Example (C_m*C_n):C_3

Group $G=(C_m \oplus C_n):C_3$ is described by the short exact sequence

\[\begin{align*} 1 \to C_m \oplus C_n \to G \to C_3 \to 1 \end{align*}\]

When $n=14, m=2$, group $G=C_7:A_4$.

\[\begin{align*} G= \langle g_{m,r}=\begin{pmatrix} \zeta_m^r & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}, g_{n,s,t}=\begin{pmatrix} \zeta_n^s & 0 & 0\\ 0 & \zeta_n^t & 0\\ 0 & 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{pmatrix} \rangle \subseteq \rm{PGL}(3,\mathbb{C}). \end{align*}\]

There are two representation $V_1,V_2$ of $G$, for $V_1$, we choose $s=3,r=t=1$ and for $V_2$, we choose $s’=5,r’=t’=1$.

As a linear group in $\rm{GL}(3,\mathbb{C})$, it has center

\[\begin{align*} C_n= \langle \begin{pmatrix} \zeta_n & 0 & 0\\ 0 & \zeta_n & 0\\ 0 & 0 & \zeta_n \end{pmatrix} \rangle. \end{align*}\]

The Magma code is as follows:

m:=2;
d:=7;
n:=m*d;


FScale:=CyclotomicField(3:Sparse:=true);
FScale:=Compositum(FScale,CyclotomicField(n:Sparse:=true));

F:=FScale;

Zm:=RootOfUnity(m);
Zn:=RootOfUnity(n);

s1:=3;
r1:=1;
t1:=1;

G1:=MatrixGroup<3,F|
[Zm^(r1),0,0, 0,1,0, 0,0,1],
[Zn^(s1),0,0, 0,Zn^(t1),0, 0,0,1],
[0,0,1, 1,0,0, 0,1,0]>;

s2:=5;
r2:=1;
t2:=1;

G2:=MatrixGroup<3,F|
[Zm^(r2),0,0, 0,1,0, 0,0,1],
[Zn^(s2),0,0, 0,Zn^(t2),0, 0,0,1],
[0,0,1, 1,0,0, 0,1,0]>;

The Burnside Symbols are

where

\[\begin{align*} C_2=\langle \begin{pmatrix} 1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 1 \end{pmatrix} \rangle, C_2\times C_{14}=\langle \begin{pmatrix} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}, \begin{pmatrix} -\zeta_7 & 0 & 0\\ 0 & -\zeta_7^5 & 0\\ 0 & 0 & 1 \end{pmatrix} \rangle \end{align*}\]

There are conjugation relations:

\[\begin{align*} &(C_2\times C_{14},1 \circlearrowright k,((1,5),(1,6)))=(C_2\times C_{14},1 \circlearrowright k,((0,3),(1,6)))=(C_2\times C_{14},1 \circlearrowright k,((0,3),(1,5))),\\ &(C_2\times C_{14},1 \circlearrowright k,((0,11),(1,9)))=(C_2\times C_{14},1 \circlearrowright k,((1,8),(1,9)))=(C_2\times C_{14},1 \circlearrowright k,((0,11),(1,8))). \end{align*}\] \[\begin{align*} [\mathbb{P}(V_2) \circlearrowleft G]=&(1,G \circlearrowright k(x,y),())\\ &+2(C_2,C_{14} \circlearrowright k(x),(1))\\ &+(C_2\times C_{14},1 \circlearrowright k,((1,5),(1,10)))+(C_2\times C_{14},1 \circlearrowright k,((0,1),(1,4))) \end{align*}\]

where

\[\begin{align*} C_2=\langle \begin{pmatrix} 1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & 1 \end{pmatrix} \rangle, C_2 \times C_{14}=\langle \begin{pmatrix} -1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}, \begin{pmatrix} -\zeta_7^6 & 0 & 0\\ 0 & -\zeta_7^4 & 0\\ 0 & 0 & 1 \end{pmatrix} \rangle \end{align*}\]

There are conjugation relations:

\[\begin{align*} &(C_2\times C_{14},1 \circlearrowright k,((1,5),(1,10)))=(C_2\times C_{14},1 \circlearrowright k,((0,13),(1,10)))=(C_2\times C_{14},1 \circlearrowright k,((0,13),(1,5))),\\ &(C_2\times C_{14},1 \circlearrowright k,((0,1),(1,4)))=(C_2\times C_{14},1 \circlearrowright k,((0,1),(1,9)))=(C_2\times C_{14},1 \circlearrowright k,((1,4),(1,9))). \end{align*}\]

Now if we fix generator for $C_2 \times C_{14}=\langle g_{m,r},g_{n,s,t}\rangle$. Two Burnside Symbols are

\[\begin{align*} [\mathbb{P}(V_1) \circlearrowleft G]=&(1,G \circlearrowright k(x,y),())\\ &+2(C_2,C_{14} \circlearrowright k(x),(1))\\ &+(C_2\times C_{14},1 \circlearrowright k,((1,2),(1,11)))+(C_2\times C_{14},1 \circlearrowright k,((1,3),(1,12))) \end{align*}\] \[\begin{align*} [\mathbb{P}(V_2) \circlearrowleft G]=&(1,G \circlearrowright k(x,y),())\\ &+2(C_2,C_{14} \circlearrowright k(x),(1))\\ &+(C_2\times C_{14},1 \circlearrowright k,((1,5),(1,10)))+(C_2\times C_{14},1 \circlearrowright k,((1,4),(1,9))) \end{align*}\]

The combinatorial Burnside group of $C_7:A_4$ is $\begin{align*} \mathcal{BC}_2(C_7:A_4)=(\mathbb{Z}/2)^{16} \times \mathbb{Z}/6 \times (\mathbb{Z}/12)^2 \times \mathbb{Z}/72 \times \mathbb{Z}^{21} \end{align*}$

Remark: we don’t consider symbols with trivial stabilizer group in the computation of $\mathcal{BC}$.

Let $T_2^{1,\dots,16},T_{6},T_{12}^{1,2},T_{72}$ and $e_i$, $i=1,\dots,21$ be the generators for $(\mathbb{Z}/2)^{16}$, $\mathbb{Z}/6$, $(\mathbb{Z}/12)^2$, $\mathbb{Z}/72$ and the torsion-free part respectively.

The difference of images of two Burnside symbols $[\mathbb{P}(V_1) \circlearrowleft G]-[\mathbb{P}(V_2) \circlearrowleft G]$ under the map $\rm{Burn}_2(G) \to \mathcal{BC}_2(G)$ is:

\[\begin{align*} 4T_{12}^1+56T_{72} \neq 0 \end{align*}\]

Thus two actions are not birational to each other.

Kaiqi Yang(杨凯奇)