I'm trying to show that none of these sets of gates are universal:

  • ${CNOT^2, H, T}$

  • ${CNOT, H^2, T}$

  • ${CNOT, H, T^2}$

For the first one $CNOT^2$ seems to be the identity gate which means that there's no way this gate set can flip the value of some bits. I'm not sure if this argument is correct/suffices.

For the second one, $H^2$ is the identity gate again(rotation of angle $2\pi$), so now this gate set can only rotate about 1 axis with the $T$ gate hence it's not a universal gate set.

The third one is the one I'm most unsure about. $T^2$ seems to be a rotation of $\pi/2$ about the $z$ axis and that does not seem to cause any problems (eg: $HT^2$) still seems to be a rotation about 2 distinct axes by an angle that's an irrational.

  • 1
    $\begingroup$ For the third one, please see the Gottesman-Knill theorem. $T^2=S$, which is a gate in the normalizer of the Pauli group (a Clifford gate). $\endgroup$ Apr 28, 2022 at 20:44

1 Answer 1


First gate set: only has single qubit gates. The group of single qubit gates is closed under composition. E.g. you can't get to CNOT because it's not a single qubit gate.

Second gate set: only has phased permutations. The group of phased permutations is closed under composition. E.g. you can't get to H because it's not a phased permutation.

Third gate set: only has Cliffords. The group of Cliffords is closed under composition. E.g. you can't get to T because it's not a Clifford.


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