The Eastin-Knill Theorem shows that the transversal gates always form a group and that moreover this group is a finite subgroup of the group of all unitaries.

For many codes, for example all self dual CSS codes, the group of transversal gates is exactly the Clifford group.


Does anyone know of codes whose group of transversal gates is significantly different from the Clifford group? For example, does anyone know a code on $ n $ qubits whose group of transversal gates is not isomorphic to a subgroup of the Clifford group on $ n $ qubits?

Or failing that, does anyone know any restrictions on which finite groups can occur as the group of transversal gates of a code?

Update: The answer to

Exotic transversal gate group

seems to show that the transversal gate group of an $ [[n,1,d]] $ stabilizer code, for $ d \geq 2 $, must be generated by Clifford gates and/or $ T_k $ gates. So by the classification of finite subgroups of $ PU_2 $ we can conclude that the transversal gate group must be either a dihedral 2 group $ D_{2^k} $ or $ A_4 $ or $ S_4 $ (cyclic transversal gate group is not possible because we have chosen to specialize to stabilizer codes and thus $ X $ and $ Z $ are both transversal and so they generate a noncyclic Klein 4 subgroup and thus the whole transversal gate group must be non cyclic).

Moreover all these groups can indeed be realized of the transversal gate group of some $ [[n,1,d]] $ , $ d \geq 2 $, stabilizer code. Each dihedral 2-group $ D_{2^k} $ arise as the transversal gate group of the corresponding $ [[2^{k+1}-1,1,3]] $ quantum Reed-Muller code. $ A_4 $ is the transversal gate group of the perfect $ [[5,1,3]] $ code see

What are the transversal gates of the [[5,1,3]] code?

And $ S_4 $ (the single qubit Clifford group) is the transversal gate group of the $ [[7,1,3]] $ Steane code see

Transversal logical gate for Stabilizer (or at least Steane code)

  • $\begingroup$ @unknown that's absolutely right! Ya I tend to think of all my transversal gates in the projective unitary group (otherwise every code has infinitely many global phase gates which are transversal), that's why I mentioned $ PU_2$ above. But it's good that you clarified since I did not! $\endgroup$ Oct 27, 2022 at 21:46

1 Answer 1


If you take the Reed-Muller code of 15 qubits, this is a distance 3 CSS code (so has transversal c-NOT, Z and X) but it also has transversal T (and transversal controlled-S and controlled-controlled-Z). What it doesn't have is transversal Hadamard. You'll find this code properly defined in a bunch of places, but, for example, here is the first one that Google threw at me!

  • $\begingroup$ Wow this is a great example! Follow-up question: Let $ Cl_n $ be the (logical) Clifford group on $ n $ (logical) qubits (say encoded with respect to the $ [7,1,3] $ code so that all Clifford gates are transversal). Any set of generators for $ Cl_n $ together with any one gate outside of $ Cl_n $ forms a universal gate set. Let $ G_n $ be the group of transversal gates on $ n $ qubits encoded with respect to the $ [15,1,3] $ code. Is it true that any set of generators of $ G_n $ together with any one gate outside of $ G_n $ forms a universal gate set? $\endgroup$ Mar 2, 2022 at 14:07
  • $\begingroup$ @IanGershonTeixeira I guess that is true, but I don't know if anyone has ever proven it. $\endgroup$
    – DaftWullie
    Mar 2, 2022 at 14:29
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    $\begingroup$ @IanGershonTeixeira Let me revoke my previous affirmation as it's clearly not true. The group of unitaries created by $G_n$ is a subgroup of the permutation matrices with phases on each entry. Other phase gates and other multi-controlled phase gates extend the subgroup, but keep it within that larger group (which essentially corresponds to reversible classical computation). I guess the condition you would need is any unitary for which at least one row has more than one non-zero entry. $\endgroup$
    – DaftWullie
    Mar 3, 2022 at 9:49
  • $\begingroup$ your answer here quantumcomputing.stackexchange.com/a/28699/19675 is nice because it highlights that every dihedral 2-group $ D_{2^k} $ can be realized as the transversal gates set of the $ [[2^{k+1}-1,1,3]] $ quantum Reed-Muller code. This follows from the fact that $ T_{k}=diag(1,e^{2 \pi i/2^k}) $ is transversal for the $ k $th quantum Reed-Muller code. It also seems to provide a restriction on transversal gate groups for $ [[n,1,d]] $ stabilizer codes, they must be generated by Cliffords and/or $ T_k $ gates. $\endgroup$ Oct 27, 2022 at 21:00

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