\chapter{Representations of finite groups}
\label{sec : GLnZ : Representation Theory}

Our goal in Chapter \ref{sec : GLnZ} is to enumerate the finite subgroups of $\generalLinear(n,\integers)$, up to $\integers$-conjugacy, and to compute their first cohomology groups. The framework of representation theory clarifies our approach to enumerating these subgroups, so we survey the general framework here. Our main reference for this section is \cite{Curtis1962}*{Sections 8--11, 14--15, and 73}.% Begin footnote.
\footnote{Also related to our study are Section 16, on modules over principal ideal domains; Sections 30--31 and 38--39, on group characters; and Sections 68--70, on splitting fields and the Schur index. Concepts in these related sections (with the possible exception of the discussion of elementary divisors in Section 16) are not critical to the application we pursue, so we do not discuss them here.}% End footnote.



\section{Algebraic structures}
\label{sec : Rep : Algebraic Structures}

Let $M,N$ be commutative groups, with their group operations written additively. A \fontDefWord{homomorphism} (of commutative groups) from $M$ to $N$ is a map $\varphi : M \rightarrow N$ that preserves the group structure; namely, for all $m_{1},m_{2} \in M$,
\begin{align*}
\varphi(m_{1} + m_{2})
=
\varphi(m_{1}) + \varphi(m_{2})
\end{align*}
Let $\Hom(M,N)$ denote the set of all such homomorphisms from $M$ to $N$. In the case $N = M$, denote $\Hom(M) = \Hom(M,M)$.

Given $\varphi_{1},\varphi_{2} \in \Hom(M,N)$, define
\begin{align}
\varphi_{1} + \varphi_{2}
:
M
&\rightarrow
N%
\nonumber
\\
m
&\mapsto
\varphi_{1}(m) + \varphi_{2}(m)%
\label{eq : GLnZ : Addition On Hom(M,N)}
\end{align}
Then $\Hom(M,N)$, equipped with this operation $+$, is a commutative group. In the case $N = M$, we may define another operation on the set of homomorphisms: Given $\varphi_{1},\varphi_{2} \in \Hom(M)$, define
\begin{align}
\varphi_{1} \cdot \varphi_{2}
:
M
&\rightarrow
M%
\nonumber
\\
m
&\mapsto
\varphi_{1}(\varphi_{2}(m))%
\label{eq : GLnZ : Multiplication on Hom(M)}
\end{align}
Then $\Hom(M)$, equipped with these operations $+$ and $\cdot$, is a ring. This ring is not commutative, in general, but it does contain a multiplicative identity, namely, the identity map $\identityMap_{M}$.

Now let $R$ be a ring with multiplicative identity $1$; and let $M,N$ be $R$-modules. A \fontDefWord{homomorphism} (of $R$-modules) from $M$ to $N$ is a map $\varphi : M \rightarrow N$ that preserves the $R$-module structure; namely, for all $r \in R$, for all $m,m_{1},m_{2} \in M$,
\begin{align*}
\varphi(m_{1} + m_{2})
&=
\varphi(m_{1}) + \varphi(m_{2})
&
\varphi(r m)
&=
r \varphi(m)
\end{align*}
Let $\Hom_{R}(M,N)$ denote the set of all such homomorphisms from $M$ to $N$. As above, in the case $N = M$, denote $\Hom_{R}(M) = \Hom_{R}(M,M)$. We may define an operation $+$ on $\Hom_{R}(M,N)$ as in \eqref{eq : GLnZ : Addition On Hom(M,N)}. Note that $\Hom_{R}(M,N)$ is a subgroup of $\Hom(M,N)$. We may also define an operation of scalar multiplication on $\Hom_{R}(M,N)$: Given $r \in R$, $\varphi \in Hom_{R}(M,N)$, define
\begin{align}
r \varphi
:
M
&\rightarrow
N%
\nonumber
\\
m
&\mapsto
r \varphi(m)
\end{align}
Then $\Hom_{R}(M,N)$, equipped with $+$ and this scalar multiplication, is an $R$-module. In the case $N = M$, we may also define the operation $\cdot$ on $\Hom_{R}(M)$ as in \eqref{eq : GLnZ : Multiplication on Hom(M)}. Then $\Hom_{R}(M)$ is both an $R$-module and a ring. Scalar multiplication and ring multiplication are related by the following rules: for all $r \in R$, for all $\varphi_{1},\varphi_{2} \in \Hom_{R}(M)$,
\begin{align*}
r (\varphi_{1} \varphi_{2})
=
(r \varphi_{1}) \varphi_{2}
=
\varphi_{1} (r \varphi_{2})
\end{align*}
Thus $\Hom_{R}(M)$ is an $R$-algebra.

