# M(5,1,3)

M(5,1,3) - $B_0(kA_5)$
Representative: $B_0(kA_5)$ $C_5$ $C_2$ 4 2 1 $\left( \begin{array}{cc} 2 & 1 \\ 1 & 3 \\ \end{array} \right)$ Yes Yes Yes $B_0(\mathcal{O} A_5)$ $\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 1 \\ 1 & 1 \\ \end{array}\right)$ 1 $\mathcal{T}(B)=C_4$ {{{PIgroup}}} Yes Yes M(5,1,2) Yes {{{coveringblocks}}} {{{coveredblocks}}} {{{pcoveringblocks}}}

## Basic algebra

Quiver: a:<1,2>, b:<2,1>, c:<2,2>

Relations w.r.t. $k$: ac=cb=ba-c^2=0

## Covering blocks and covered blocks

Let $N \triangleleft G$ with $p'$-index and let $B$ be a block of $\mathcal{O} G$ covering a block $b$ of $\mathcal{O} N$.

If $b$ lies in M(5,1,3), then $B$ must lie in M(5,1,3) or M(5,1,5). Examples needed.

## Projective indecomposable modules

Labelling the simple $B$-modules by $S_1, S_2$, the projective indecomposable modules have Loewy structure as follows:

$\begin{array}{cc} \begin{array}{c} S_1 \\ S_2 \\ S_1 \\ \end{array}, & \begin{array}{ccc} & S_2 & \\ S_1 & & S_2 \\ & S_2 & \\ \end{array} \end{array}$

## Irreducible characters

All irreducible characters have height zero.