{Q}(\sqrt{-2})/\mathbb{Q}}(x\transpose{\bar{y}})$.
Thus the group $\Aut_{\mathbb{Q}(\sqrt{-2})}(M_n)$ is the subgroup of $\Aut(\sqrt{-2} L'_n \perp L_n, I_{2^{n+1}})$ which commutes with $\sqrt{-2}$. Since $n \ne 3$ is assumed, the automorphism groups of $L_n$ and $L'_n$ equal $H_n$ \cite[Theorem 3.2]{Wall}. Hence with respect to appropriate $\mathbb{Z}$-bases,  $\Aut_{\mathbb{Q}(\sqrt{-2})}(M_n)$ contains a subgroup $G_n$ of index at most two where
$$G_n = \left\lbrace \begin{pmatrix} g_1 & 0 \\ 0 & g_2\end{pmatrix} \mid g_1, g_2 \in H_n \right\rbrace \cap \Aut_{\mathbb{Q}(\sqrt{-2})}(M_n)\:.$$
Now $\sqrt{-2}$ interchanges $\sqrt{-2}L'_n$ and $L_n$, i.e. it operates as a block matrix $\left(\begin{smallmatrix} 0 & w \\ -2w^{-1} & 0\end{smallmatrix}\right)$ for some $w \in \textnormal{GL}_{2^n}(\mathbb{Q})$. So $\left(\begin{smallmatrix} g_1 & 0 \\ 0 & g_2\end{smallmatrix}\right) \in G_n$ if and only if $g_2=g_1^w$. Therefore the subgroup $G_n< \Aut_{\mathbb{Q}(\sqrt{-2})}(M_n)$ is conjugate to $H_n$ in $\textnormal{GL}_{2^n}(\mathbb{Q}(\sqrt{-2}))$.
The result follows since $\frac{1}{\sqrt{-2}} h_n$ acts on $M_1 \otimes M_{n-1} = M_n$.
\end{proof}

\begin{theorem}\label{e2plus}
If $n \ge 2$, then ${}_{\sqrt{-2}}[2_+^{1+2n}.(O^+_{2n}(2) \! : \! 2)]_{2^n}$ is the (up to conjugacy) unique s.p.i.m.f.\ subgroup of $\textnormal{Sp}_{2^{n+1}}(\mathbb{Q})$ with Fitting group $2_+^{1+2n}$. In particular, $\mathcal{B}^o(T_n)$ is conjugate to $H_n$.
\end{theorem}
\begin{proof}
Since $\mathbb{Z}[\sqrt{-2}]$ is a PID, $M_n$ has a $\mathbb{Z}[\sqrt{-2}]$-basis. With respect to this basis, its automorphism group $G_n$ is a finite subgroup of $\textnormal{GL}_{2^n}(\mathbb{Z}[\sqrt{-2}])$. By explicit calculations in {\sc MAGMA}\ \cite{Magma} one checks that for $n\in\{2,3\}$ the $\mathbb{Z}$-span $\left\langle G_n \right\rangle_\mathbb{Z}$ equals $\mathbb{Z}[\sqrt{-2}]^{2^n \times 2^n}$. For $n \ge 4$ it follows from $M_n = M_{n-2} \tensordown{\mathbb{Z}[\sqrt{-2}]}M_2$ that $G_2 \otimes G_{n-2} \subset G_{n}$ (using appropriate bases). By induction, one gets
$$\mathbb{Z}[\sqrt{-2}]^{2^n \times 2^n} = \left\langle G_{n-2} \right\rangle_\mathbb{Z} \tensordown{\mathbb{Z}[\sqrt{-2}]} \left\langle G_{2} \right\rangle_\mathbb{Z} \subseteq \left\langle G_{n} \right\rangle_\mathbb{Z} \subseteq \mathbb{Z}[\sqrt{-2}]^{2^n \times 2^n}\:.$$
In particular, each $G_n$-invariant $\mathbb{Z}[\sqrt{-2}]$-lattice is a multiple of $M_n$, since $\mathbb{Z}[\sqrt{-2}]$ has class number 1. Thus $G_n$ is a maximal finite subgroup of $\textnormal{GL}_{2^n}(\mathbb{Q}(\sqrt{-2}))$. But any finite symplectic supergroup of $G:= {}_{\sqrt{-2}}[2_+^{1+2n}.(O^+_{2n}(2) \! : \! 2)]_{2^n}$ comes from a finite supergroup of $\mathcal{H}_n\simeq G_n < \textnormal{GL}_{2^n}(\mathbb{Q}(\sqrt{-2}))$. Thus $G$ is s.i.m.f.{}.

As explained in the beginning of this section, to prove that $\mathcal{B}^o(T_n)$ is conjugate to $H_n$, it suffices to show that $T_n \unlhd \mathcal{B}^o(T_n)$. But this is clear from Lemma \ref{lemmabravais} since (a rational constituent of) $F(G)$ is conjugate to $T_n$.

Suppose now $S < \textnormal{Sp}_{2^{n+1}}(\mathbb{Q})$ is s.p.i.m.f.\ such that $F(S) \simeq 2^{1+2n}$. Then by Lemma \ref{mainlemma}, $F(S)$ must be conjugate to $U$, so one can assume $U < S$. But then $\mathcal{B}^o(U) \unlhd S$.  Now $\abs{\Out(2^{1+2n})} = 2 [\mathcal{B}^o(U): U]$ shows that $G$ and $S$ contain $\mathcal{B}^o(U)$ with index $2$. Since $\End(U) \simeq \mathbb{Q}^{2\times 2}$ it follows from Theorem \ref{QA} that $G$ is the (up to conjugacy) unique symplectic extension of $\mathcal{B}^o(U)$ by $C_2$.
\end{proof}

\subsection{The group $p_+^{1+2n}$}

In this section, let $p$ be an odd prime. In this section, a family of irreducible symplectic matrix groups in dimension $p^n(p-1)$ is described which will be maximal finite in the case that $p$ is a Fermat prime, i.e. $p-1$ is a power of two.