The invertible elements of the ring $\Hom_{R}(M)$ form a subgroup (under the operation of multiplication, that is, of function composition), called the \fontDefWord{general linear group of $M$}, denoted $\generalLinear(M)$.

\begin{remark}
Let $K$ be a field. Taking $R = K$ in the above discussion, an $R$-module is a $K$-vector space, and $\Hom_{K}(M)$ is a $K$-algebra.
\end{remark}

Now let $R$ be a commutative ring with multiplicative identity, and let $M$ be a free $R$-module of finite rank $n$. A choice of basis $B = (m_{1},\ldots,m_{n})$ of $M$ induces an isomorphism $\fontMatrixRep{M}_{B} : M \rightarrow R^{n}$ of $R$-modules. More precisely, given $m \in M$, express $m$ as its unique linear combination $m = \sum_{i = 1}^{n} r_{i} m_{i}$ of the basis vectors, and map $m$ to this corresponding vector of coefficients:
\begin{align*}
\fontMatrixRep{M}_{B}(m)
=
(r_{1},\ldots,r_{n})
\end{align*}
Similarly, fixing a basis $B$ of $M$ induces an isomorphism $\fontMatrixRep{M}_{B,B} : \Hom_{R}(M) \rightarrow \matrixAlgebra(n \times n,R)$, both of $R$-modules and of rings. More precisely, given $T \in \Hom_{R}(M)$, for each basis vector $m_{i} \in B$, express $T(m_{i})$ as its unique linear combination $T(m_{i}) = \sum_{j = 1}^{n} r_{j,i} m_{j}$ of the basis vectors, and define
\begin{align}
\fontMatrixRep{M}_{B,B}(T)
=
(r_{j,i})_{i,j}%
\label{eq : GLnZ : HomRM To Matrix}
\end{align}
where the right side is the $n \times n$ matrix with entries in $R$, whose column $i$ is the vector $\fontMatrixRep{M}_{B}(T(m_{i}))$.

In what follows, we will often suppress the choice of basis $B$ and denote the matrix $\fontMatrixRep{M}_{B,B}(T)$ by boldface $\fontMatrixRep{T}$.

Let $T \in \Hom_{R}(M)$. A change of basis of $M$ corresponds to conjugation by an element of $\generalLinear(M)$. More precisely, let $B = (m_{1},\ldots,m_{n})$ and $B' = (m'_{1},\ldots,m'_{n})$ be bases of $M$. Using the procedure above, define the matrix $\fontMatrixRep{S} = \fontMatrixRep{M}_{B,B'}(\identityMap_{M})$, whose column $i$ is the vector $\fontMatrixRep{M}_{B'}(m_{i})$. Direct computation shows that $\fontMatrixRep{S}^{-1} = \fontMatrixRep{M}_{B',B}(\identityMap_{M})$ and
\begin{align}
\fontMatrixRep{T}'
=
\fontMatrixRep{S} \fontMatrixRep{T} \fontMatrixRep{S}^{-1}%
\label{eq : GLnZ : Change Of Basis Equivalence}
\end{align}
where $\fontMatrixRep{T}' = \fontMatrixRep{M}_{B',B'}(T)$.

We remark that $T \in \generalLinear(M)$ if and only if $\det T$ (equal to the matrix determinant $\det \fontMatrixRep{T}$, for any basis $B$ of $M$) is a unit in $R$, that is, if and only if $\det T \in R^{\times}$.



\section{Representations}

Throughout this section, let $G$ be a finite group, with its group operation written multiplicatively; let $R$ be a commutative ring with multiplicative identity, denote it $1$; and let $M$ be a free $R$-module of finite rank $n$.

A \fontDefWord{(group) representation of $G$ with representation space $M$} is a group homomorphism $T : G \rightarrow \generalLinear(M)$. The \fontDefWord{degree} of $T$, denoted $\deg T$, is the rank of $M$ as an $R$-module.

\begin{remark}
More generally, because $\Hom_{R}(M)$ admits the structure of an $R$-al\-ge\-bra, we may require our representation to preserve any substructure---group, ring, $R$-module, $R$-algebra---depending on the object we wish to represent and the structure we wish to preserve. For example, let $A$ be an $R$-algebra with multiplicative identity $1_{A}$, and let $M$ be an $R$-module. An \fontDefWord{(algebra) representation of $A$ with representation space $M$} is an algebra homomorphism $T : A \rightarrow \Hom_{R}(M)$. That is, the map $T$ satisfies, for all $a,a_{1},a_{2} \in A$, for all $r \in R$,
\begin{align*}
T(a_{1} + a_{2})
&=
T(a_{1}) + T(a_{2})
&
T(r a)
&=
r T(a)
\\
T(a_{1} a_{2})
&=
T(a_{1}) T(a_{2})
&
T(1_{A})
&=
\identityMap_{M}
\end{align*}
\end{remark}

For $i \in \{1,2\}$, let $M_{i}$ be a free $R$-module of finite rank $n_{i}$, and let $T_{i} : G \rightarrow \generalLinear(M_{i})$ be a group representation of $G$ with representation space $M_{i}$. The representations $T_{1}$ and $T_{2}$ are \fontDefWord{equivalent (as representations on $R$-modules)}, denoted $T_{1} \sim_{R} T_{2}$, if there exists an isomorphism $S \in \Hom_{R}(M_{1},M_{2})$ such that, for all $g \in G$, the following diagram commutes:
\begin{align}
\begin{tikzpicture}
% Nodes:
\node (TL) {$M_{1}$};
\node (TR) [right = of TL] {$M_{1}$};
\node (BL) [below = of TL] {$M_{2}$};
\node (BR) [below = of TR] {$M_{2}$};
% Arrows:
\draw[-stealth] (BL) -- node [left] {$S^{-1}$} (TL);
\draw[-stealth] (TR) -- node [right] {$S$} (BR);
\draw[-stealth] (TL) -- node [above] {$T_{1}(g)$} (TR);
\draw[-stealth] (BL) -- node [above] {$T_{2}(g)$} (BR);
\end{tikzpicture}%
\label{eq : GLnZ : Equivalent Representations : Diagram}
\end{align}
That is, for all $g \in G$,
\begin{align*}
T_{2}(g)
=
S T_{1}(g) S^{-1}
\end{align*}
We emphasize that the isomorphism $S$ is the same for all $g \in G$.

