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\title{Lecture 4-1: Irreducible and indecomposable representations}
\date{April 1, 2024}
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Last time we saw that there are exactly $p$ equivalence classes of indecomposable representations of the cyclic group $C_p$ of prime order $p$ over any field $k$ of characteristic $p$; these correspond to the Jordan blocks of size at most $p\times p$ and eigenvalue 1.  We express this situation by saying that the cyclic group $C_p$ has \alert{finite representation type} over any field of characteristic $p$.  Indecomposable representations of the {\sl infinite} cyclic group $\mathbb Z$ over any field $k$ correspond up to equivalence to companion matrices of powers of irreducible polynomials other than $x$ over $k$; thus there are infinitely many of them, but they can be parametrized in a nice way.  We say that $\mathbb Z$ has \alert{tame representation type}.
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In stark contrast, indecomposable representations of the Klein four-group $C_2\times C_2$ do not admit any reasonable parametrization; entire Ph.D. theses and papers are written on such representations.  We say that $C_2\times C_2$ has \alert{wild representation type}.  We can get a better handle on representations if we replace indecomposability by a stronger hypothesis. 
\begin{alertblock} {Definition:  DF p. 847}
The representation $V$of $G$ is called {\sl irreducible} or {\sl simple} if it does not admit any proper subrepresentation $W$.
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For example, any one-dimensional representation is trivially irreducible.  For any irreducible polynomial $p$ over a field $k$, we have already observed that the quotient $R=k[x]/(p)$ does not admit any proper subspace stable under multiplication by $x$, since any such subspace would correspond to a proper ideal of $R$ as a ring.  Thus in particular the representation $\mathbb Q[x]/(\Phi_n)$ of the cyclic group $C_n$ over $\mathbb Q$ is irreducible, where $\Phi_n$ is the $n$th cyclotomic polynomial.  On the other hand, any representation of $C_n$ over a field $k$ for which the polynomial $x^n-1$ is a product of distinct linear factors is a direct sum of one-dimensional subrepresentations, so that the only irreducible representations of $C_n$ over $k$ are one-dimensional, with the generator $g$ of $C_n$ acting by an $n$th root of unity in $k$.
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More generally, we have
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Let $A$ be a finite abelian group of order $n$ and $k$ an algebraically closed field of characteristic not dividing $n$.  Then any representation $V$ of $A$ is a direct sum of one-dimensional representations, so that every irreducible representation is one-dimensional.  There are $n$ inequivalent irreducible representations of $A$.
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\begin{proof}
We know that $A$ is isomorphic to a direct sum $\bigoplus\limits_{i=1}^m C_{n_i}$ of cyclic groups.  Letting $g_i$ be a generator of $C_{n_i}$, we have seen that $V$ is the direct sum of eigenspaces $E$ of $g_i$, each with eigenvalue $e_i$, an $n_i$th root of 1 in $k$.  Since $A$ is abelian, every eigenspace $E$ is stable under the action of the generators $g_j$ with $j\ne i$ of the other cyclic factors $C_{n_j}$ of $A$.  By induction on the number of cyclic factors, we deduce that $V$ is the direct sum of one-dimensional simultaneous eigenspaces of all generators $g_i$, each of which is an irreducible subrepresentation.  Thus all irreducible representations are one-dimensional.  Conversely, given a one-dimensional space $W$ over $k$, we make it into a representation by decreeing that each generator $g_i$ act by a scalar equal to a suitable $e_i$; as there $n_i$ choices for each $e_i$ and the product of the $n_i$ is $n$, the result follows.
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In fact \alert{the set $\widehat A$ of equivalence classes of irreducible representations of $A$ has a group structure and $\widehat A\cong A$}.  This holds because such classes correspond to homomorphisms from $A$ to $GL_1(k) = k^*$ and the product $\pi_1\pi_2$ of two such homomorphisms $\pi_1,\pi_2$ (sending $a\in A$ to $\pi(1)\pi_2(a)$) is another homomorphism.  Since (as noted above) any $\pi$ sends a generator $g_i$ of a cyclic factor of $A$ to an $n_i$th root of 1 in $k$, the isomorphism follows.  Note however that there is no {\sl canonical} homomorphism from $\hat A$ to $A$, since the isomorphism between these groups depends on the choice of a particular primitive $n_i$th root of 1 in $k$ for all $i$.