Let $T_n^{(p)} \simeq p_+^{1+2n}$ be the $n$-fold tensor product of $T_1^{(p)}$ where $T_1^{(p)}$ is the subgroup of $\textnormal{GL}_p(\mathbb{Q}(\zeta_p))$ generated by the diagonal matrix $\Diag(1, \zeta_p, \dots, \zeta_p^{p-1})$ and the permutation matrix corresponding to the $p$-cycle $(1,\dots,p)$. Further let $H_n^{(p)} = N_{U_{p^n}(\mathbb{Q}(\zeta_p))}(T_n^{(p)})$. By \cite{Winter} $H_n^{(p)}$ is isomorphic to a subgroup of $C_2 \times p_+^{1+2n}.\textnormal{Sp}_{2n}(p)$, since the group of outer automorphisms which act trivially on the center of $T_n^{(p)}$ is isomorphic to $\textnormal{Sp}_{2n}(p)$.

In \cite[Section 4]{Wall} Wall constructs a $(p-1)p^n$ dimensional rational lattice on which $C_2 \times p_+^{1+2n}.\textnormal{Sp}_{2n}(p)$ acts. Thus $H_n^{(p)} = \mathcal{B}^o(T_n^{(p)}) \simeq C_2 \times p_+^{1+2n}.\textnormal{Sp}_{2n}(p)$ (if $T_n^{(p)} \unlhd \mathcal{B}^o(T_n^{(p)})$ it assumed). Now $H_n^{(p)}$ gives rise to a finite subgroup of $\textnormal{Sp}_{p^n(p-1)}(\mathbb{Q})$ which will be denoted by ${}_{\zeta_p}[\pm p_+^{1+2n}.\textnormal{Sp}_{2n}(p)]_{p^n}$.

\begin{theorem}\label{Fermat}
If $p$ is a Fermat prime, then ${}_{\zeta_{p}}[\pm p_+^{1+2n}.\textnormal{Sp}_{2n}(p)]_{p^n}$ is a s.i.m.f.\ subgroup of $\textnormal{Sp}_{p^n(p-1)}(\mathbb{Q})$.
\end{theorem}
\begin{proof}
The commuting algebra $C$ of ${}_{\zeta_{p}}[\pm p_+^{1+2n}.\textnormal{Sp}_{2n}(p)]_{p^n}$ is isomorphic to $\mathbb{Q}(\zeta_p)$. Since $p$ is a Fermat prime, $C$ has only one maximal subfield, which must be totally real. Thus any symplectic supergroup must come from a finite subgroup of $\textnormal{GL}_{p^n}(\mathbb{Q}(\zeta_p))$. But $H_n^{(p)}$ is maximal finite in $\textnormal{GL}_{p^n}(\mathbb{Q}(\zeta_p))$ according to \cite[Theorem 7.3]{cliff}.
\end{proof}

\section{Dimension 2p}\label{sec2p}

Let $p\ge 5$ be prime. The structure of s.p.i.m.f.\ subgroups of $\textnormal{Sp}_{2p}(\mathbb{Q})$ is rather restricted.

\begin{lemma}
If $G < \textnormal{Sp}_{2p}(\mathbb{Q})$ is s.p.i.m.f.\ then $F(G)$ is cyclic of order $2, 4$ or $6$.
\end{lemma}
\begin{proof}
Clearly, if $q \ge 5$ is a prime such that $O_q(G) \ne 1$ then $q=2p+1$ by Lemma \ref{mainlemma}. In particular, $q\equiv -1 \mod 4$ and $q \ge 11$. This contradicts Theorem \ref{pminus1}. So $F(G) = O_2(G)O_3(G)$. If $G$ would be cyclic, then $G=F(G)= O_2(G)O_3(G)$ is reducible. So $G$ embeds into $\textnormal{GL}_p(K)$ for some imaginary quadratic number field $K$. Hence the result follows from Hall's classification (see Theorem \ref{Hall}).
\end{proof}

Hence the generalized Fitting subgroup $F^*(G)$ of a s.p.i.m.f.\ group $G < \textnormal{Sp}_{2p}(\mathbb{Q})$ cannot be soluble. Thus $F^*(G)$ is either reducible with $\mathbb{Q}^{2\times 2}$ as commuting algebra or irreducible and its commuting algebra is an imaginary quadratic number field. Thus $G$ can easily be recovered from $F^*(G)$ by the methods in the next section.

\section{Some methods}\label{secmet}

\subsection{Normal subgroups of index $2^k$}\label{subsecind2}

If $G < \textnormal{Sp}_{2n}(\mathbb{Q})$ is maximal finite, then $G/\mathcal{B}^o(F^*(G))$ is very often a $2$-group. Thus two criteria are given that eliminate some candidates for $F^*(G)$ in these cases.

The first lemma is an analogon to \cite[Corollary III.4]{FRMG}.

\begin{lemma}
Let $N$ be a subgroup of a s.p.i.m.f.\ group $G < \textnormal{Sp}_{2n}(\mathbb{Q})$ of index $2$. If $N$ is rationally reducible then $\End(N) \simeq L^{2\times 2}$ for some totally real number field $L$.
\end{lemma}
\begin{proof}
Let $g \in G - N$.
By Cifford theory, the natural representation $\triangle$ of $G$ splits into $\triangle_1 + \triangle_1^g$ for some irreducible $\triangle_1 \colon N \to \textnormal{GL}_n(\mathbb{Q})$. Thus $G$ is conjugate to $\left \langle I_2 \otimes \triangle_1(N), \left( \begin{smallmatrix} 0 & I_n \\ \triangle_1(g^2) & 0 \end{smallmatrix} \right) \right\rangle \le \triangle_1(N)\wr C_2$. Since $G$ is symplectic primitive and $\triangle_1(N)$ is rationally irreducible this implies $\dim_\mathbb{Q}(\mathcal{F}_{skew}(\triangle_1(N))) = 0$.