A \fontDefWord{matrix representation of $G$ of degree $n$ over $R$} is a group homomorphism $\fontMatrixRep{T} : G \rightarrow \generalLinear(n,R)$. For $i \in \{1,2\}$, let $\fontMatrixRep{T}_{i} : G \rightarrow \generalLinear(n_{i},R)$ be a matrix representation of $G$ of degree $n_{i}$ over $R$. The matrix representations $\fontMatrixRep{T}_{1}$ and $\fontMatrixRep{T}_{2}$ are \fontDefWord{equivalent (as matrix representations over $R$)}, denoted $\fontMatrixRep{T}_{1} \sim_{R} \fontMatrixRep{T}_{2}$, if (1) $n_{1} = n_{2}$, denote it $n$; and (2) there exists an $\fontMatrixRep{S} \in \generalLinear(n,R)$ such that, for all $g \in G$,
\begin{align}
\fontMatrixRep{T}_{2}(g)
=
\fontMatrixRep{S} \fontMatrixRep{T}_{1}(g) \fontMatrixRep{S}^{-1}%
\label{eq : GLnZ : Equivalent Matrix Representations : Equation}
\end{align}
Condition \eqref{eq : GLnZ : Equivalent Matrix Representations : Equation} is simply a translation of diagram \eqref{eq : GLnZ : Equivalent Representations : Diagram} to the language of matrix representations. As there, we emphasize that the matrix $\fontMatrixRep{S}$ is the same for all $g \in G$.

As before, let $M$ be a free $R$-module of finite rank $n$. The isomorphism $\fontMatrixRep{M}_{-,-} : \Hom_{R}(M) \rightarrow \matrixAlgebra(n \times n,R)$ defined in \eqref{eq : GLnZ : HomRM To Matrix} connects the two representations we have discussed. More precisely, let $T : G \rightarrow \generalLinear(M)$ be a (group) representation of $G$ with representation space $M$, and let $B$ be a basis of $M$. For each $g \in G$, we have $\fontMatrixRep{M}_{B,B}(T(g)) \in \generalLinear(n,R)$. Aggregating these data, the map
\begin{align*}
\fontMatrixRep{T}
:
G
&\rightarrow
\generalLinear(n,R)
\\
g
&\mapsto
\fontMatrixRep{M}_{B,B}(T(g))
\end{align*}
defines a matrix representation of $G$ of degree $n$ over $R$, called the \fontDefWord{matrix representation induced by $T$ and $B$}.% Begin footnote.
\footnote{We say that $T$ \fontDefWord{affords} the matrix representation $\fontMatrixRep{T}$, in the sense that with respect to the basis $B$, $T$ induces the matrix representation $\fontMatrixRep{T}$. Of course, $T$ may afford other matrix representations, arising from other choices of basis. In light of this, the notation $\fontMatrixRep{M}_{B,B}(T)$ is more precise than $\fontMatrixRep{T}$.} % End footnote.
For a fixed representation $T$, different bases $B$ of $M$ may induce different matrix representations $\fontMatrixRep{T}$, all of which are equivalent by \eqref{eq : GLnZ : Equivalent Matrix Representations : Equation} and \eqref{eq : GLnZ : Change Of Basis Equivalence}.

\begin{remark}
To define a representation $T$ of $G$, it suffices to define $T(g_{i})$ for a set $\{g_{i}\}_{i}$ of generators of $G$. If any of these generators satisfy a (nontrivial) relation, then because $T$ is a group homomorphism, the corresponding images $T(g_{i})$ must satisfy the same relation. The definition of $T$ on the generators $g_{i}$ may then be extended to all of $G$ by using the property that $T$ is a group homomorphism.
\end{remark}

Fix a representation $T : G \rightarrow \generalLinear(M)$. A \fontDefWord{$G$-submodule} (or, in the case when $R$ is a field, a \fontDefWord{$G$-subspace}) of $M$ is a submodule $N \subspaceeq M$ such that, for all $g \in G$, $T(g) N \subseteq N$.% Begin footnote.
\footnote{In fact, if this inclusion holds, then it holds with equality, because $T(g) \in \generalLinear(M)$. Note that the definition of $G$-submodule requires a representation $T$ of $G$, which the terminology ``$G$-submodule'' leaves implicit.} % End footnote.
Let $N \submoduleeq M$ be a $G$-submodule. Then the map
\begin{align*}
T_{1}
:
G
&\rightarrow
\generalLinear(N)
\\
g
&\mapsto
T(g)|_{N}
\end{align*}
is a representation of $G$ with representation space $N$.

