Group action in sage
I want to define a group action in sage. The group is a direct product of two general linear groups. The set under action is of matrices and the action is $(A,B) (M)=A^{-1}MB$. Assume all matrices have compatible sizes in order for multiplication. I might need a stabelizer later so I was thinking doing it in GAP but could not figure out how. Any suggestions?
The group $G_A$ of all "$A$-matices" is commuting (elementwise) with the corresponding $G_B$? Is $G_A$ commutative?
Since we expect (for a left action): $$(A_1A_2)^{-1}(B_1B2)\ M =(A_1A_2,B_1B_2)\ M=((A_1,B_1)(A_2,B_2))\ M$$ $$\qquad =(A_1,B_1)\ ((A_2,B_2)M)=(A_1,B_1)\ A_2^{-1}B_2M =A_1^{-1}B_1\ A_2^{-1}B_2M $$
Is it ok to get the $G_A$-stabilizer of the $G_B$-stabilizer of the relevant set $X_M$ of considered "$M$-matrices" ?
@dan_fulea $G_{A}$ is not commutative . Sorry I made a mistake while posting. The definition of action is now corrected. I have defined a function action of the direct product group on space of matrices which produces the required output. Now if you can guide how to get that stabilizer that would be great.
Please explain again which is the frame and the point of the question:
(1) Which are exactly the matrix spaces? All matrices $A,B,M$ are of the shape $n\times n$? Or $M$ is of shape $k\times n$ and $A,B$ square matrices of type $k\times k$ and $n\times n$?
(2) Mathematical issue: Is the action a left action? (Using usual matrix multiplications, not the "opposite ones".)
(3) We fix $A,B$ and look for all matrices $M$ satisfying $$ A^{-1}MB=M\ ?$$ If yes, do we really need the group action, as the question was mainly stated?
(4) The above stabilizing condition is equivalent to $MB=AM$. This is a linear equation that we may solve in
sage, but maybe $A,B$ should be given explicitly. Is this the question, the only one (i imagine) we can attack usingsage.@dan_fulea M is of shape k×n and A,B square matrices of type k×k and n×n. The group is $GL(k,F_{q})$ direct product with $GL(n,F_{q})$ which acts on space of k×n matrices as defined above. I need a stabelizer i.e all the pairs $(A,B) \in GL(k,F_{q}) \times GL(n,F_{q})$ for a fixed M. I think what you stated with points 3 and 4 gives different thing than this.
Thanks, this is a statement we can live with! WLOG $k\le n$. Then we reduce $M$, of rank $r\le k\le n$ to a block matrix $X=X(r)$ of the shape $(r+(k-r))\times(r+(k-r)+(n-k))$
because for some invertible matrices $S,T$ we have $M=SXT$, so that $AM=MB$ becomes $ASXT=SXTB$, i.e. $$\underbrace{S^{-1}AS}_{A'}\ X=X\ \underbrace{TBT^{-1}}_{B'}$$and we replace $(A,B)$ by $(A',B')$, getting a controlled conjugated group. So let $M=X$. Then we write $A,B$ in blocks such that we can multiply
and
... so $A_{11}=B_{11}$, $A_{21}=0$, $B_{12}=0$, $B_{13 ...(more)