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Imagine that I have a two-qubit system, and one of the qubits performs single-qubit rotations from a subgroup of SU(2), specifically 2O. In this subgroup, the single-qubit rotations are represented by H, S,X, Y, Z, and I, where I denotes no rotation. Additionally, I have a CNOT gate for this system. I am wondering how to show numerically that this subgroup and the CNOT gate together form a finite group.

P.S. I understand that this may seem like a trivial question, but I am also trying to grasp the methodology for solving these types of problems, particularly in studying groups and determining whether they are infinite or finite. If someone can provide a comprehensive answer to this question, it will help me understand the approach and process for handling more complex groups as well. I kindly request that any response refrain from phrases such as "with H and T gate, they perform a universal gate set, so if you include them this will perform infinite group" or "it is obvious this will not give us a universal gate set without the T gate and H, so it is finite.

quest
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    look into https://www.gap-system.org/ ...as long your groups are "algebraic" (no floating point definitions) it's a pretty powerful software. – unknown Jun 30 '23 at 15:33
  • Thank you! Will look at that. Can you give some tips about how to solve that kind of problem? – quest Jul 01 '23 at 01:07
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    you define the set of matrices that generate the group; then you ask GAP to calculate the size of the group. It will give back the size of the group or infinity... – unknown Jul 01 '23 at 02:50
  • Excellen! Thank you! And for the interaction of CNOT, then I should include the CNOT gate matrices in the set – quest Jul 01 '23 at 03:59
  • well, actually, I found what to do!! Thank you for referencing the GAP :) – quest Jul 01 '23 at 06:24
  • $SU(2)$ contains only the $1$-qubit gates. In particular, CNOT is not in $SU(2)$. – Condo Jul 07 '23 at 20:57
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    @quest if you solve the problem, kindly post the solution here. This is a good problem. – Abdullah Khalid Jul 08 '23 at 03:38

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