Defect and Brauer pairs for a4 in characteristic three

The four characters $\chi :V\to \{1,-1\}\subset k^{\times }$ give $q_{\chi }=\frac{1}{4}{\sum }_{v\in V}\chi (v)v$. For characters $\chi ,\psi$, the coefficient calculation in their product uses ${\sum }_{v\in V}\chi (v)\psi (v)=4$ if $\chi =\psi$, and $0$ otherwise: a nontrivial sign character has two values of each sign. Hence $q_{\chi }{q}_{\psi }={\delta }_{\chi \psi }{q}_{\chi }$ and their sum is $1$. Each $kV{q}_{\chi }$ is one-dimensional, since $v{q}_{\chi }=\chi (v)q_{\chi }$. As $4=1$ in $k$, the trivial character idempotent is $e$. Conjugation by $t$ cycles the three nontrivial characters and fixes $e$. Thus $e,f$ are orthogonal central idempotents of $kG$.

The algebra $ekG$ has basis $e,et,e{t}^2$ and multiplication $(e{t}^i)(e{t}^j)=e{t}^{i+j}$. It is $k{C}_3\cong k[u]\mathrm{/}((u-1{)}^3)$ and is local: writing $x=\lambda +n$ with $n$ in the nilpotent ideal $(u-1)$, it is a unit if $\lambda \mathrm{\ne }0$ by the finite geometric inverse, and is nilpotent otherwise. Its only idempotents are $0,1$, since for an idempotent one of it and its complement is a unit. Therefore $e$ is primitive central.

Choose a nontrivial character idempotent $q$ and put $E_{ij}=t^{i}q{t}^{-j}$ for $0\le i,j<3$. For $j\mathrm{\ne }l$, $q{t}^{l-j}q=q({}^{t^{l-j}}q)t^{l-j}=0$; for $j=l$ the middle product is $q$. Thus $E_{ij}{E}_{lm}={\delta }_{jl}{E}_{im}$ and $\sum E_{ii}=f$. These nine nonzero elements are linearly independent: multiplying a relation on left and right by suitable matrix units isolates each coefficient times a nonzero unit. The dimension of $fkG$ is $3\dim (fkV)=9$, so these units give $fkG\cong M_3(k)$. Its centre is $kf$ (commuting with the diagonal and then off-diagonal units forces a scalar diagonal), so $f$ is primitive central. Since $e+f=1$, there are no further blocks.

The eight $3$-cycles partition into four order-three subgroups, and these and $1$ are all $3$-subgroups of $A_4$. The centralizer of a $3$-cycle in $S_4$ consists of its three powers, since a commuting permutation must preserve its three-point orbit and fixed point; hence $C_G(P)=P$. None of $a,b,c$ lies in it. Coefficient projection therefore gives ${\mathrm{Br}}_P(e)=1$ and ${\mathrm{Br}}_P(f)=0$. The algebra $kP$ is local by the same calculation as step 2.1, so its only block is $1$. At the identity subgroup projection is the identity. Thus the complete pair lists are as stated.

By [F1], the four nonzero-support Sylow subgroups are the defect groups of $e$, and $1$ is the defect group of $f$. By [F2], $(1,e)\le (P,1)$ because the computed product is $1$, while no such pair exists for $f$. Different order-three subgroups are incomparable. By [F3] precisely the four displayed $P$-pairs and the sole $f$-pair are maximal.