Assume that $M \neq \{0\}$. $T$ is \fontDefWord{reducible} if there exists a nontrivial, proper $G$-submodule $N \submodule M$. $T$ is \fontDefWord{irreducible} if it is not reducible. $T$ is \fontDefWord{completely reducible} if for each $G$-submodule $N \submoduleeq M$, there exists a companion $G$-submodule $N' \submoduleeq M$ such that $M = N \directSum N'$.

Now let $K$ be a field; let $M$ be a nontrivial, finite-dimensional $K$-vector space, and denote $n = \dim_{K} M$; and let $T : G \rightarrow \generalLinear(M)$ be a representation of $G$ with representation space $M$. Suppose that $T$ is reducible, so there exists a nontrivial, proper $G$-subspace $N \submodule M$. Let $B_{N} = (m_{1},\ldots,m_{n_{1}})$ be a basis of $N$; extend this to a basis $B = (m_{1},\ldots,m_{n})$ of $M$. Let $\fontMatrixRep{T} : G \rightarrow \generalLinear(n,K)$ be the matrix representation induced by $T$ and $B$. For each $g \in G$, $\fontMatrixRep{T}(g)$ has the block upper-triangular form
\begin{align}
\fontMatrixRep{T}(g)
=
\begin{bmatrix}
\fontMatrixRep{T}_{1}(g)	&	\fontMatrixRep{V}(g)		\\
\fontMatrixRep{0}			&	\fontMatrixRep{T}_{2}(g)
\end{bmatrix}%
\label{eq : GLnZ : Matrix Representation Reducible}
\end{align}
where $\fontMatrixRep{T}_{1} : G \rightarrow \generalLinear(N)$ is the matrix representation induced by $T_{1}$ (as defined above) and $B_{N}$. Consider the map
\begin{align*}
T_{2}
:
G
&\rightarrow
\generalLinear(M/N)
\\
g
&\mapsto
\left(m + N \mapsto (T(g))(m) + N\right)
\end{align*}
It is straightforward to verify that
\begin{enumerate}
\item $T_{2}$ is well defined (because $N$ is a $G$-subspace of $M$),
\item $B_{2} = (m_{n_{1} + 1} + N,\ldots,m_{n} + N)$ is a $K$-basis of $M/N$, and
\item $\fontMatrixRep{T}_{2}$ is the matrix representation induced by $T_{2}$ and $B_{2}$.
\end{enumerate}
Let $n_{2} = n - n_{1}$. Then for each $g \in G$, the matrix $\fontMatrixRep{V}(g)$ in \eqref{eq : GLnZ : Matrix Representation Reducible} is in $\matrixAlgebra(n_{1} \times n_{2},K)$. We may aggregate these data into a map $\fontMatrixRep{V} : G \rightarrow \matrixAlgebra(n_{1} \times n_{2},K)$, which we call the \fontDefWord{binding function} for the pair $(\fontMatrixRep{T}_{1},\fontMatrixRep{T}_{2})$.% Begin footnote.
\footnote{Binding functions are discussed in more detail in Section \ref{sec : GLnZ : Matrix Representations Over Z}.}% End footnote.

Let $B'$ be another basis of $M$, and let $\fontMatrixRep{T}'$ be the matrix representation of $G$ induced by $T$ and $B'$. Then $\fontMatrixRep{T}' \sim_{K} \fontMatrixRep{T}$, and there exists some $\fontMatrixRep{S} \in \generalLinear(n,K)$ such that, for all $g \in G$,
\begin{align*}
\fontMatrixRep{S} \fontMatrixRep{T}'(g) \fontMatrixRep{S}^{-1}
=
\fontMatrixRep{T}(g)
\end{align*}
Thus, given a reducible representation $T$ and any matrix representation $\fontMatrixRep{T}'$ that it affords, we may always find a change of basis that transforms $\fontMatrixRep{T}'$ into the block upper-triangular form in \eqref{eq : GLnZ : Matrix Representation Reducible}.

Now suppose that $T$ is completely reducible. Let $N \subspaceeq M$ be a $G$-subspace of $M$.% Begin footnote.
\footnote{In what follows, it may be helpful to assume that the $G$-subspace $N \subspaceeq M$ is nontrivial and proper. However, this is not required by our definitions. If $T$ is irreducible, then the matrix in \eqref{eq : GLnZ : Matrix Representation Reducible} consists of a single block.} % End footnote.
As previously, we may choose a basis $B_{N} = (m_{1},\ldots,m_{n_{1}})$ of $N$, and extend this to a basis $B = (m_{1},\ldots,m_{n})$ of $M$. In this case, we may choose the vectors $m_{n_{1} + 1},\ldots,m_{n}$ such that, defining $N' = \Span_{k}(m_{n_{1} + 1},\ldots,m_{n})$ and $B_{N'} = (m_{n_{1} + 1},\ldots,m_{n})$, we have
\begin{align*}
M
=
N \directSum N'
\end{align*}
In this setting, the matrix representation \eqref{eq : GLnZ : Matrix Representation Reducible} of $G$ induced by $T$ and $B$ has the block diagonal form
\begin{align}
\fontMatrixRep{T}(g)
=
\begin{bmatrix}
\fontMatrixRep{T}_{1}(g)	&	\fontMatrixRep{0}			\\
\fontMatrixRep{0}			&	\fontMatrixRep{T}_{2}(g)
\end{bmatrix}%
\label{eq : GLnZ : Matrix Representation Completely Reducible}
\end{align}
That is, for all $g \in G$, $\fontMatrixRep{V}(g)$ is the $n_{1} \times n_{2}$ zero matrix. An argument analogous to that in the preceding paragraph shows that when $T$ is completely reducible, every matrix representation induced by $T$ is equivalent to one in block diagonal form.