Let $\delta\colon N \to \textnormal{GL}_n(\mathbb{R}), h \mapsto \triangle_1(h)$ decompose into $\mathbb{R}$-irreducible representations $\delta_1, \dots, \delta_s$. It follows from the above that $\mathcal{F}(\delta(N)) = \mathcal{F}_{sym}(\delta(N))$. In particular, this implies that the $\delta_i$ are pairwise nonisomorphic and $\End(\delta_i(N)) \simeq \mathbb{R}$.
Hence $L \otimes_\mathbb{Q} \mathbb{R} \simeq \oplus_{i=1}^s\mathbb{R}$ and therefore $L$ is a totally real number field.
\end{proof}

The smallest example of such a group $N$ is $N = \left\langle -I_2 \right\rangle \unlhd \left\langle \left( \begin{smallmatrix} 0 & 1 \\ -1 & 0 \end{smallmatrix} \right) \right\rangle < \textnormal{Sp}_{4}(\mathbb{Q})$ which has $\mathbb{Q}^{2\times 2}$ as commuting algebra.

\begin{example}
The Fitting subgroup of a s.p.i.m.f.\ group $G < \textnormal{Sp}_8(\mathbb{Q})$ is not isomorphic to $Q_8$.
\end{example}
\begin{proof}
 Suppose otherwise. Then $\End(F(G)) \simeq Q_{\infty,2}^{2\times 2}$ where $Q_{\infty,2}$ denotes the quaternion algebra over $\mathbb{Q}$ ramified only at $2$ and $\infty$. Thus $E(G)=1$ by \cite{HissMalle}. Hence $G/F(G) \le \Out(F(G)) \simeq S_3$. Since $\mathcal{B}^o(F(G))\simeq \SL_2(3)$ one gets $[G: \mathcal{B}^o(F(G))] \le 2$. This contradicts the result above since $\End(F(G))  = \End(\mathcal{B}^o(F(G))) \simeq Q_{\infty, 2}^{2\times 2}$.
\end{proof}

If $n$ is not a power of $2$, the following lemma can be used to rule out several candidates for normal subgroups.

\begin{lemma}
Let $G< \textnormal{Sp}_{2n}(\mathbb{Q})$ be rationally irreducible and symplectic primitive. Let $N \unlhd G$ such that $[G:N] =2^k$. If $\triangle$ denotes the natural representation of $N$ then $\triangle = 2^\ell \cdot \delta$ for some irreducible representation $\delta$ of $N$ and some $0 \le \ell \le k$.
\end{lemma}
\begin{proof}
The proof given in \cite[Lemma III.4]{ERMG} applies mutatis mutandis.
\end{proof}

\begin{example}
Let $G < \textnormal{Sp}_{12}(\mathbb{Q})$ be s.p.i.m.f.{}. If $F(G)=O_2(G)$, then $O_2(G)$ is isomorphic to $C_2, C_4$ or $D_8$ and $E(G) \ne 1$.
\end{example}
\begin{proof}
Let $U = \{Q_8,\; C_8,\; D_{16},\; QD_{16},\; D_8 \otimes C_4,\; 2^{1+4}_+\}$. Hall's theorem (see Theorem \ref{Hall}) shows that $O_2(G) \in \{C_2, C_4, D_8\} \cup U$.
Suppose first that $E(G) = 1$. Then $G/\mathcal{B}^o(F(G)) \le \Out(F(G))$ is a $2$-group. By the previous result this implies that the natural representation of $\mathcal{B}^o(F(G))$ must be a multiple of a faithful rationally irreducible representation of degree $3, 6$ or $12$. But $F(G) = O_2(G)$ and thus $\mathcal{B}^o(F(G))$ does not admit such representations. This implies $E(G) \ne 1$.
Suppose now $O_2(G) \in U$. Then $\End(O_2(G)) \simeq Q^{3 \times 3}$ where $Q$ is isomorphic to $\mathbb{Q}$, $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt{\pm 2})$ or $Q_{\infty, 2}$. But no quasisimple group embeds in to $Q^{3 \times 3}$ (see \cite{HissMalle}). Thus $E(G)=1$ gives the desired contradiction.
\end{proof}

\subsection{m-parameter argument}\label{subsecmpar}

Suppose $U < \textnormal{GL}_{n}(\mathbb{Q})$ s finite such that $E:= \End(U)$ is a number field of absolute degree $m$ and let $F_0 \in \Forms_{>0}(U)$. In particular, this applies to irreducible cyclic matrix groups. If $E^+$ denotes the maximal totally real subfield of $E$ then $\Forms_{>0}(U) = E^+ \cdot F_0$. The thesis \cite{ERMG} gives an algorithm which uses this parametrization and ideal arithmetic in $E^+$ to show that every finite supergroup $G\ge U$ fixes (up to conjugacy) one of the forms $cF$ where $c$ runs through some finite set.

Two more definitions are needed to  state the result.

\begin{definition}
For a totally real number field $K$, denote by $\Pi(K)$ a finite set of prime numbers such that
\begin{itemize}
\item For each $x \in K$ there exists some $y \in \mathbb{Z}_K$ whose norm is supported at $\Pi(K)$ such that $xy$ is totally positive.
\item In each ideal class of $\mathbb{Z}_K$ there exists some integral ideal that contains a natural number whose prime factors come from $\Pi(K)$.
\end{itemize}
Furthermore, if $\ell \in \mathbb{Z}$ set
$$\tilde{\Pi}(\ell, K) = \{p \mid\mbox{$p$ is a prime divisor of $\ell$ or $p \in \Pi(k)$ for some subfield $k$ of $K$}\}\:.$$
\end{definition}

\begin{definition}
Let $L$ be a lattice in $\mathbb{Q}^{1\times n}$ of full rank and let $F \in \mathbb{Q}^{n\times n}$ be symmetric positive definite.
\begin{enumerate}
\item $L^{\#, F} = \{x \in \mathbb{Q}^{n \times 1} \mid xF\transpose{y} \in \mathbb{Z} \mbox{ for all } y \in L\}$
denotes the dual of $L$ with respect to $F$.
\item The pair $(L, F)$ is said to be integral if $L \subset L^{\#, F}$.
\item The integral pair $(L, F)$ is said to be normalized if the abelian group $L^{\#, F} / L$ has squarefree exponent and its rank is at most $n/2$.
\end{enumerate}
\end{definition}

\begin{theorem}[\protect{\cite[Korollar III.3]{ERMG}}]\label{mparam}
Suppose $U < \textnormal{GL}_{n}(\mathbb{Q})$ is finite such that $E:= \End(U)$ is a field and denote by $E^+$ its maximal totally real subfield. Let $U < G <\textnormal{GL}_n(\mathbb{Q})$ be finite and fix some $L \in \mathcal{Z}(G)$. Then there exists some $F \in \Forms_{>0}(U)$ such that $(L, F)$ is integral and the prime divisors of $\det(L, F)$ are contained in $\tilde{\Pi}(\abs{G}, E^+)$.
\end{theorem}

To find all (conjugacy classes of) s.i.m.f.\ matrix groups $G$ that contain a conjugate copy of $U$, one has to consider the groups $\Aut_K(L, F)$ for some minimal totally complex subfield $K$ of $E$ and some integral pair $(L, F)$ where the prime divisors of $\det(L, F)$ are bounded by $\tilde{\Pi}((n+1)!, E^+)$ (see Minkowski's bound).