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Of course most finite groups $G$ are nonabelian; accordingly most irreducible representations of such groups have degree larger than one.  The following key result reduces the study of such representations to the irreducible case, under a mild restriction on $k$.  
\begin{alertblock}{Maschke's Theorem:  DF, p. 849}
If the characteristic of $k$ does not divide the order $n$ of $G$, then any representation $V$ of $G$ is semisimple; that is, it is the direct sum of irreducible subrepresentations.
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\begin{proof}
There is nothing to prove if $V$ is irreducible, so assume not and let $W$ be a proper submodule.  It is enough to show that there is a a submodule $W'$ complementary to $W$ in $V$, so that $V$ is the direct sum of $W$ and $W'$, for then by induction on dimension both $W$ and $W'$ are direct sums of irreducible submodules, whence so is $V$.  To construct $W'$.  let $f$ be any linear projection of $V$ onto $W$ (so that $f$ maps $V$ onto $W$ and the restriction of $f$ to $W$ is the identity).  Set $\tilde f(v) = \frac 1 n\sum\limits_{g\in G} g^{-1}f(gv)$.  Then for $h\in G$ we have $h^{-1}\tilde f(hv) =\frac 1 n \sum\limits_{g\in G} h^{-1}g^{-1}f(ghv)=\tilde f(v)$, so that $\tilde f$ is a $G$-module homomorphism from $V$ to $W$, which is the identity on $W$, since $f$ is.  The kernel of $\tilde f$ is then a submodule $W'$ intersecting $W$ trivially; computing its dimension we see that $V$ is the direct sum of $W$ and $W'$, as desired.
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The technique of this proof is called \alert{averaging over $G$} and occurs frequently in the study of finite (or more generally compact) groups.  We now study irreducible modules in a rather roundabout way, first investigating homomorphisms between them rather than the modules themselves.  Note first that the set $\hom(M,M')$ of $G$-module homomorphisms from one module $M$ to another one $M'$ is clearly a vector space over the basefield $k$.
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\begin{alertblock}{Theorem: Schur's Lemma; see DF, Exercise 18, p. 853} 
Assume that $k$ is algebraically closed.  Let $V,W$ be irreducible $G$-modules.  If $V$ is not isomorphic to $W$, then $\hom(V,W)=0$.  If $V$ is isomorphic to $W$, then $\hom(V,W)\cong k$.
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\begin{proof}
The kernel and image of any $G$-module homomorphism are both $G$-submodules, so if $V\not\cong W$, then any module homomorphism from $V$ to $W$ necessarily has kernel $V$ and image 0, by irreducibility.  If $V\cong W$ let $f$ be an isomorphism and $g$ a homomorphism between them.  The homomorphism $f^{-1}g$ from $V$ to $V$, as a linear map, must have an eigenvalue $\lambda$; but then its $\lambda$-eigenspace, being the kernel of $g-\lambda f$, must be a nonzero $G$-submodule of $V$ and thus all of $V$.  Hence $g=\lambda f$, as claimed.
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This result says in particular that the only linear maps from $V$ to itself commuting with the action of $G$ are the scalars.
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We now introduce a ring such that modules over this ring (for fixed basefield $k$) are the same things as $G$-modules.  
\begin{alertblock}{Definition:  DF, p. 840}
Given $k$ and $G$, the {\sl group algebra} $kG$ consists of all formal linear combinations $\sum\limits_{g\in G}k_g g$ of elements of $G$ with coefficients in $k$.  The group elements $g$ are regarded as linearly independent, so that two such combinations agree if and only if their coefficients match up term by term.  We add two such combinations and multiply by elements of $k$ in the obvious way.  We take the product $\sum\limits_{g\in G}k_g g\sum\limits_{g\in G}\ell_g g$ of two elements of $kG$ to be $\sum\limits_{g,h\in G} k_g\ell_h gh$, collecting coefficients in this last sum to make the group elements appearing in it distinct.
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Given any $G$-module $V$ over $k$ we then make $V$ into a $kG$-module by decreeing that $(\sum\limits_{g\in G}k_gg)v =\sum\limits_{g\in G}k_g(gv)$.  Equivalently, any homomorphism $\pi:G\rightarrow GL(V)$ extends uniquely to a $k$-algebra homomorphism from $kG$ to the ring $M(V)$ of all linear transformations from $V$ to itself, where $V$ is a finite-dimensional vector space over $k$.  In particular, $kG$ itself, clearly being a $kG$-module, is also a $G$-module.  We call it the \alert{regular representation} of $G$.
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