**Proposition 3.1.1:***
The sum or product of two characters of the finite group
$G$ is a character.
*

**Proof. **(Click to Expand/Collapse)

If $(\pi, V)$ and $(\pi', V')$ are representations, let us choose bases for
$V$ and $V'$, so that we can identify $V =\mathbb{C}^d$ and $V'
=\mathbb{C}^{d'}$ for $d = \dim (V)$, $d' = \dim (V')$. Then $\pi (g)$ and
$\pi' (g)$ may be identified with their matrices
\[ \Pi (g) = \left(\begin{array}{cc}
\pi (g) & 0\\
0 & \pi' (g)
\end{array}\right) \hspace{2em} \text{(block matrix)} \]
gives a representation with character $\chi_{\pi} + \chi_{\pi'}$. This shows
that a sum of characters is a character.

where as usual $\pi' (g)^t$ is the transpose matrix. If $X$ has entries
$x_{i j}$, $\pi (g)$ has entries $a_{i j}$ and $\pi' (g)$ has entries $b_{i
j}$, then $\Pi' (g) X$ has $i, j$-th entry
\[ \sum_{k, l} a_{i k} x_{k l} b_{j l} = \sum_{k, l} \left( a_{i k} b_{j l}
\right) x_{k l} \]
where the left-hand side comes from simple matrix multiplication remembering
to take the transpose of $\pi' (g)$. The right-hand side is obviously the
same, but we have rewritten it in order to facilitate thinking about the
matrix representing this linear transformation. If we refer this to a matrix
by taking the $d d'$ entries $x_{i, j}$ in some order, the entries in this
matrix will be the $(d d')^2$ coefficients $a_{i k} b_{j l}$, and the
diagonal terms will be the contribution of $i = k$ and $j = l$, that is
\[ \operatorname{tr} \; \Pi' (g) = \sum_{i, j} a_{i i} b_{j j} = \left( \sum_i a_{i
i} \right) \left( \sum_j b_{j j} \right) = \chi_{\pi} (g) \chi_{\pi'} (g)
. \]
This shows that the product of two characters is a character.

On the other hand, let $W$ be the $d d'$-dimensional vector space of $d \times d'$ rectangular matrices. Define an action $\Pi' : G \longrightarrow \operatorname{GL} (W)$ by

\[ \Pi' (g) X = \pi (g) X \pi' (g)^t, \hspace{2em} X \in W, \] | (3.1.1) |

It is not true that the characters form a ring, since although they are closed
under addition and multiplication, they are not closed under subtraction.
However *generalized characters* do form a ring. We recall that a
*generalized character* is a difference of two characters. If $f$ is
any class function on a finite group $G$, and if $\chi_1, \cdots, \chi_h$ are
the irreducible characters, we know that $f$ can be written uniquely as $\sum
c_i \chi_i$. Clearly:

- The class function $f$ is a character if and only if the $c_i$ are nonnegative integers.
- The class function $f$ is a generalized character if and only if the $c_i$ are integers.

**Proposition 3.1.2:***
The generalized characters form a ring, called the character ring.
*

**Proof. **(Click to Expand/Collapse)

It is clear from Proposition 3.1.1 that the sum or product of
two generalized characters is a generalized character; since the difference
of two generalized characters is a generalized character, they are indeed
closed under the ring operations and form a subring of the ring of class
functions.

**Exercise 3.1.1:***
Let $G = S_3$, and let $\chi_1, \chi_2, \chi_3$ be the irreducible
characters, so $\chi_3 (1) = 2$. Determine the decomposition of $\chi_3^n$
into irreducibles when $n \le 10$. That is, write $\chi_3^n = a_n
\chi_1 + b_n \chi_2 + c_n \chi_3$ and compute $a_n, b_n$ and $c_n$.
*

**Exercise 3.1.2:***
Let $G$ and $H$ be groups, and let $\chi, \sigma$ be characters of $G$ and
$H$. Prove that $\chi \otimes \sigma$ is a character, where $\chi \otimes
\sigma$ is defined by
\[ (\chi \otimes \sigma) (g, h) = \chi (g) \sigma (h) . \]
Hint: Let $\pi : G \longrightarrow \operatorname{GL} (n, \mathbb{C})$
and $\pi' : H \longrightarrow \operatorname{GL} (m, \mathbb{C})$ be representations
with characters $\chi$ and $\sigma$. Let $W$ be the vector space of $n
\times m$ rectangular complex matrices, and define $\Pi : G \times H
\longrightarrow \operatorname{GL} (W)$ by
\[ \Pi (g, h) X = \pi (g) X \pi' (h)^t, \hspace{2em} (X \in W) . \]
Compute the character.
*

**Exercise 3.1.3:***
In the last exercise, if $\chi$ and $\sigma$ are irreducible characters of
$G$ and $H$, prove that $\chi \otimes \sigma$ is an irreducible character of
$G \times H$. ( Hint: one way is to compute $\left\langle \chi
\otimes \sigma, \chi \otimes \sigma \right\rangle_{G \times H}$.
*

**Exercise 3.1.4:***
Prove that if $\chi_1, \cdots, \chi_h$ are distinct irreducible characters
of $G$, and $\sum_i \chi_i (1)^2 = |G|$ then every irreducible character is
one of the $\chi_i$.
*

**Exercise 3.1.5:***
Prove that every irreducible character of $G \times H$ is of the form $\chi
\otimes \sigma$ for some irreducible characters $\chi$ and $\sigma$ of $G$
and $H$. ( Hint: use the criterion in the last exercise.)
*