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Consider a group $G=\langle P, Z, Q \rangle$ generated $P,Z,Q$ where $$Z=\left[ {\begin{array}{ccc} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & -1\\ \end{array} } \right] $$ $$ \quad P = \frac{1}{2} \left [ {\begin{array}{ccc} 1 & \tau^{-1} & -\tau\\ \tau^{-1} & \tau & 1\\ \tau & -1 & \tau^{-1}\\ \end{array} } \right]$$ where $\tau=\frac{1+\sqrt{5}}{2}.$ and $$Q=\left[ {\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & \rho ^2\\ 0 & -\rho & 0\\ \end{array} } \right]$$ where $\rho = e^{\frac{2\pi}{3}}$ is a third root of unity,

I want to understand the elements of $G$ that are of order $3$. Is there any known algorithm to find such elements?

I have shown that the center of $G$ is $SL_3(\mathbb{C})\cap \mathbb{C}^*.Id = C_3$, i.e. the scalar matrices where the scalars are third roots of unity. So I would like to know which elements not in the center are of order 3.

I know the following about $G$, in case it might be useful:

The order of $G$ is $1080$ and $G$ is a $3:1$ central extension of the alternating group $A_6.$ That is following sequence is exact. $$0 \to C_3 \to G \to A_6 \to 0.$$

Since I don't know much about the group structure of $G,$ I am having hard time to answer the question above. Any help will be appreciated.

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  • $\begingroup$ Well there are either 22 or 242. $\endgroup$ – Matt Samuel Jan 29 '15 at 3:10
  • $\begingroup$ 22 if $G$ contains an element of order 9, and 242 if not. $\endgroup$ – Matt Samuel Jan 29 '15 at 3:12
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The group is (by explicit computation in GAP) the (unique) perfect group of order 1080. Its elements of order 3 are the elements in the centre, C3, as well as two conjugacy classes of order 120 each (they consist of elements that in the A_6 image are 3-cycles, respectively double 3-cycles). Representatives of these two classes are given by $PZ$ and $PQP$. A nice matrix representative for one of the classes is the matrix $$\left(\begin{array}{ccc}0&0&1\\ 1&0&0\\ 0&1&0\\ \end{array}\right),$$ but the other class has only more messy representatives.

In case anyone is interested in the calculation itself, I defined the group:

z:=[[-1,0,0],[0,1,0],[0,0,-1]];
tau:=(1+ER(5))/2;
p:=1/2*[[1,tau^-1,-tau],[tau^-1,tau,1],[tau,-1,tau^-1]];
rho:=E(3);
q:=[[1,0,0],[0,0,rho^2],[0,-rho,0]];
G:=Group(z,p,q);

and determined classes whose representatives have order 3:

cl:=ConjugacyClasses(G);;
cl:=Filtered(cl,x->Order(Representative(x))=3);;
List(cl,Size);
[ 1, 1, 120, 120 ]

Concerning the question for elements of order 9:

s:=SylowSubgroup(G,3);;
AbelianInvariants(s);
[ 3, 3 ]
Exponent(s);
3

tells us that the 3-Sylow subgroup is the nonabelian group of order 27 and exponent 3.

To find diagonal matrices I would look at the stabilizer of the three 1-dimensional subspaces spanned by the standard basis vectors:

s:=Stabilizer(G,[1,0,0],OnLines);
s:=Stabilizer(s,[0,1,0],OnLines);
s:=Stabilizer(s,[0,0,1],OnLines);

and you get the subgroup generated by the diagonal matrices for [1,-1,-1] and E(3)*[-1,-1,1]. You could ask for Elements(s) to get a list.

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  • $\begingroup$ @ahulpke: some explanation on how you arrived at two representatives would be helpful to me, because I do not know anything about GAP yet. Thanks! $\endgroup$ – user166467 Jan 29 '15 at 3:40
  • $\begingroup$ @monomorphic: I added a brief description of the commands used. $\endgroup$ – ahulpke Jan 29 '15 at 5:00
  • $\begingroup$ Questions: 1) How does one read the statement $[1,1,120,120]?$ 2) Is the question about order $9$ related to my original question? Or were you answering for the earlier comments? $\endgroup$ – user166467 Jan 29 '15 at 8:10
  • $\begingroup$ Oh I get the comment: If $G$ does not contain an element of order $9,$ then there are $1+1+120+120 = 242$ elements of order $3$. Correct? But I don't get how you showed that there are no elements of order $9$. $\endgroup$ – user166467 Jan 29 '15 at 8:17
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    $\begingroup$ @monomorphic I added a description $\endgroup$ – ahulpke Apr 14 '15 at 21:20

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