For matrix representations, we may rephrase the notions of reducibility in terms of matrices. Let $\fontMatrixRep{T}' : G \rightarrow \generalLinear(n,K)$ be a matrix representation. $\fontMatrixRep{T}'$ is \fontDefWord{reducible} if it equivalent to a matrix representation in the block upper-triangular form \eqref{eq : GLnZ : Matrix Representation Reducible}, with $n_{1},n_{2} > 0$. $\fontMatrixRep{T}'$ is \fontDefWord{irreducible} if it is not reducible. $\fontMatrixRep{T}'$ is \fontDefWord{completely reducible} if, whenever $\fontMatrixRep{T}'$ is equivalent to a matrix representation in the block upper-triangular form \eqref{eq : GLnZ : Matrix Representation Reducible}, it is also equivalent to a matrix representation in the block diagonal form \eqref{eq : GLnZ : Matrix Representation Completely Reducible}.



\section{From representations to modules}
\label{sec : Rep : From Representations To Modules}

Throughout this section, let $G$ be a finite group, with its group operation written multiplicatively, and with identity element $e$; let $R$ be a commutative ring with multiplicative identity, denote it $1$; let $M$ be a free $R$-module of finite rank $n$; and let $T : G \rightarrow \generalLinear(M)$ be a group representation of $G$ with representation space $M$.

The \fontDefWord{group ring of $G$ over $R$}, denoted $R G$, is the set of all formal sums
\begin{align*}
\sum_{g \in G} r_{g} g
\end{align*}
equipped with two operations: addition, defined ``componentwise'', that is,
\begin{align*}
\left(\sum_{g \in G} r_{g} g\right) + \left(\sum_{g \in G} r'_{g} g\right)
=
\sum_{g \in G} (r_{g} + r'_{g}) g
\end{align*}
and multiplication, defined by convolution, that is,
\begin{align*}
\left(\sum_{g_{1} \in G} r_{g_{1}} g_{1}\right) \left(\sum_{g_{2} \in G} r_{g_{2}} g_{2}\right)
=
\sum_{g \in G} \left(\underset{g_{1} g_{2} = g}{\sum_{g_{1},g_{2} \in G}} r_{g_{1}} r_{g_{2}}\right) g
\end{align*}
With these operations, $R G$ forms a ring with multiplicative identity $1 e$. We may equip $R G$ with a third operation: scalar multiplication by elements of $R$, defined by, for all $r \in R$,
\begin{align*}
r \sum_{g \in G} r_{g} g
=
\sum_{g \in G} (r r_{g}) g
\end{align*}
These three operations give $R G$ the structure of an $R$-algebra.

We now use the representation $T : G \rightarrow \generalLinear(M)$ to give $M$ the structure of an $R G$-module, which encodes all the data of $T$. Leave the structure of $M$ as a commutative group unchanged. Extend the extra structure of $M$ as an $R$-module to that of an $R G$-module as follows: for all $\sum_{g \in G} r_{g} g \in R G$, for all $m \in M$, define
\begin{align*}
\left(\sum_{g \in G} r_{g} g\right) m
=
\sum_{g \in G} r_{g} (T(g))(m)
\end{align*}
where the sum on the right is evaluated within the given structure of $M$ as an $R$-module.

Conversely, given an $R G$-module $M$, we may recover a unique representation $T : G \rightarrow \generalLinear(M)$ by defining, for each $g \in G$,
\begin{align*}
T(g)
:
M
&\rightarrow
M
\\
m
&\mapsto
(1 g) m
\end{align*}
Note that, to define each map $T(g) \in \Hom_{R}(M)$, it suffices to compute $(1 g) m_{i}$ for each $m_{i}$ in a basis $(m_{1},\ldots,m_{n})$ of $M$, then extend by linearity to all of $M$.

Thus we may translate between the theory of group representations and the theory of modules.



\section{Chain conditions and composition series}
\label{sec : Rep : Chain Conditions And Composition Series}

As before, let $R$ be a commutative ring with multiplicative identity, and let $M$ be a free $R$-module of finite rank $n$.

The $R$-module $M$ satisfies the \fontDefWord{descending chain condition (DCC)} if every descending chain of submodules stabilizes, that is, if for every chain $M_{1} \supseteq M_{2} \supseteq \ldots$ of submodules $M_{i} \submoduleeq M$, there exists an index $i_{0}$ such that for all $i \geq i_{0}$, $M_{i} = M_{i_{0}}$. $M$ satisfies the \fontDefWord{ascending chain condition (ACC)} if every ascending chain of submodules stabilizes. In this case, we call $M$ \fontDefWord{noetherian}. A ring $R$ is \fontDefWord{(left) noetherian} if its left regular module ${}_{R}R$ is noetherian.% Begin footnote.
\footnote{The \fontDefWord{left regular module} ${}_{R}R$ is the left $R$-module $M$ formed by taking $M = R$, that is, taking the commutative group of the module to be $R$ equipped with its addition operation, and defining scalar multiplication using the ring multiplication in $R$.}% End footnote.

