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Consider the Modulor group $M_{16}$ of order $16$ given by the presentation
$$ M_{16}=\langle a,b:a^8=b^2=1,ba=ab^5\rangle $$ Is there a known realization of $M_{16}$ by matrices?

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    $\begingroup$ Every finite group can be represented as matrices. Use Cayley's theorem if need be. $\endgroup$
    – Randall
    May 6, 2022 at 16:02
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    $\begingroup$ To be more helpful: is there a particular kind of matrix representation you're looking for? $\endgroup$
    – Randall
    May 6, 2022 at 16:06
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    $\begingroup$ Title: it is "Modular group", not Modulor. See here for the name and some useful facts. $\endgroup$ May 6, 2022 at 16:45
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    $\begingroup$ @march But there are two conjugate faithful complex representations of degree $2$. The minimum degree of a real representation is $4$. $\endgroup$
    – Derek Holt
    May 7, 2022 at 21:50
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    $\begingroup$ @march As the centre is cyclic there is a faithful irreducible representation of $M_{16}$. The character degrees have to be $1$ and $2$ because the sum of the squares is 16. Thus it must embed in $\mathrm{GL}_2(\mathbb C)$. $\endgroup$ May 7, 2022 at 21:55

2 Answers 2

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There is a theorem saying that a $p$-group has a faithful irreducible complex character if and only if the center is cyclic. The modular group has a cyclic center and all irreducible character has degree 1 or 2. Hence, you can represent this group by complex $2\times 2$-matrices.

To be more precise: Let $\zeta\in\mathbb{C}$ be a primitive $8$-th root of unity, $a=\mathrm{diag}(\zeta,\zeta^5)$ and $b=\bigl(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\bigr)$.

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One automatic way to get a faithful matrix representation of a group is to construct the regular representation, where we assign a basis vector $v_g$ to each group element $g$, and construct the linear transformation $T_h$ of $h$ by left-multiplication by $h$, i.e. $$ T_hv_g = v_{hg}\,. $$ Then, one can see that $$ T_{h_1}T_{h_2}v_g = T_{h_1}v_{h_2g} = v_{h_1h_2g} = T_{h_1h_2}v_{g}\,. $$ Ordering the group elements $\{g_1=e, g_2, g_3, \dots, g_{16}\}$, we can construct the Cayley table, and from each column $c_j$ of the table, we can construct a corresponding permutation matrix as $$ P_{i,c_i} = 1\,, $$ with all other matrix elements equal to zero.


Here's an implementation in Mathematica. Using the permutation representation from the list of groups of order 16, which is $$ M_{16} = \langle (1, 2, 3, 4, 5, 6, 7, 8), (1, 5)(3,7)\rangle\,, $$ we can construct the group in Mathematica as

group = PermutationGroup[{Cycles[{Range[8]}], Cycles[{{1, 5}, {3, 7}}]}];

The group multiplication table is given by

GroupMultiplicationTable[group] // TableForm

enter image description here

and so we can construct the set of permutation matrices via

matrixRep = IdentityMatrix[16][[#]] & /@ Transpose@GroupMultiplicationTable[group];

This set of matrices matches the group multiplication. For example, multiplying the 2nd and third elements and finding the position of the result in the original list yields

PermutationProduct[elements[[2]], elements[[3]]];
First@Position[elements, %]
(* Cycles[{{1, 2, 7, 8, 5, 6, 3, 4}}] *)
(* {4} )*

whereas multiplying the corresponding matrices yields

matrixRep[[2]].matrixRep[[3]];
First@Position[matrixRep, %]
(* {4} *)

If you want a list of all of the matrices, I can include them elsewhere.


Finally, I do not know the minimum dimension required to get a faithful matrix representation of this group.

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  • $\begingroup$ Thanks @march for your answer $\endgroup$
    – boaz
    May 7, 2022 at 17:31

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