# M(16,14,10)

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M(16,14,10) - $B_0(k(A_5 \times A_5))$
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Representative: $B_0(k(A_5 \times A_5))$ $(C_2)^4$ $C_3 \times C_3$ 16 9 1 $\left( \begin{array}{ccccccccc} 16 & 8 & 8 & 8 & 8 & 4 & 4 & 4 & 4 \\ 8 & 8 & 4 & 4 & 4 & 4 & 2 & 2 & 4 \\ 8 & 4 & 8 & 4 & 4 & 2 & 4 & 4 & 2 \\ 8 & 4 & 4 & 8 & 4 & 4 & 2 & 4 & 2 \\ 8 & 4 & 4 & 4 & 8 & 2 & 4 & 2 & 4 \\ 4 & 4 & 2 & 4 & 2 & 4 & 1 & 2 & 2 \\ 4 & 2 & 4 & 2 & 4 & 1 & 4 & 2 & 2 \\ 4 & 2 & 4 & 4 & 2 & 2 & 2 & 4 & 1 \\ 4 & 4 & 2 & 2 & 4 & 2 & 2 & 1 & 4 \end{array} \right)$ Yes Yes Yes $B_0(\mathcal{O} (A_5 \times A_5))$ See below 1 No Yes M(16,14,8), M(16,14,9) Yes

## Covering blocks and covered blocks

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

If $b$ is in M(16,14,10), then $B$ is also in M(16,14,10).

## Irreducible characters

All irreducible characters have height zero.

## Decomposition matrix

$\left( \begin{array}{ccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 0 \\ 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \end{array}\right)$