The category of finite dimensional right $KG$-modules is equivalent to the category of finite dimensional representations of a quiver $Q$ Let $K$ be an algebraically closed field and $G$ be a finite group such that $|G|$ is not divisible by the characteristic of $K$ (so that Maschke's theorem can be applied). Let $Q$ be the quiver consisting of $n$ vertices and no arrows, where $n$ is the number of distinct conjugacy classes of $G$. Consider the category $\text{rep}(Q)$ of finite dimensional representations of $Q$. How can we show that $\text{rep}(Q)$ is equivalent to the category of finite dimensional right $KG$-modules, where $KG$ is the group algebra? I know that $\text{rep}(Q)$ is equivalent to the category of finite dimensional right $KQ$-modules where $KQ$ is the quiver algebra (though I'm not sure that this fact is relevant). I can't see how to define a functor from $\text{rep}(Q)$ to $\text{mod}$-$KQ$. Thanks in advance.
 A: The number $n$ of conjugacy classes does coincide with the number of isomorphism classes of irreducible, finite-dimensional $G$-representations, because $K$ is algebraically closed und Maschke’s theorem applies.¹
(Maybe we also need that $\operatorname{char}(K) = 0$?)
Let $S_1, \dotsc, S_n$ be a set of representatives for the isomorphism classes of irreducible, finite-dimensional $G$-representations.
Then by Maschke’s theorem every finite-dimensional $G$-representation is isomorphic to a representation of the form
$$
  S_1^{r_1} \oplus \dotsb \oplus S_n^{r_n}
$$
for some unique natural numbers $r_1, \dotsc, r_n$.
Let $\mathcal{A}$ be the full subcategory of $\mathbf{rep}_K(G)$ whose objects are these direct sums.
Then the inclusion from $\mathcal{A}$ to $\mathbf{rep}_K(G)$ is both fully faithful and essentially surjective, and therefore an equivalence.
We need to better understand morphisms in $\mathcal{A}$.
For this we recall the matrix calculus for morphisms between direct sums:

Let $R$ be some ring and let $M_1, \dotsc, M_s$ and $N_1, \dotsc, N_r$ be $R$-modules.²
Let $f_{ij} \colon M_j \to N_i$ be a homomorphism.
Then the map
\begin{align*}
    f
    \colon
    M_1 \oplus \dotsb \oplus M_r
    &\to
    N_1 \oplus \dotsb \oplus N_s \,,
    \\
    (m_1, \dotsc, m_r)
    &\mapsto
    \left(
      \sum_{j=1}^r f_{1j}(m_j) ,
      \dotsc,
      \sum_{j=1}^r f_{sj}(m_j) ,
    \right)
  \end{align*}
is again a homomorphism.
If we write the elements of $M_1 \oplus \dotsb \oplus M_s$ as column vectors
$$
    \begin{bmatrix}
      m_1 \\ \vdots \\ m_s
    \end{bmatrix}
  $$
then we can write the homomorphism $f$ as a matrix
$$
    f
    =
    \begin{bmatrix}
      f_{11} & \cdots & f_{1s} \\
      \vdots & \ddots & \vdots \\
      f_{r1} & \cdots & f_{rs}
    \end{bmatrix} \,.
  $$
The application of $f$ to an element of $M_1 \oplus \dotsb \oplus M_s$ is then just the usual matrix-vector-multiplication.
If on the other hand $f$ is any homomorphism from $M_1 \oplus \dotsb \oplus M_s$ to $N_1 \oplus \dotsb \oplus N_r$, then the compositions
$$
    f_{ij}
    \colon
    M_j
    \hookrightarrow
    M_1 \oplus \dotsb \oplus M_s
    \xrightarrow{\;f\;}
    N_1 \oplus \dotsb \oplus N_r
    \twoheadrightarrow
    N_i
  $$
are again homomorphism.
These two constructions are mutually inverse, and result in a bijection between the set
$$
    \operatorname{Hom}_R( M_1 \oplus \dotsb \oplus M_s, N_1 \oplus \dotsb N_r )
  $$
and the set of all matrices $[ f_{ij} ]_{i,j}$ consisting of homomorphisms $f_{ij} \colon M_j \to N_i$.
Given two such homomorphisms $f = [f_{jk}]_{jk}$ and $g = [g_{ij}]_{ij}$, their composition $g \circ f$ (if it is defined) can be computed via the usual rules of matrix multiplication, i.e. we have
$$
    g \circ f
    =
    \left[ \sum_j g_{ij} \circ f_{jk} \right]_{ik} \,.
  $$
(This holds because the application of $f$ and $g$ can be computed via matrix-vector-multiplication.)