The following ideas reduce the number of pairs $(L,F)$ that one has to consider to a finite number (see \cite[Chapter VII]{ERMG} and \cite[Remark 2.2.13]{MK} for details).
\begin{itemize}
\item
If $G$ fixes some integral $(L, F)$ which is not normalized, then exists some prime $p$ dividing $\det(L, F)$ such that  $(L \cap p^{-1}L^{\#, F}, F)$ is integral. Iterating this process, one finds some normalized pair $(L', F) \in \mathcal{Z}(G)\times \Forms_{>0}(G)$.
\item
Suppose $L_1, \dots, L_r$ represent the isomorphism classes of $\left\langle U \right\rangle_\mathbb{Z}$-invariant lattices in $\mathbb{Q}^{n \times n}$.
Then $L'$ is isomorphic to some $L_i$ i.e. there exists some $x \in \textnormal{GL}_n(\mathbb{Q})$ such that $L' = L_i x$.
\item
Since $G= \Aut_K(L', F) = \Aut_K(L_ix, F) = x\Aut_K(L_i, x^{-1} F \mtranspose{x})x^{-1}$ one only has to consider the normalized pairs $(L_i, F)$ with $F \in \Forms_{>0}(U)$ such that the prime divisors of $\det(L_i, F)$ are contained in $\tilde{\Pi}((n+1)!, E^+)$.
\item
If $y \in E$ such that $L_i y = L_i$ then $\Aut_K(L_iy, F)$ is conjugate to $\Aut_K(L_i, y^{-1}F\mtranspose{y}) = \Aut_K(L_i, \Nr_{E/E^+}(y^{-1})F)$.
\end{itemize}

\subsection{Quaternion algebras as endomorphism rings}\label{subsetquat}

Suppose $N < \textnormal{Sp}_{2n}(\mathbb{Q})$ is finite such that $Q:= \End(N)$ is a quaternion algebra, i.e. a central simple algebra of degree $4$ over its center $K$.
The next theorem explains how to find (up to conjugacy) all s.i.m.f.\ supergroups $G$ of $N$ such that $N \unlhd G$. After replacing $N$ by $\mathcal{B}^o(N)$, one may assume that $N=\mathcal{B}^o(N)$.

\begin{theorem}\label{QA}
Suppose $x \in \textnormal{GL}_{2n}(\mathbb{Q})$ commutes with $K$ and acts on $N$ such that $x^2 \in N$.
\begin{enumerate}
\item $\End(\left\langle N, x \right\rangle) \simeq K[X] / (X^2-\nr(a))$ for some $a \in Q$ where $\nr\colon Q \to K$ denotes the reduced norm.
\item If $y \in \textnormal{GL}_{2n}(\mathbb{Q})$ induces the same outer automorphism on $N$ as $x$ then $\End(\left\langle N, y \right\rangle) \simeq K[X] / (X^2-u\nr(a))$ for some torsion unit $u \in K^* \cap \nr(Q^*)$.
\item If $\End(\left\langle N, x \right\rangle)$ is a field and $u \in (K^*)^2$ then $\left\langle N, x \right\rangle$ and $\left\langle N, y \right\rangle$ are conjugate.
\end{enumerate}
\end{theorem}
\begin{proof}
\begin{enumerate}
\item For any $z \in Q$ one has $z^x \in \End(N^x) =  \End(N) = Q$. Hence $x$ induces a $K$-automorphism on $Q$. By the Skolem-Noether theorem there exists some $a \in Q^*$ such that $z^a = z^x$ for all $z \in Q$.  Since $x$ does not centralize $Q$ (otherwise $x \in \mathcal{B}^o(N)= N$ by Lemma \ref{lemmabravais}) it follows that
$$K \subsetneq K[a] \subseteq C_Q(a) = \End(\left\langle N, x \right\rangle)) \subsetneq Q\:. $$
Thus $K[a] = \End(\left\langle N, x \right\rangle)$. Since $x^2 \in N$, it induces the identity on $Q$. Hence $a^2 \in K$ and therefore $K[a] \cong K[X]/(X^2-\nr(a))$.
\item Since $yx^{-1}$ induces an inner automorphism on $N$, it is contained in $N Q^*$. Say $yx^{-1} = g c$ with $g \in N$ and $c \in Q$. Then
$$\End({\left\langle N, y\right\rangle} )= \End({\left\langle N, cx \right\rangle} ) \simeq K[X] / (X^2 - \nr(ca))\:.$$
Further $cxcx^{-1} \in (g^{-1}y)^2 x^{-2} \in N$ has finite order. Hence the reduced norm $\nr(c)^2 = \nr(c \cdot (xcx^{-1})) \in K^*$ also has finite order. Thus $u:= \nr(c)$ is a torsion unit as claimed.
\item Let $v\in K$ such that $v^2 = u$. Since $v$ has finite order it is contained in $\mathcal{B}^o(N) = N$ by Lemma \ref{lemmabravais}. The elements $a^2$ and $(v^{-1}ca)^2$ are both contained in $K$. So $\nr(v^{-1}c)=1$ implies that $a$ and $v^{-1}ca$ have the same minimal polynomial over $K$. It follows from the Skolem-Noether theorem that there exists some $t\in Q^*$ such that $a^t=v^{-1}ca$. Finally
$$x^t = t^{-1} xtx^{-1}x = t^{-1} ata^{-1} x = v^{-1} ca a^{-1} x = (gv)^{-1} y $$
shows $\left\langle N, x \right\rangle^t = \left\langle N, (gv)^{-1} y \right\rangle = \left\langle N, y \right\rangle$.
\end{enumerate}
\end{proof}

In most cases, the structure of $Q$ allows some simplifications:
\begin{enumerate}
\item If $K \simeq \mathbb{Q}$ then obviously every $x \in \textnormal{GL}_{2n}(\mathbb{Q})$ commutes with $K$.
\item If $Q$ is totally definite then $\End(\left\langle N, x \right\rangle)$ must be a field and there exists no nontrivial torsion unit in $K^* \cap \nr(Q^*)$. I.e. each class of outer automorphisms gives rise to at most one conjugacy class  $\left\langle N, x \right\rangle$.
\item Suppose one wants to find all s.p.i.m.f.\ supergroups $G$ of $N$ satisfying $N = \mathcal{B}^o(F^*(G))$. If $Q$ is totally definite and $K \simeq \mathbb{Q}$ then the proof of \cite[Theorem 4]{PrimStructure} shows that the exponent of $G/N$ is at most $2$. Hence one gets all such groups $G$ (up to conjugacy) with the above result and the $2$-parameter argument.
\end{enumerate}

Here is a simple example.

\begin{example}
Suppose $G < \textnormal{Sp}_4(\mathbb{Q})$ is s.p.i.m.f.\ and it contains a normal subgroup $U$ isomorphic to $Q_8$. (Note that the character of $U$ is uniquely determined by Lemma \ref{mainlemma}.)

