TITLE: Computing Brauer characters of a finite group
QUESTION [9 upvotes]: I am studying character theory from the book "Character Theory of Finite Groups" by Martin Isaac. (I am not too familiar with valuations and algebraic number theory.)
In the last chapter on modular representation theory, Brauer characters, blocks, and defect groups are introduced. 
My question is this: How do we find the irreducible Brauer characters and blocks, given a group and a prime?
For instance, let's say we have $p=3$ and the group $G = S_5$, the symmetric group.
An example of the precise calculation or method used to determine the characters would be very helpful. Thanks.

REPLY [3 votes]: In your case, $G = S_5$ and $p = 3$, all irreducible Brauer characters are easily derived by the ordinary character table of $G$.
The ordinary character table of $G$ is given by:
$\begin{array}{c|c|c|c|c|c|c|c}
S_5 & () & (12) & (12)(34) & (1234) & (12345) & (123) & (123)(45)\\
\small{\#} & \small{1} & \small{10} & \small{15} & \small{30} & \small{24} & \small{20} & \small{20}\\
\hline
\chi_1 & 1 & 1 & 1 & 1 & 1 & 1 & 1\\
\chi_2 & 1 & -1 & 1 & -1 & 1 & 1 & -1\\
\chi_3 & 4 & 2 & 0 & 0 & -1 & 1 & -1\\
\chi_4 & 4 & -2 & 0 & 0 & -1 & 1 & 1\\
\chi_5 & 5 & -1 & 1 & 1 & 0 & -1 & -1\\
\chi_6 & 5 & 1 & 1 & -1 & 0 & -1 & 1\\
\chi_7 & 6 & 0 & -2 & 0 & 1 & 0 & 0\\
\end{array}$
By dropping the 3-irregular conjugacy classes, we immediately get the following Brauer characters:
$\begin{array}{c|c|c|c|c|c}
S_5 & () & (12) & (12)(34) & (1234) & (12345)\\
\hline
\beta_1 & 1 & 1 & 1 & 1 & 1\\
\beta_2 & 1 & -1 & 1 & -1 & 1\\
\beta_3 & 4 & 2 & 0 & 0 & -1\\
\beta_4 & 4 & -2 & 0 & 0 & -1\\
\beta_5 & 5 & -1 & 1 & 1 & 0\\
\beta_6 & 5 & 1 & 1 & -1 & 0\\
\beta_7 & 6 & 0 & -2 & 0 & 1\\
\end{array}$
Of course $\beta_1$ and $\beta_2$ are (being linear characters) irreducible. Moreover, since $S_5 / A_5 \cong C_2$, these are all linear Brauer characters. Since the integer $|G|/\chi_7(1)$ is not divisible by 3, we get that $\beta_7$ is irreducible by a general theorem. On the other hand, we have $\beta_5 = \beta_4 + \beta_2$ and $\beta_6 = \beta_3 + \beta_1$. So both $\beta_5$ and $\beta_6$ are reducible. So we get to the following table, which already has the right size of a Brauer character table:
$\begin{array}{c|c|c|c|c|c}
S_5 & () & (12) & (12)(34) & (1234) & (12345)\\
\hline
\beta_1 & 1 & 1 & 1 & 1 & 1\\
\beta_2 & 1 & -1 & 1 & -1 & 1\\
\beta_3 & 4 & 2 & 0 & 0 & -1\\
\beta_4 & 4 & -2 & 0 & 0 & -1\\
\beta_7 & 6 & 0 & -2 & 0 & 1\\
\end{array}$
Indeed, the remaining Brauer characters $\beta_3$, $\beta_4$ are irreducible. To begin with, we show that these characters have no linear constituents. Let $U = \langle (1,2,3,4,5) \rangle$ be the subgroup of $S_5$ generated by a 5-cycle. Since $|U|$ is not divisible by 3, the restriction of any Brauer character of $S_5$ to $U$ is an ordinary character, so we can use the usual inner product on characters. In particular, we may compute
$\langle (\beta_3)_U, (\beta_1)_U \rangle = \langle (\beta_3)_U, (\beta_2)_U \rangle  = \langle (\beta_4)_U, (\beta_1)_U \rangle  = \langle (\beta_4)_U, (\beta_2)_U \rangle = 0$,
which shows that neither $\beta_3$ nor $\beta_4$ has a linear constituent. It remains to show that neither $\beta_3$ nor $\beta_4$ is a sum of two irreducible Brauer characters of degree 2. Suppose this is the case, say $\beta_3 = \gamma + \delta$ (which we can assume without loss of generality since $\beta_4 = \beta_3 \cdot \beta_2$). Since $\beta_3$ attains the value $-1$ (and since values of Brauer characters are algebraic integers), we see that $\gamma \neq \delta$. Since $\beta_4 = \beta_2 \cdot \beta_3 \neq \beta_3$, we get that $\beta_4 = \beta_2 \cdot \gamma + \beta_2 \cdot \delta$ decomposes into further irreducible Brauer characters, which we haven't considered so far. But this is a contradiction since $S_5$ has exactly 5 irreducible Brauer characters in characteristics 3.

Note that I am asking the same question in the case $p=2$. The case $p=5$ is also fun to consider. In that case, the Brauer character table can be obtained by similar thoughts as here. But the case $p=5$ is more interesting, since then there are irreducible Brauer characters which are not obtained by ordinary characters.