In our situation we have $R = K[G]$ and can apply Schur’s lemma to compute $\operatorname{Hom}(S_i, S_j)$ for any two indices $i, j$:
We find that
$$
  \operatorname{Hom}(S_i, S_j) = 0
$$
for all $i \neq j$, and
$$
  \operatorname{Hom}(S_i, S_i) = K
$$
for every index $i$, because $K$ is algebraically closed.
We hence find that every morphisms in $\mathcal{A}$ can be described by a block-diagonal matrix with entries in $K$, giving us a bijection
$$
  \operatorname{Hom}_{\mathcal{A}}
  (
    S_1^{s_1} \oplus \dotsb \oplus S_n^{s_n},
    S_1^{r_1} \oplus \dotsb \oplus S_n^{r_n}
  )
  \to
  \operatorname{M}(r_1 \times s_1, K)
  \times \dotsb \times
  \operatorname{M}(r_n \times s_n, K) \,.
$$
We want to emphasize that this bijection does not rely on any additional choices.
(One may call it “canonical”.)
The composition of morphisms in $\mathcal{A}$ corresponds to the multiplication of block-diagonal matrices, which in turns corresponds under the above bijection to the componentwise multiplication.
We have now shown that the category $\mathcal{A}$ is isomorphic to the category $\mathcal{B}$ which is given as follows:

*

*The objects of $\mathcal{B}$ are tupels $(r_1, \dotsc, r_n)$ of natural numbers.

*The morphism sets of $\mathcal{B}$ are given by
$$
    \operatorname{Hom}_{\mathcal{B}}( (s_1, \dotsc, s_n), (r_1, \dotsc, r_n) )
    =
    \operatorname{M}(r_1 \times s_1, K)
    \times \dotsb \times
    \operatorname{M}(r_n \times s_n, K) \,.
  $$

*The composition of morphisms in $\mathcal{B}$ is given by
$$
    (A_1, \dotsc, A_n) \circ (B_1, \dotsc, B_n)
    =
    (A_1 B_1, \dotsc, A_n B_n) \,.
  $$
Let us now consider the category $\mathbf{rep}_K(Q)$.
Let $\mathcal{C}$ be the full subcategory of $\mathbf{rep}_K(Q)$ whose objects are those representations $(V_1, \dotsc, V_n)$ whose vector spaces $V_i$ are of the form $V_i = K^{r_i}$ for some natural numbers $r_i$.
The inclusion from $\mathcal{C}$ to $\mathbf{rep}_K(Q)$ is fully faithful und essentially surjective, and therefore an equivalence.
A morphism in $\mathcal{C}$ from $(K^{s_1}, \dotsc, K^{s_n})$ to $(K^{r_1}, \dotsc, K^{r_n})$ is just an arbitrary tupel $(f_1, \dotsc, f_n)$ of linear maps $f_i$ from $K^{s_i}$ to $K^{r_i}$.
(These linear maps don’t need to satisfy any compatability conditions because the quiver $Q$ doesn’t have any arrows.)
Each linear map $f_i$ is then given by multiplication with a unique matrix of size $r_i \times s_i$ with coefficients in $K$.
We see from this that the category $\mathcal{C}$ is also isomorphic to the category $\mathcal{B}$.
We have overall shown that
$$
  \mathbf{rep}_K(G)
  \simeq
  \mathcal{A}
  \cong
  \mathcal{B}
  \cong
  \mathcal{C}
  \simeq
  \mathbf{rep}_K(Q)
$$
and thus overall $\mathbf{rep}_K(G) \simeq \mathbf{rep}_K(Q)$.

¹ Every irreducible $G$-representation is automatically finite-dimensional because $G$ is finite.
² One can do the same construction in any additive category instead of just $R\text{-}\operatorname{Mod}$