Let $N:= \mathcal{B}^o(U) \simeq \SL_2(3)$. Then $\End(U) = \End(N) \simeq Q_{\infty,2}$ is the quaternion algebra over $\mathbb{Q}$ ramified only at $2$ and $\infty$. If there exists some $x \in C_G(N) - N$ then $x$ is contained in some maximal order of $Q$. These orders are all conjugate and have $\SL_2(3)$ as group of torsion units. Thus one can assume that the order of $x$ equals $3$ or $4$. In both cases, the commuting algebra of $H:= \left \langle N ,x \right \rangle$ is an imaginary quadratic number field. If $x$ has order $3$ then $H$ is s.i.m.f.. In the other case, it follows from the $2$-parameter argument that $H$ is only contained in the s.i.m.f.\ group $\mathcal{B}^o(D_8 \otimes C_4) = (D_8 \otimes C_4).S_3$.
Suppose now $N$ is self-centralizing in $G$. One immediately finds the 4-dimensional faithful representation of $\textnormal{GL}_2(3)$ with commuting algebra $\mathbb{Q}(\sqrt{-2})$ as one possibility for $G$. By the above result, $\textnormal{GL}_2(3)$ is (up to conjugacy) the only possibility since $\Out(N) \simeq C_2$ and $\mathbb{Q}^* \cap \nr(Q_{\infty, 2})\subseteq \mathbb{Q}_{>0}$ contains no nontrivial torsion unit.
\end{example}

\section{Tables}\label{sectab}

A database for the the conjugacy classes of s.p.i.m.f.\ subgroups of $\textnormal{GL}_{2n}(\mathbb{Q})$ up to $n=11$ will be available in future versions of {\sc MAGMA}. In the mean time, the database is
available from \url{http://www.math.rwth-aachen.de/~Markus.Kirschmer/symplectic/} .

Loading the file in {\sc MAGMA}\ via the {\tt load} command generates an associative array called {\tt SGroups}.
The data for dimension $2n$ is stored in {\tt SGroups[2n]} as a sequence of triples {\tt <FF, o, s>}.
\begin{enumerate}
\item {\tt FF} is a sequence of two forms $F,S$ such that
\begin{align*}
A &= \Aut_{\mathbb{Q}[SF^{-1}]}(\mathbb{Z}^{1\times 2n}, F)\\
  &= \Aut(\mathbb{Z}^{1\times 2n}, \{F,S\}) := \{g \in \textnormal{GL}_{2n}(\mathbb{Z})\mid gF\transpose{g}=F \mbox { and } gS\transpose{g}=S\}
\end{align*}
represents the corresponding conjugacy class.
\item The forms $F,S$ satisfy the following properties:
\begin{enumerate}
\item $F\in \Forms_{>0}(A)$ and $S \in \mathcal{F}_{skew}(A)$.
\item The pair $(\mathbb{Z}^{1\times 2n}, F) \in \mathcal{Z}(A)\times \Forms_{>0}(A)$ is normalized and has minimal determinant among all normalized $(L, F') \in  \mathcal{Z}(A)\times \Forms_{>0}(A)$.
\item $SF^{-1}$ is a primitive element of the field $\End(A)$ such that its minimal polynomial $\mu(SF^{-1}, X)$ is one of
\begin{itemize}
\item $X^2+d$ for some squarefree $d \in \mathbb{N}$.
\item $\mu(\zeta_k-\zeta_k^{-1}, X)$ for some even $k \in \mathbb{Z}_{\ge 6}$.
\item $\mu(\zeta_{26} + \zeta_{26}^3 + \zeta_{26}^9, X)$.
\item $\mu(\sqrt{k}\cdot(\zeta_{\ell}-\zeta_{\ell}^{-1}), X)$ where $(k, \ell) \in \{(2, 10),\, (3,10),\, (3,16)\}$.
\item $\mu(\sqrt{-1}+\sqrt{-3}+\sqrt{-5}, X)$.
\end{itemize}
\end{enumerate}
\item {\tt o} is the order of the groups in the conjugacy class of $A$
\item {\tt s} is a string which contains the name of the class of $A$ given in my thesis ($\otimes$ is abbreviated t and so on...).
\end{enumerate}

\begin{example}
Suppose one wants to access the group $\textnormal{GL}_2(\mathbb{F}_3) < \textnormal{GL}_4(\mathbb{Q})$ in the database. From
\begin{verbatim}
  [x[3] : x in SGroups[4]];
\end{verbatim}
which yields the list:
\begin{verbatim}
  > [ D8tC4.S3, C4tA2, GL23, SL23oC3, C10 ]
\end{verbatim}
one sees that it is the third entry of SGroups[4]. So
\begin{verbatim}
  A:= AutomorphismGroup(SGroups[4,3,1]);
\end{verbatim}
generates a representative {\tt A} of that class.
\end{example}

\begin{remark}
If $G < \textnormal{GL}_{2n}(\mathbb{Q})$ is s.p.i.m.f., one can use the information in the database to recover the conjugacy class of $G$ as follows.

