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.
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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 $|G\_1|=n\_1^2$.
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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).
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**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.



Image of Comultiplication on Finite Quantum Groups/Hopf Algebras
I guess what you are looking for is the inclusion matrix for the unital inclusion of finite dimensional ${\rm C}^{\star}$-algebras $\Delta(A) \subset A \otimes A$. It is given by the fusion rules for $Rep(A)$, see Proposition 7.4 in my preprint [arXiv:1704.00745v5](https://arxiv.org/pdf/1704.00745v5.pdf). 


*Example*: consider the finite dimensional Hopf ${\rm C}^{\star}$-algebra $A=\mathbb{C}S\_3$ with the symmetric group $S\_3$. Then as an algebra $$A \simeq \mathbb{C} \oplus \mathbb{C} \oplus M\_2(\mathbb{C}).$$
Now, $Rep(A)=Rep(S\_3)$ and the fusion rules $(n\_{ij}^k)$ are given by the following matrices:  

$$\left(\begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 
\end{matrix} \right), 
\left(\begin{matrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1 
\end{matrix} \right), 
\left(\begin{matrix}
0 & 0 & 1 \\
0 & 0 & 1 \\
1 & 1 & 1 
\end{matrix} \right) $$


This provides a counter-example for your questions because $n\_{33}^3 = 1 \neq 0$.  

[If you want, you can find the explicit formulas for the comultiplication computed [here](https://math.stackexchange.com/a/902411/84284).]


You can get many counter-examples from some non-abelian finite groups.  

See the following fusion rules for $Rep(A\_5)$: 


$$\left(\begin{smallmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1 \end{smallmatrix} \right), 
\left(\begin{smallmatrix}0&1&0&0&0\\1&1&0&0&1\\0&0&0&1&1\\0&0&1&1&1\\0&1&1&1&1 \end{smallmatrix} \right), 
\left(\begin{smallmatrix}0&0&1&0&0\\0&0&0&1&1\\1&0&1&0&1\\0&1&0&1&1\\0&1&1&1&1 \end{smallmatrix} \right), 
\left(\begin{smallmatrix}0&0&0&1&0\\0&0&1&1&1\\0&1&0&1&1\\1&1&1&1&1\\0&1&1&1&2 \end{smallmatrix} \right), 
\left(\begin{smallmatrix}0&0&0&0&1\\0&1&1&1&1\\0&1&1&1&1\\0&1&1&1&2\\1&1&1&2&2 \end{smallmatrix} \right)$$


**Counter-examples (for your questions) which are neither commutative nor cocommutative**  

 Let $\mathcal{C}$ be the class of non-abelian finite groups $G$ having an irreducible complex representation $V$ with $\dim(V)>1$ and which is an irreducible component of $V \otimes V$ (the groups $S\_3$ and $A\_5$ are of class $\mathcal{C}$, as shown above). Let $G$ be of class $\mathcal{C}$, and let $H$ be any non-abelian finite group. Let $A,B$ be the finite dimensional Hopf ${\rm C}^{\star}$-algebras $\mathbb{C}G$ and $\mathbb{C}H$ (respectively), then the usual tensor product of Hopf algebras $A \otimes B^\*$ is a non-commutative and non-cocommutative counter-example of your questions.  

In addition, for any twist $J$ of $A$, $A\_J$ has the same algebra structure than $A$. Moreover, $J$ is an invertible element of $A \otimes A$ and $\Delta\_J = J^{-1} \Delta J$, so the inclusions $\Delta(A) \subset A \otimes A$ and $\Delta\_J(A\_J) \subset A\_J \otimes A\_J$ have the same inclusion matrices. Thus, $Rep(A\_J)$ and $Rep(A)$ have the same fusion rules. Take a group $G$ of class $\mathcal{C}$, which admits a non-trivial twisting $(\mathbb{C}G)\_J$ as for $G=A\_5$ (by a result of Nikshych). It follows that $(\mathbb{C}G)\_J$ is also a non-commutative and non-cocommutative counter-example of your questions.