\begin{remark}
\label{rmk : Finite Dimensional Vector Space ACC DCC}
Consider the case $R = K$ is a field, so that $M$ is a finite-dimensional $K$-vector space. Then $M$ satisfies both the ACC and DCC for $K$-subspaces.
\end{remark}

The following are equivalent:
\begin{enumerate}[label=(\arabic*)]
\item\label{itm : Noetherian Module Equivalence : Noetherian} $M$ satisfies the ACC (in other words, $M$ is noetherian).
\item\label{itm : Noetherian Module Equivalence : Finitely Generated} Each submodule of $M$ is finitely generated.
\item\label{itm : Noetherian Module Equivalence : Maximal Element} Let $\mathcal{N}$ be a nonempty collection of submodules $N \submoduleeq M$. Then $\mathcal{N}$ contains a maximal element, that is, there exists some $N_{0} \in \mathcal{N}$ such that, for all $N \in \mathcal{N}$ with $N \neq N_{0}$, $N_{0} \not\submoduleeq N$.
\end{enumerate}
An equivalence analogous to ``\ref{itm : Noetherian Module Equivalence : Noetherian} is equivalent to \ref{itm : Noetherian Module Equivalence : Maximal Element}'' is true for the DCC and minimal elements of $\mathcal{N}$.

If $M$ is finitely generated (which we are assuming in this section), then every submodule of $M$ is finitely generated.

Let $M$ be nontrivial. $M$ is \fontDefWord{reducible} if $M$ contains a nontrivial, proper submodule. $M$ is \fontDefWord{irreducible} if it is not reducible. Let $\mathcal{N}$ be the collection of all submodules of $M$, equipped with the partial order of set inclusion. (Note that the hypothesis that $M$ is nontrivial implies that $\mathcal{N}$ contains at least two distinct elements, $(0)$ and $M$.) Let $N$ be a submodule of $M$. $N$ is \fontDefWord{maximal} if $N$ is a maximal element of $\mathcal{N} \setminus \{M\}$. $N$ is \fontDefWord{minimal} if $N$ is a minimal element of $\mathcal{N} \setminus \{(0)\}$. Note that $N$ is maximal if and only if $M/N$ is irreducible. $N$ is minimal if and only if $N$ is irreducible. These equivalences may be illuminated by considering the sublattice of $\mathcal{N}$ above and below $N$, respectively (with $N$ included in both sublattices).

A \fontDefWord{composition series} $\mathcal{M}$ of $M$ is a strictly descending chain
\begin{align*}
M
=
M_{z}
\supset
M_{z - 1}
\supset
\ldots
\supset
M_{0}
=
(0)
\end{align*}
of submodules $M_{i} \submoduleeq M$, starting with $M$ and ending with $(0)$, such that for each $i$, $M_{i}/M_{i - 1}$ is irreducible. The quotients $M_{i}/M_{i - 1}$ of consecutive submodules are called the \fontDefWord{factors} of $\mathcal{M}$. The \fontDefWord{length} of $\mathcal{M}$ is its number of factors. Let $\mathcal{M},\mathcal{M}'$ be composition series of $M$. $\mathcal{M}$ and $\mathcal{M}'$ are \fontDefWord{equivalent}, denoted $\mathcal{M} \sim \mathcal{M}'$, if (1) they have the same length and (2) the factors of $\mathcal{M}$ and $\mathcal{M}'$ can be paired such that the factors within each pair $(M_{i},M_{j}')$ are isomorphic as $R$-modules.

\begin{lemma}
\label{lem : Composition Series ACC DCC}
$M$ has a composition series if and only if $M$ satisfies both the ACC and DCC.
\end{lemma}

\begin{lemma}
Let $M$ have a composition series, denote its length $z$. Then any composition series of $M$ has the same length $z$. Moreover, any strictly descending chain of submodules of $M$ can be augmented to a composition series of $M$ by inserting submodules into the chain.
\end{lemma}



\section{Key results}

In our application to matrix representations in Section \ref{sec : GLnZ : Matrix Representations Over Z}, we use three key results from the general theory of representations: the Jordan--H\"{o}lder theorem, the Krull--Schmidt theorem, and Maschke's theorem. We present these results here. Throughout this section, unless noted otherwise, let $R$ be a ring with multiplicative identity, and let $M$ be an $R$-module.



\subsection{The {J}ordan--{H}{\"{o}}lder theorem}
\label{sec : Rep : The Jordan Holder Theorem}

The first key result is the Jordan--H\"{o}lder theorem.

\begin{theorem}[Jordan--H\"{o}lder]
Let $M$ be an $R$-module with a composition series. Then any two composition series for $M$ are equivalent.
\end{theorem}