If $L_1,\dots,L_s$ represent the isomorphism classes of $G$-invariant lattices, then there exist only finitely many forms $F \in \Forms_{>0}(G)$ such that $(L_i, F)$ is normalized and of minimal determinant. After a change of base (i.e. conjugating $G$ with a basis matrix of $L_i$), one may assume that $L_i = \mathbb{Z}^{1\times 2n}$. Then for each $F$ there exist at most $[\End(G):\mathbb{Q}]$ skewsymmetric forms $S \in \mathcal{F}(G)$ such that $(F,S)$ satisfies the properties from above. For each such pair, one checks whether there exists some simultaneous isometry between $(F,S)$ and one of the pair of forms $(F', S')$ in the database. I.e. one tests (using the approach described in \cite{Souvignier} which is available in {\sc MAGMA}\ via the {\tt IsIsometric} command) whether there exists some $T \in \textnormal{GL}_{2n}(\mathbb{Z})$ such that $TF\transpose{T} = F'$ and $TS\transpose{T} = S'$.
Such a matrix $T$ exists if and only if $G$ and $A:= \Aut(\mathbb{Z}^{1\times 2n}, \{F',S'\})$ are conjugate. If so, then clearly $\Aut(\mathbb{Z}^{1\times 2n}, \{F,S\}) = A^T$. So after reverting the base change from above, one ends up with a conjugating element between $G$ and $A$.
\end{remark}

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As you may have, uh, heard a thing or two about, Britain is . . . in the midst of some extreme turmoil, putting it mildly, following the vote last week to leave the European Union. While the Queen has not made an official statement following Brexit, she has now made her first official appearance, as she met Monday with Northern Ireland deputy minister Martin McGuinness. (The Queen and Prince Philip are currently in the midst of a two-day trip to Northern Ireland.)
The Queen—per the Daily Mail—"seemed in good spirits," especially considering the wild events of the past week in the country. Asked by McGuinness how she was, she responded, drily, "I'm still alive, anyway." She then explained she'd been "quite busy," continuing, "There's been a lot going on. I've had two birthdays so we've been quite busy." With this, and her recent scolding of Prince William (in which she told him to stand up at the Trooping the Colour ceremony), the Queen would seem to be, well, speaking with slightly more candor and humor. Honestly, the thought of having to make the amount of small talk she does, on a daily basis, is enough to make one want to crawl inside a small storage closet and never emerge again; so the fact that she still happily engages in this sort of banter (or, at least, doesn't let on that she isn't happy about it) is almost unfathomably impressive.
Queen Elizabeth's 90th Birthday Celebration
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Prince William Surprises a Princess Diana Vigil at Kensington Palace
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Gabriele de Seta – The no-venue underground: Sounding Hong Kong's lack of performance spaces
It's 10 October 2012. I've been living in Hong Kong for less than two months, and Dennis Wong is waiting for me to arrive at the performance venue and do my sound check. Fellow experimental musicians from Shanghai suggested Dennis as one of the local organizers with whom I should get in touch.
A few weeks before, Dennis asked me if I wanted to play a show at a newly-opened local venue called C.I.A. (Cultural Industries Association—the acronym is also a play on the covert and secret operations of the U.S. CIA). "Sure," I said, "when and where?" Dennis gave me an address and asked me if I needed anything else besides a guitar amplifier. I didn't.
Now I stand outside the Kwai Hing MTR station, which, according to Google Maps, is the closest station to the venue. As I carry my hollow-body guitar and a backpack full of effect pedals, I try to find my bearings via GPS. The air is still heavy and humid, although Hong Kong is cooler in the fall than during the summer, and there aren't many useful reference points around the station: a convenience store, a small noodle restaurant, and an overpass leading to the entrance of a shopping mall. Behind me stands a dense cluster of residential towers and a public housing estate. In front of me I see a wall of industrial buildings, the overpass disappearing into a small crevice between two of them.
Google Street View of the industrial buildings on Wah Sing Street, where art gallery/performance venue C.I.A (Cultural Industries Association) was located. Source: Google Maps, June 2011.
I spend ten minutes circling around the block, following the directions provided by Google Maps, only to realize that there is no way of crossing the six lanes of Hing Fong Road. I memorize the address of C.I.A. and put my smartphone back in my pocket: Unit 7, 8th floor, Block B, Wah Tat Industrial Centre, 8-10 Wah Sing Street, Kwai Hing, Kowloon, Hong Kong. I head back to the station, walk through the overpass, an elevated courtyard, another overpass, down an escalator, and venture into an alleyway between two industrial buildings. I end up in a maze of streets without shops, cars, or pedestrians; only container trucks come in and out of garage doors and loading ramps. Once I find the Wah Tat Industrial Centre, I still have to figure out how to reach Block B and how to get to the eighth floor. When I reach the elevator, the security guard instructs me: "that is the cargo elevator. People go in that other one." The only signal that helps me find C.I.A.'s metal door is the bass frequencies reverberating in the damp corridors. I slide a heavy metal door open and I'm greeted by Dennis and the other musicians doing their sound checks and preparing for the show in a large whitewashed room with no windows.
