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Beginning with the ordinary character table of the symmetric group $S_5$, one immediately gets the following Brauer characters in characteristic two:

$\begin{array}{c|c|c|c} S_5 & () & (12345) & (123) \\ \hline \beta_1 & 1 & 1 & 1\\ \beta_2 & 4 & -1 & 1 \\ \beta_3 & 5 & 0 & -1 \\ \end{array}$

It is easy to see that both $\beta_1$ and $\beta_2$ are irreducible. Moreover, the third (and last) irreducible 2-Brauer character must be given either by $\beta_3$ or by $\beta_3 - \beta_1$. Indeed, by a computer search, I found a representation $S_5 \to \mathrm{GL}(4,2)$ affording the Brauer character $\beta_3 - \beta_1$. So $\beta_3$ is not irreducible.

Is it possible to come to the same conclusion by character theoretic arguments?


I actually found a purely character theoretic solution to my question, which requires some work though. I am still very interested in seeing other solutions!

Starting with the character table of the alternating group $A_5$ it is possible to derive the table of its irreducible 2-Brauer characters in a straightforward way (this is carried out here).

$\begin{array}{c|c|c|c|c} A_5 & () & (12345) & (13524) & (123) \\ \hline \gamma_1 & 1 & 1 & 1 & 1\\ \gamma_2 & 2 & \varphi-1 & \psi -1 & -1 \\ \gamma_3 & 2 & \psi-1 & \varphi-1 & -1\\ \gamma_4 & 4 & -1 & -1 & 1\\ \end{array}$

($\varphi$ is the golden ratio, and $\psi$ is its algebraic conjugate)

At this point, it is easy to see that $\beta_3-\beta_1 = (\gamma_2)^{S_5}$ is induced by a Brauer character of $A_5$, and so is a Brauer character as well.

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  • $\begingroup$ You might want to compare to the case $p=3$, which seems easier. $\endgroup$ – Dune Aug 14 '18 at 13:12
  • $\begingroup$ Not an answer, just something possibly worth considering. Doesn't the identification of $A_5$ and $SL(2,2^2)$ give the four Brauer characters at once? [The $2^n$ $2$-modular irreducibles of $SL(2,2^n)$ are the obvious $2^n$ modules got by taking tensor products of distinct algebraic conjugates of the standard representation of dimension $2$.] $\endgroup$ – ancientmathematician Aug 16 '18 at 15:58
  • $\begingroup$ @ancientmathematician: Thanks for pointing that out! In particular, any isomorphism $A_5 \cong \mathrm{SL}(2,4)$ immediately gives one of the Brauer characters $\gamma_2$ or $\gamma_3$. $\endgroup$ – Dune Aug 17 '18 at 11:53
  • $\begingroup$ Does identifying $S_5$ with $\Sigma L(2,4)$ [that's right, isn't it?] also help? $\endgroup$ – ancientmathematician Aug 17 '18 at 14:48
  • $\begingroup$ @ancientmathematician: I wasn't aware of that coincidence. But to be honest, I was hoping to see conceptual solutions which may also apply to other examples. I expected to see applications of (Brauer's) orthogonality relations or block theory. $\endgroup$ – Dune Aug 19 '18 at 13:27

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