be seen in the next two examples of $S_n^{\varepsilon}$. Note that when Banica, Bichon and others investigate graphs which have no quantum symmetry [@BanBicQAG; @BBGNoSym; @BanSurvey], this is exactly the same question: They ask whether or not $C(S_n^{\Gamma_{\varepsilon}})$ is commutative. Such investigations and concrete examples may also be found in [@QAGThesis Thm. 5.6.1 and Thm. 6.4.1].

Recall from Section \[SectEpsSphere\] that we may view the full group $C^*$-algebra associated to $\mathbb Z_2*\mathbb Z_2$ as a universal $C^*$-algebra. We now give a well-known alternative presentation.

\[LemPQ\] The following universal $C^*$-algebras are isomorphic and noncommutative.

-   $C^*(\mathbb Z_2 *\mathbb Z_2)=C^*(z_1,z_2,1\;|\; z_i=z_i^*, z_i^2=1, i=1,2)$

-   $C^*(p,q,1\;|\; p=p^*=p^2,q=q^*=q^2)$

The isomorphism between (a) and (b) is given by: $$z_1\mapsto 2p-1,\qquad z_2\mapsto 2q-1,\qquad 1\mapsto 1$$ Noncommutativity follows from Lemma \[RepSphere\](a).

The following example has also been treated by Bichon in [@BicQAG Prop. 3.3].

\[ExKey\] Let ${\varepsilon}\in M_4(\{0,1\})$ be as in Example \[ExEps\](d) given by: $${\varepsilon}=\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}$$ The $C^*$-algebra $C(S_4^{\varepsilon})$ is noncommutative and hence $S_4^{\varepsilon}$ is a quantum group which is not a group.

The following matrix (using Lemma \[LemPQ\]) in $M_4(C^*(\mathbb Z_2*\mathbb Z_2))$ gives rise to a representation of $C(S_4^{\varepsilon})$ as may be verified directly: $$\begin{pmatrix}p&1-p&0&0\\1-p&p&0&0\\0&0&q&1-q\\0&0&1-q&q\end{pmatrix}$$ We thus have a surjection of $C(S_4^{\varepsilon})$ onto $C^*(\mathbb Z_2*\mathbb Z_2)$ 