A few weeks later, my Shanghainese friend Huang Lei tells me he's been invited to perform in Hong Kong and asks me if I can help him organize another show. I get in touch with Dennis, who kindly agrees to set up a show for Huang Lei in his "NOISE to SIGNAL" series. We coordinate the details of the event through Facebook messages and I help him designing a flyer.
This time the venue, Strategic Sounds, is located on the 10th floor of the High Win Factory Building in Kwun Tong District, an industrial area on the other side of the Kowloon peninsula. Huang Lei, Dennis, and I end up playing an improvised set together, a twenty-minute mess of prepared guitar, crackling electronics, and distorted feedback echoing down the grimy ventilation shaft right outside the venue.
Throughout the following year, I was generously invited to play a few more shows at Strategic Sounds and C.I.A., meeting several of the experimental musicians active in the city at the time. Later, both venues closed down under the pressure of increasing rents and the challenges of sustaining an independent performance space in Hong Kong, a difficult enterprise even when eased by the relatively cheaper rents of vacant units in industrial buildings.
The author improvising with Shanghai-based musician Huang Lei (as 大小) and local organizer Dennis Wong (as Sin:ned) at Strategic Sounds, Hong Kong, November 2012. Video by Rolf.
Of the eight venues where I had the pleasure to play experimental music during my three years in Hong Kong, seven were located in industrial buildings. Besides C.I.A. and Strategic Sounds, they ranged from Dimension+, a small makerspace attached to an artist studio, to Hidden Agenda, a long-standing live-music house well-known among local independent music audiences. During these years, I also enjoyed live performances by some of my favorite musicians and bands, such as Makoto Kawabata, Hijokaidan, and Laibach, which played in shows organized by the same people I hung out with in former factory premises.
The two factors that pushed Hong Kong's experimental musicians to find spaces in the post-industrial peripheries of the city were the infamously high cost of Hong Kong's real estate market and the lack of suitable and welcoming performance venues in more central areas. The pubs and clubs hosting live music in central Kowloon and Hong Kong Island predominantly featured DJs and cover bands that catered to the tastes of the commercial audiences that they rely on for financial stability. Given the scarcity of spaces like garages, squats, cellars, and warehouses, experimental musicians turned to the industrial buildings hollowed out by the recent delocalization of factories. Other local artists and creative enterprises had also started taking advantage of these empty spaces.
"Performance area" signage on a mezzanine of the A.C.O. stairwell delimiting KWC's live set during Sound-On-Site: Space Oddity #1: From Below show, January 2015.
The first Sound-On-Site show organized by booking agency Twenty Alpha and record label Re-Records at A.C.O. (Art & Culture Outreach)—performed at the end of January 2015 and aptly titled "From Below"—poignantly exemplifies the lack of performance spaces for the local underground and experimental music scenes in Hong Kong. Rather than playing in the small bookstore located on the 14th floor of the Foo Tak Building, the three performing musicians decided to set up their equipment on different landings of the stairwell: laptops on small stools, with amplifiers turned on their side to fit the constraining spaces, cables dangling between floors, and performance areas marked off by improvised signage. Puzzled audiences moved up and down the stairwell, trying to figure out how to reach the floor from which the sounds were coming, or sat on the concrete steps listening to droning frequencies reverberating through the building, which itself became an essential architectural component of musicking. "This reminds me of Beijing," commented a friend visiting Hong Kong from Mainland China, "but where in Beijing musicians are taking back hutong alleyways and old housing, here it's all about industrial infrastructure."
Jonathan Solomon, Ciara Wong, and Adam Frampton have defined Hong Kong as a "city without ground," lacking the concept of "ground" both physically and culturally. It is somehow ironically appropriate that in this city without ground, underground musicians find themselves relegated to a precarious "overground" actively created out of fleeting spaces strewn across the upper floors of post-industrial peripheries. These precarious venues appear and disappear following the inexorable inflation of property prices, leaving local show organizers to work in the present tense with whatever space is available at the moment.
Predominantly sustained by personal passions and practices cultivated in spare time carved out of full-time non-musical careers, Hong Kong's experimental music scene finds its most reliable spaces on Facebook, Twitter, YouTube, and other social media platforms, where shows are organized, promoted, documented, and inscribed in the event geography of the city. Riffing on a locution coined by Rob Hayler, the British musician Jon Marshall describes the experimental music scene of Western Russia as a "no-audience underground," a fluid social milieu in which performers also double as organizers, promoters, critics, and audiences. The situation in Hong Kong isn't very different. Yet, despite the recurring lamentations of local musicians and organizers, diminutive audiences are not the biggest issue faced by the community. The precariousness and ephemerality of performance spaces determines the elastic and resilient fabric of the local experimental music scene, which could accurately be called a "no-venue underground."
Categories: Asia, Experimental music | Tags: Asia, China, Experimental music, Hong Kong | Permalink.


