# Realization of the Modulor group $M_{16}$ of order $16$. [closed]

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?

• Every finite group can be represented as matrices. Use Cayley's theorem if need be. May 6, 2022 at 16:02
• To be more helpful: is there a particular kind of matrix representation you're looking for? May 6, 2022 at 16:06
• Title: it is "Modular group", not Modulor. See here for the name and some useful facts. May 6, 2022 at 16:45
• @march But there are two conjugate faithful complex representations of degree $2$. The minimum degree of a real representation is $4$. May 7, 2022 at 21:50
• @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 $\mathrm{GL}_2(\mathbb C)$. May 7, 2022 at 21:55

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 $$b=\bigl(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\bigr)$$.

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


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.