# How can one argue that the $S$-gate is Clifford while $T$-gate is not?

How can one argue that $$S$$-gate is Clifford while $$T$$-gate is not?

By definition, conjugation by a Clifford gate preserves the Pauli group $$G = \langle X, Y, Z\rangle$$. It is easy to check that $$SXS^\dagger = Y, SYS^\dagger = -X$$ and $$SZS^\dagger = Z$$. Since $$G = \langle -X, Y, Z\rangle$$ we see that $$S$$ is Clifford. On the other hand, $$TXT^\dagger = \begin{pmatrix} & e^{-i\pi/4}\\ e^{i\pi/4}&\end{pmatrix} \notin G$$, so $$T$$ is not.
Geometrically: For a single qubit, we have the Bloch sphere and the stabiliser states span an octahedron inside it. Unitaries act in the adjoint representation as $$SO(3)$$, i.e. they induce rotations of the Bloch sphere.
It is easy to see, that only rotations about the $$X,Y,Z$$ axis with angles $$\theta=\pi/2,\pi,3\pi/2,2\pi$$ preserve the octahedron. This exactly correspond to unitaries of the form $$U = \exp(i \theta/2 P), \quad P=X,Y,Z$$ Now, $$S$$ is a $$\theta=\pi/2$$ rotation, while $$T=\sqrt{S}$$ is a $$\pi/4$$ rotation.