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Psychologists Reveal How You Know You've Found 'The One'
By Amy Lewis	19 February 2020
Relationships, we can probably all agree, are a tricky business at the best of times. Getting from a swipe right on Tinder, to making it past the first date and then sailing into 'official relationship' territory, well it can all feel a lot like running a gauntlet that you've been incredibly poorly prepared for. But if you do manage to make it successfully into coupledom — slaps on backs and clinking glasses from all of us here – the next question you'll need to ask yourself is this; are they really 'The One'?
Now regardless of whether you're onboard with the idea that there's just one single person out there, among the 7.6 billion people on Earth, that you're destined to spend the rest of your life with, the fact remains that some people are just better suited to each other. Here's a cute Answers post talking about that magic.
iStock / LittleBee80
It's also true that, once you've presumably found this adorably compatible creature, knowing whether you're meant to be together long-term or something akin to lusty convenience, is again, well, a bit tricky.
Do you feel calm, at peace, and genuinely happy? That is a great indicator.
"When you've found The One, the relationship just flows. Things are fairly easy," says the admirably optimistic Jeannie Assimos, Chief of Advice at online dating site eHarmony. "You understand each other's viewpoints and perceptions, and either accept them or feel the same way. If a relationship is characterized by conflict, strife or butting heads on a regular basis, that likely tells you the compatibility is not there."
"A huge indicator that you've found The One? It's just easy being with this person," she says. "You feel at home, totally comfortable, and are able to be yourself. Paying attention to how we feel when we're around someone is important. Do you feel calm, at peace, and genuinely happy? That is a great indicator."
iStock / PeopleImages
Trusting your gut feeling, however, can feel like a leap of faith. So how about a checklist of science-backed indicators instead?
Luckily, there's a cohort of top psychologists and relationship experts out there who have made it their mission to uncover the complexities and subtleties of love's effect on the brain. From changes in your vocabulary to tell-tale Instagram behavior, here's the expert take on whether you're dealing with a fling or the real thing.
Your Brain Changes
A sure-fire indicator of a relationship being the real deal is that you don't experience that 'out of sight, out of mind' phenomenon when your partner isn't around. Instead, you'll tend to think about them a lot – almost all the time, in fact.
Love and genuine attachment actually alter the biochemical reactions taking place in your brain
A 2005 study carried out by researchers at New York's Stony Brook University suggests this is because real love and genuine attachment actually alter the biochemical reactions taking place in your brain.
When you think about The One, you'll get a surge of happiness-boosting neurotransmitter dopamine plus a lighting up of the brain's reward centers.
iStock / nd3000
All of that makes us feel warm and fuzzy, which is why we're more likely to keep indulging in these happy thoughts regularly. Especially in the earlier stages of a strong relationship, when the effects are at their most potent.
Your Pronouns Change
As Assimos quite rightly points out: "The One is not going to try and change you. They'll accept you for who you are, and become your biggest supporter in life."
Those who feel deeply connected to their partner are more likely to use plural pronouns such as 'we' and 'us'
Broadly speaking, that's true. However, one way in which they will inevitably, albeit unintentionally, change you is by influencing your daily pronoun use.
Shutterstock / Milan Ilic Photographer
In a 2002 study, psychologists at the University of Texas at Austin, found that those who feel deeply connected to their partner are more likely to use plural pronouns such as 'we' and 'us', rather than the singular 'I' or 'me'.
The findings have since been confirmed by every person who's ever had to attend a couples' dinner as a singleton. And we feel for you.
You're Willing To Fight
Old Willy Shakes was right on the money when he declared the course of true love to be an often rocky road. But as research psychologist Luis Rubén de Borbón observes, a willingness to fight for the success of your relationship is what really sets The One apart. Even more so than how 'compatible' two people might be on paper.
A successful relationship… hangs on by the sheer will power and want to stay in a relationship.
"Everyone who is unhappy [in their relationship] naturally blames it on the facade of compatibility," he writes. "They fail to realize and comprehend that a successful relationship does not hinge its posterity on how alike you are, instead it hangs on by the sheer will power and want to stay in a relationship."
Shutterstock / Artem Tymoshenko
Leading life coach, Olga Levancuka, agrees: "You need to keep in mind that finding The One doesn't mean finding a clone of you who shares exactly the same passions or meets your exact expectations," she says.
"It means finding someone who is prepared to make your relationship work and you're prepared to do the same. Relationships aren't all about romantic getaways and butterflies, they can be hard work and you both have to be prepared to build solid foundations."
You Happily Sacrifice
As Levancuka describes, if you've met The One, you'll both be cool with making certain sacrifices for each other, even if historically, you've always been more about looking out for yourself as number one.
Psychologists call this 'costly commitment signals' – when we carry out acts or behave in ways that are likely to 'cost' us either in terms of our time, money or our emotions, simply because we want to help our significant other.
Shutterstock / nd3000
A 2015 study carried out by psychologists at Japan's Kobe University, found that relationships where this kind of behavior was either absent or infrequent, were less likely to go the distance.
You Don't Overshare
Those folks who post loved-up couple selfies on the regular? They're likely trying their best to compensate for a bond that, in all reality, just isn't there. According to Levancuka, if your partner isn't plastering your relationship all over their social media feeds, it's actually a strong indicator that all's well in Couplesville.
If they are oversharing it could be a sign of insecurity
"Unfortunately, social media is now an essential part of our daily lives, to the point where every moment needs to be documented and recorded. Happy couples who take their relationship seriously, however, tend not to share much information about their relationship online," she explains.
Shutterstock / Monkey Business Images
"It's perfectly normal for your partner to share the occasional picture of you, but happy couples are busy enjoying each other's company in the present. This means that they're not going to stop enjoying each other's company just to post a status or snap a selfie.
"If they are oversharing it could be a sign of insecurity, or that they're more concerned about how you add to their social image rather than being focused on how you enrich their day-to-day life."
You've Polled Your Friends and Family
According to psychologists, what those closest to us think of our partner can have a huge influence over how a relationship turns out.
When you've found The One, you want everyone in your life to meet them
A 2014 study also found that since those around us can have such an impact on our love lives, it's common to want to 'marshall support' for your budding romance.
So, if you've not only introduced your partner to your friends and family, but also found yourself soliciting feedback from all parties concerned while simultaneously highlighting all your partner's virtues, it's highly likely you've found The One – especially if everyone gelled on the night.
Shutterstock / Syda Productions
"When you've found The One, you want everyone in your life to meet them, and get to know them," says Assimos. "You are genuinely excited about the prospect of being with this person, and you're no longer are looking around to see what else is out there."
They Are Your #1 Cheerleader
According to famed psychological researcher Dr. John Gottman, supporting each other's 'life dreams', aka your long-term goals and ambitions, is a key component of a healthy and lasting relationship.
Couples who focused on building something together, whether it be a business or working towards a shared goal, tended to stay together the longest.
For Gottman, The One should "look up to you, admire you, and respect you." In other words, they'll become your biggest champion and an ever-reliable source of support and encouragement.
Levancuka takes the same view: "The happiest relationships are built on mutual respect, so even if you both have different interests you want to pursue, The One will be there to encourage and support you."
Shutterstock / George Rudy
You'll find yourself heavily invested in helping to build them up, too. In fact, Gottman's years of research found that couples who focused on building something together, whether it be a business or working towards a shared goal, tended to stay together the longest.
Bedroom Gymnastics Isn't Your Only Hobby
Naturally, sex plays a huge part in a healthy relationship, and being physically attracted to your partner as well as enjoying their company is a vital checkpoint.
If all your plans somehow incorporate time alone where you might be able to have sex then the alarm bells should start ringing.
But if sex is the only thing you do together when you've managed to carve out some downtime, we're very sorry to point out that this may not be a good sign.
Shutterstock / Y Photo Studio
"Sex is great and you should be having lots of it, but it's important to build a stronger, emotional bond beyond that," says Levancuka. "If all your plans somehow incorporate time alone where you might be able to have sex then the alarm bells should start ringing. With The One, you need to be able to have fun around each other without the need for sex."
You're In Sync
"When you're synced with your partner you are fully present with that person – mind and body. In a romantic relationship it's important to feel connected to someone in this way. When that connection is there, we feel seen, heard, valued, respected, and cherished," says Levancuka.
But if you thought this whole business of synchronicity was more pseudoscience than anything concrete, a 2016 paper published by research psychologists is here to prove you wrong.
As Levancuka quite rightly asserts: "Scientists have found that some couples are so in tune that their brains begin to work in sync. That means they've reached a state in which their nervous systems are ticking over in harmony, helping them to read each other's thoughts and emotions."
According to the researchers, just sitting next to The One for 15 minutes is enough to sync-up.
Now you've got the checklist, better start paying a closer eye to what's going on up top.
The Best Summer Accessories For Men And How To Wear Them Every finishing touch your style needs this season Plastic Surgery For Men: Everything You Need To Know The why, how, and how much of nips and tucks from your nose to your nutsack How Long Does A Tan Last (And How To Make It Last Longer) Getting one is easy. Holding on to it? Not so much


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As most of you know, at the end of each year, I choose a book with different thoughts for the upcoming 365-days. Giving me different things to ponder on my meditation cushion every day.
Every year, I choose a 365-day book to read on my cushion in the mornings with the my hot water and lemon.
My last entry was about living a life with soft edges… I recently spent three days up in the mountains on a solo retreat practicing all things T'ai Chi (sitting / moving / awareness).
One of my Core Desired Feelings (CDFs) for 2015 is IT (Intuitive Trust). I don't know about you, but at times, I reach and grasp for answers outside of myself. This year, I am really focusing on trusting my intuition. Being more open and recognizing when I listen, acknowledging when I don't, or when I simply ignore it. How do I feel, when I listen, how do I feel when I ignore it?