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.

  • 4
    $\begingroup$ 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
    May 1, 2017 at 7:19
  • 1
    $\begingroup$ 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
    May 1, 2017 at 8:09
  • 1
    $\begingroup$ 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
    May 1, 2017 at 8:58
  • 1
    $\begingroup$ 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
    May 1, 2017 at 17:36

1 Answer 1


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);
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);
  • 1
    $\begingroup$ 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
    May 1, 2017 at 9:50
  • 2
    $\begingroup$ 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
    May 1, 2017 at 10:33
  • $\begingroup$ Does your understanding match the interpretation of my second comment? $\endgroup$
    – YCor
    May 1, 2017 at 10:34
  • 1
    $\begingroup$ @YCor They do not match exactly, because you were just taking two geenrators, 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
    May 1, 2017 at 10:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.