In the setting of matrix representations, the Jordan--H\"{o}lder theorem implies a powerful equivalence. Let $G$ be a finite group, let $K$ be a field, let $M$ be a finite-dimensional $K$-vector space, and let $T : G \rightarrow \generalLinear(M)$ be a group representation of $G$ with representation space $M$. Form the corresponding group algebra $K G$, then use the theory developed above to view $M$ as a $(K G)$-module. It follows from Remark \ref{rmk : Finite Dimensional Vector Space ACC DCC} that $M$ satisfies both the ACC and the DCC for $(K G)$-submodules. Thus $M$ has a composition series, denote it $\mathcal{M}$, say
\begin{align*}
M
=
M_{z}
\supset
\ldots
\supset
M_{0}
=
(0)
\end{align*}
where each $M_{i}$ is a $(K G)$-submodule of $M$, or equivalently, a $G$-subspace of $M$. By iterating the method we used to construct the basis yielding the block upper-triangular matrix \eqref{eq : GLnZ : Matrix Representation Reducible}, we may construct a $K$-basis $B$ of $M$ such that the matrix representation $\fontMatrixRep{T}$ of $G$ induced by $T$ and $B$ has, for all $g \in G$, the block upper-triangular form
\begin{align}
\fontMatrixRep{T}(g)
=
\begin{bmatrix}
\fontMatrixRep{T}_{z}(g)	&			&	\fontMatrixRep{*}			\\
				&	\ddots	&					\\
\fontMatrixRep{0}			&			&	\fontMatrixRep{T}_{1}(g)
\end{bmatrix}%
\label{eq : GLnZ : Jordan Holder Matrix Representation}
\end{align}
where each $\fontMatrixRep{T}_{i}$ is an irreducible matrix representation of $G$ induced by the representation $T_{i} : G \rightarrow \generalLinear(M_{i}/M_{i - 1})$ and a basis corresponding to the relevant subsequence of $B$. Let $\fontMatrixRep{T}' : G \rightarrow \generalLinear(M)$ be another matrix representation induced by $T$ and another $K$-basis $B'$ of $M$. Then, as we have seen, $\fontMatrixRep{T}' \sim_{K} \fontMatrixRep{T}$. That is, letting $n = \dim_{K} M$, there exists a fixed matrix $\fontMatrixRep{S} \in \generalLinear(n,K)$ such that, for all $g \in G$,
\begin{align}
\fontMatrixRep{S} \fontMatrixRep{T}'(g) \fontMatrixRep{S}^{-1}
=
\fontMatrixRep{T}(g)%
\label{eq : GLnZ : Jordan Holder Matrix Equivalence}
\end{align}
In particular, if for all $g \in G$, $\fontMatrixRep{T}'$ also has the block upper-triangular form \eqref{eq : GLnZ : Jordan Holder Matrix Representation}---a priori, with a different number $z'$ of diagonal blocks and different on-diagonal matrix representations $\fontMatrixRep{T}_{i}'$---then the basis $B'$ corresponds to a composition series $\mathcal{M}'$ of $M$. By the Jordan--H\"{o}lder theorem, $\mathcal{M}' \sim \mathcal{M}$, and hence $z' = z$, and we may order the on-diagonal matrix representations $(\fontMatrixRep{T}_{i}')_{i}$ and $(\fontMatrixRep{T}_{i})_{i}$ such that, for each $i$, $\fontMatrixRep{T}_{i}' \sim_{K} \fontMatrixRep{T}_{i}$.



\subsection{The {K}rull--{S}chmidt theorem}

The second key result is the Krull--Schmidt theorem. We first introduce the notion of an indecomposable module.

$M$ is \fontDefWord{indecomposable} if $M \neq (0)$ and $M$ cannot be written as a direct sum of two nontrivial, proper submodules. If $M$ satisfies the DCC, then $M$ can be written as a finite direct sum of indecomposable submodules. A direct-sum decomposition
\begin{align*}
M
=
\bigoplus_{i = 1}^{z} M_{i}
\end{align*}
is equivalent to the existence of projection maps $\pi_{i} \in \Hom_{R}(M)$ such that
\begin{align*}
\sum_{i = 1}^{z} \pi_{i}
=
\identityMap_{M}
\end{align*}
and, for all indices $i,j$ (with $i \neq j$),
\begin{align*}
\pi_{i}^{2}
&=
\pi_{i}
&
\pi_{i} \circ \pi_{j}
&=
0
=
\pi_{j} \circ \pi_{i}
&
\image \pi_{i}
&=
M_{i}
\end{align*}

Comparing the definitions of irreducible and indecomposable modules, we note that irreducible is stronger---it forbids the existence of \emph{any} nontrivial, proper submodules of $M$---and implies indecomposable. As an example, let $R = \integers$ and $M = \integers$, with scalar multiplication defined by multiplication in $\integers$. Then $M$ is reducible yet indecomposable.% Begin footnote.
\footnote{To see that the $\integers$-module $M = \integers$ is reducible, let $n \in \integers$ such that $\abs{n} \geq 2$. Then $N = n M$ is a nontrivial, proper submodule of $M$. To see that $M$ is indecomposable, let $N_{1},N_{2}$ be nontrivial, proper submodules of $M$ such that $M = N_{1} + N_{2}$. Then there exist $n_{i} \in N_{i}$ such that $1 = n_{1} + n_{2}$. By hypothesis on the $N_{i}$, for both $i$, $n_{i} \neq 1$, and therefore also $n_{i} \neq 0$. It follows that $n_{1} n_{2}$ is a nonzero element in $N_{1} \cap N_{2}$, and hence the sum $M = N_{1} + N_{2}$ is not direct.}% End footnote.

