Let $A=:F(G)$ be the algebra of functions on a finite quantum groups aka a finite dimensional C*-Hopf Algebra.

Suppose that $F(G)$ is neither commutative nor cocommutative.

In their 1966 paper Kac and Paljutkin, show that when we write $F(G)$ as a multimatrix algebra, one of which must be one-dimensional (to account for the counit), that we can decompose into one-dimensional factors and $n_j>1$ dimensional factors: $$F(G)=\left(\bigoplus_{g\in G_1}\mathbb{C}\delta_g\right)\oplus\left( \bigoplus_{j=1}^mM_{n_j}(\mathbb{C})\right):=A_1\oplus B,\qquad (\star)$$

where $G_1$ is a finite group.

Kac and Paljutkin show that:

$$\Delta(A_1)\subseteq A_1\otimes A_1+B\otimes B.$$

Assumption 1:Kac and Paljutkin then assume that $m=1$ so that $B=M_{n_1}(\mathbb{C})$ contains a single summand.

They show that: $$\Delta(B)\subseteq A_1\otimes B+B\otimes A_1+B\otimes B.$$

Let $f\in B$ and write:

$$\Delta(f)=\underbrace{T(f)}_{\in A_1\otimes B+B\otimes A_1}+\underbrace{K(f)}_{\in B\otimes B}.$$

Assumption 2:Then they assume further that $|G_1|=n_1^2$.

From there they show that $K(f)=0$ so that in fact

$$\Delta(B)\subseteq A_1\otimes B+B\otimes A_1.$$

I am not sure if the assumptions were made to write down a quicker/easier proof for the purposes of illustration... or if there is a finite quantum group such that this inclusion does not hold.

I am wondering in the 50 odd years hence has anyone working in Finite Quantum Groups (finite dimensional Hopf *-algebras $H$ with $f^*f\neq 0$ for all $f\in H$) showed that with respect to the decomposition $(\star)$ that $$\Delta(B)\subseteq A_1\otimes B+B\otimes A_1?$$

Question 1:Does this inclusion hold? If not, what is a counterexample?

Question 2:Is it possible that $\Delta(M_{n_j}(\mathbb{C}))$ never falls into $M_{n_j}(\mathbb{C})\otimes M_{n_j}(\mathbb{C})$? It might be easier to show this more specific result (that must also be useful for me).

**Some Efforts:** Let $E^j_{mn}$ be a matrix unit in $M_{n_j}(\mathbb{C})$. We want to show that
$$(I_{n_j}\otimes I_{n_j})\Delta(E^j_{mn})=0.$$
This would be sufficient for my needs. I have the following assumption about how the antipode relates to the multi-matrix structure but I cannot find any suggestion that it is true. The only thing that seems remotely consistent with it is that $S^2=I_{F(G)}$. **The assumption is that on the $m$ non-commutative factors, the antipode is the transpose.** This allows a partial result to be proved using the antipodal property.

Take a matrix unit. We have that $\varepsilon(E_{mn}^j)=0$ and this implies further that $$m\circ (S\otimes I_{F(G)})\circ\Delta(E^j_{mn})=0=m\circ (I_{F(G)}\otimes S)\circ \Delta(E^j_{mn}).$$

Think about the $A_1\otimes B$ and $B\otimes A_1$ parts of $\Delta(E^j_{mn})$ (there is no $A_1\otimes A_1$ part by Kac and Paljutkin). We know that $S(A_1)\subset A_1$, $S(B)\subset B$, and moreover $A_1B=0$, and so when we multiply $m\circ (A_1\otimes B)$ we get zero.

Now choose lots of structure constants (for a fixed $j$) $\alpha^{mn}_{xy,zw}\in\mathbb{C}$: $$(I_{n_j}\otimes I_{n_j})\Delta(E^j_{mn})=\sum_{x,y,z,w=1}^{n^j}\alpha_{xy,zw}^{mn}(E^{j}_{xy}\otimes E^j_{zw}).$$ There are probably lots of conditions on the $\alpha$, but under the assumption that the antipode is the transpose, and the linear independence of the matrix units, it is the case that for any $x,y,w=1,\dots,n_j$, $\alpha_{xy,wy}=0$ and $\alpha_{xy,xw}=0$.

I am hoping that the homomorphism property of $\Delta$ might 'finish off' the other structure constants.