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Let $G$ be the subgroup of $\mathrm{GL}_9(\mathbf{Q})$ defined as follows: Letting $(a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8,a_9)$ be the canonical basis of $\mathbf{Q}^9$, $G$ is generated by:

  • All permutation matrices permuting the basis elements $(a_1,\dots,a_8)$ (fixing $a_9$)
  • The additional transformation $c$ defined by $a_9\mapsto a_1+a_2$; $a_1\mapsto \frac12(a_1-a_2+a_9)$; $a_2\mapsto \frac12(a_2-a_1+a_9)$; $a_{i>2}\mapsto a_i+\frac12(a_1+a_2-a_9)$

I am interested to identify what is this group and more simply whether it is finite and if so, to determine its order.

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    $\begingroup$ Could you change to a slightly more informative title? Also I don't know what you mean. From the first 2 sentences I understand that you want to identify some group of permutations on 9 elements, but this is incompatible with the last assumption. You seem to mean a subgroup of $\mathrm{GL}_9$? the sentence "the entries $a_1$ to $a_8$ can be permuted" sounds unclear anyway. And "discrete group" is also unclear (every group can be endowed with the discrete topology). $\endgroup$
    – YCor
    Commented May 1, 2017 at 7:19
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    $\begingroup$ so it's my guess that you want to describe the subgroup of $GL_9$ generated by two elements: the matrix $u$ induced by the 8-cycle permuting $(a_1,\dots,a_8)$ (fixing $a_9$) and the matrix $v$ defined in your second item. (If this latter matrix has finite order you should say it and say what you have checked, e.g., whether $uv$ has finite order, etc) $\endgroup$
    – YCor
    Commented May 1, 2017 at 8:09
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    $\begingroup$ Or possibly all permutations of $a_1,\ldots,a_8$ are to be included? The group generated is finite in either case. I can answer the question once it has been clarified. $\endgroup$
    – Derek Holt
    Commented May 1, 2017 at 8:58
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    $\begingroup$ Since Stefan Kohl seems to have guessed the correct meaning of the question, I edited to make the question meaningful, which you should have done yourself. $\endgroup$
    – YCor
    Commented May 1, 2017 at 17:36

1 Answer 1

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If I understand your question right, your group $G$ has order $5160960$, and it has an elementary abelian normal subgroup $N$ of order $2^7$ such that $G/N \cong {\rm S}_8$. This can be found with GAP as follows:

gap> A := PermutationMat((1,2),9);;
gap> B := PermutationMat((1,2,3,4,5,6,7,8),9);;
gap> C := [[1,-1,0,0,0,0,0,0,1]/2,
>          [-1,1,0,0,0,0,0,0,1]/2,
>          [1,1,2,0,0,0,0,0,-1]/2,
>          [1,1,0,2,0,0,0,0,-1]/2,
>          [1,1,0,0,2,0,0,0,-1]/2,
>          [1,1,0,0,0,2,0,0,-1]/2,
>          [1,1,0,0,0,0,2,0,-1]/2,
>          [1,1,0,0,0,0,0,2,-1]/2,
>          [1,1,0,0,0,0,0,0,0]];;
gap> G := Group(A,B,C);
<matrix group with 3 generators>
gap> Size(G);
5160960
gap> N := NormalSubgroups(G);
[ Group([  ]), <matrix group of size 2 with 1 generators>, 
  <matrix group of size 128 with 7 generators>, 
  <matrix group of size 2580480 with 9 generators>, 
  <matrix group of size 5160960 with 3 generators> ]
gap> StructureDescription(N[3]);
"C2 x C2 x C2 x C2 x C2 x C2 x C2"
gap> Q := G/N[3];
Group([ (16,24)(17,21)(18,22)(19,23)(20,25)(27,28), (1,6,19,21,12,20,24,13)
(2,16,23,11)(3,5,9,17,22,27,25,14)(4,7,8,10,18,26,28,15), (16,24)(17,21)
(18,22)(19,23)(20,25)(27,28) ])
gap> StructureDescription(Q);
"S8"
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    $\begingroup$ Yes that was also the result of my calculation. It is a split extension of the elementary abelian subgroup by $S_8$. $\endgroup$
    – Derek Holt
    Commented May 1, 2017 at 9:50
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    $\begingroup$ In other words, the Weyl group of $D_8$, with OP's coordinates probably related to the extended Dynkin diagram of $D_8$. $\endgroup$
    – Gro-Tsen
    Commented May 1, 2017 at 10:33
  • $\begingroup$ Does your understanding match the interpretation of my second comment? $\endgroup$
    – YCor
    Commented May 1, 2017 at 10:34
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    $\begingroup$ @YCor They do not match exactly, because you were just taking two generators, an $8$-cycle and the OP's additional transformation, whereas in this answer there are three generators, the extra one being the transposition $(a_1,a_2)$, to generate the whole of $S_8$ on $a_1,\ldots,a_8$. The group generated by your two generators is smaller and is $C_2 \times S_8$. $\endgroup$
    – Derek Holt
    Commented May 1, 2017 at 10:49

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