\begin{theorem}[Krull--Schmidt]
Let $M$ satisfy both the ACC and DCC, and let
\begin{align*}
M
&=
\bigoplus_{i = 1}^{z} M_{i}
&
M
&=
\bigoplus_{i = 1}^{z'} M_{i}'
\end{align*}
be two direct-sum decompositions of $M$ into indecomposable submodules. Then $z = z'$, and we can reorder the sequences $(M_{i})_{i}$ and $(M_{i}')_{i}$ such that, for each $i$, $M_{i} \isomorphism M_{i}'$ as $R$-modules.
\end{theorem}

$M$ is \fontDefWord{completely reducible} if, for every submodule $N \submoduleeq M$, there exists a companion submodule $N' \submoduleeq M$ such that $M = N \directSum N'$. The following are equivalent:
\begin{enumerate}[label=(\arabic*)]
\item $M$ is completely reducible.
\item $M$ can be written as a direct sum of irreducible submodules.
\item $M$ can be written as a sum (not necessarily direct) of irreducible submodules.
\end{enumerate}

\begin{lemma}
Let $M$ be completely reducible. Then $M$ satisfies the ACC if and only if $M$ satisfies the DCC.
\end{lemma}

\begin{corollary}
Let $M$ satisfy either the ACC or the DCC. Then $M$ is completely reducible if and only if $M$ can be written as a finite direct sum of irreducible submodules.
\end{corollary}



\subsection{{M}aschke's theorem}
\label{sec : Rep : Maschke Theorem}

The third key result is Maschke's theorem. Let $G$ be a finite group, let $K$ be a field, let $M$ be a finite-dimensional $K$-vector space, and let $T : G \rightarrow \generalLinear(M)$ be a representation.

\begin{theorem}[Maschke's theorem]
If $\characteristic(K) \notDivides \order(G)$, then $T$ is completely reducible.
\end{theorem}

In the framework of modules, we may translate Maschke's theorem thus: Let $M$ be a (left) $(K G)$-module. If $\characteristic(K) \notDivides \order(G)$, then $M$ is completely reducible.

If $T$ is completely reducible---for example, if the hypotheses of Maschke's theorem are satisfied---then $M$ can be written as a direct sum of irreducible $G$-subspaces.

Applying these results to matrix representations, we get the following.

\begin{corollary}
\label{cor : GLnZ : Maschke Theorem Matrix Representation}
Let $\fontMatrixRep{T} : G \rightarrow \generalLinear(n,K)$, and let $\characteristic(K) \notDivides \order(G)$. Then there exists a fixed matrix $\fontMatrixRep{S} \in \generalLinear(n,K)$ such that, for all $g \in G$,
\begin{align}
\fontMatrixRep{S} \fontMatrixRep{T}(g) \fontMatrixRep{S}^{-1}
=
\begin{bmatrix}
\fontMatrixRep{T}_{1}(g)	&			&	\fontMatrixRep{0}			\\
				&	\ddots	&					\\
\fontMatrixRep{0}			&			&	\fontMatrixRep{T}_{z}(g)
\end{bmatrix}%
\label{eq : GLnZ : Maschke Theorem Matrix Representation}
\end{align}
where each $\fontMatrixRep{T}_{i} : G \rightarrow \generalLinear(n_{i},K)$ is an irreducible matrix representation.
\end{corollary}



\subsection{Preview}

We end with a preview of our ultimate application of this theory. Let $G$ be a finite group, and let $\fontMatrixRep{T} : G \rightarrow \generalLinear(n,\rationals) \isomorphism \generalLinear(\rationals^{n})$ be a matrix representation.

First, $\characteristic \rationals = 0 \notDivides \order(G)$. Hence by Corollary \ref{cor : GLnZ : Maschke Theorem Matrix Representation} of Maschke's theorem, there exists a matrix representation $\fontMatrixRep{T}' : G \rightarrow \generalLinear(n,\rationals)$ of the block diagonal form \eqref{eq : GLnZ : Maschke Theorem Matrix Representation} such that the on-diagonal matrix representations $\fontMatrixRep{T}_{i}'$ are irreducible, and $\fontMatrixRep{T}' \sim_{\rationals} \fontMatrixRep{T}$. Note that it is possible that the matrix representation $\fontMatrixRep{T}'$ consists of a single block, that is, that $\fontMatrixRep{T}'$ (and hence $\fontMatrixRep{T}$) is irreducible.

Second, $\rationals^{n}$ is a finite-dimensional $\rationals$-vector space, so by Remark \ref{rmk : Finite Dimensional Vector Space ACC DCC} and Lemma \ref{lem : Composition Series ACC DCC}, $\rationals^{n}$ has a composition series. Hence the Jordan--H\"{o}lder theorem applies. In particular, it follows that any matrix representation $\fontMatrixRep{T}$ of $G$ with representation space $\rationals^{n}$ is equivalent to the block diagonal one $\fontMatrixRep{T}'$. Thus, up to order and $\rationals$-equivalence, the on-diagonal representations $\fontMatrixRep{T}_{i}'$ are invariants for the group $G$ and the representation space $\rationals^{n}$. We call the sequence $(\fontMatrixRep{T}_{i}')_{i}$ the \fontDefWord{irreducible constituents} of $G$ on $\rationals^{n}$.