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Stefan Kohl
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Your representation $p$ is not faithful, since we have $$ ABA^{-1}BA^{-1}BAB^{-1}ABA^{-1}BA^{-1}BAB^{-1}ABA^{-1}BA^{-1}BAB^{-1} \ = \ 1. $$

Stefan Kohl
  • 19.6k
  • 21
  • 75
  • 137