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let $(H,\Delta,m,s)$ be a Hopf algebra. To this Hopf algebra one can associate two obvious linear maps $T_H, S_H: H \to H $ with $T_H=m\circ \Delta,\quad S_H=s$.

Are there two finite dimensional Hopf algebras $H,H'$ which are not isomorphic Hopf algebras but they are equivalents in the following sense:

Each coefficient of characteristic polynomial $Det (T_H-\lambda I)$ equals to the corresponding coefficient of $Det (T_{H'}-\lambda I)$. Similarly, Each coefficients of $Det (S_H-\lambda I)$ equals to the corresponding coefficients in $Det (S_{H'}-\lambda I)$.That is:$Det(T_H)=Det(T_{H'}),\ldots,trace(T_H)=traceT_{H'}$ and likewise for $S_H$ and $S_{H'}$.

In the other words, to what extent the set of these coefficients can determine the nature of Hopf algebras. Are there some infinite dimensional analogy for these quantities?

Are there two finite groups $G,G'$ of the same orders which are not isomorphic groups but their corresponding Hopf algebras $CG, CG'$ have the same quantities described above?

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    $\begingroup$ Ali Taghavi, just out of curiosity: what was the motivation? I mean what kind of problem were you studying while these quantities came out ? $\endgroup$ Commented Aug 21, 2019 at 16:50
  • $\begingroup$ @KonstantinosKanakoglou To be honest no problem I was working which lead me to this question. I asked myself this question Spontaneously(suddenly) without any pre-motivation. then I presented at MO. $\endgroup$ Commented Aug 21, 2019 at 17:47
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    $\begingroup$ @KonstantinosKanakoglou But may be the following concept was in my subconscious mind; Hyperrigidity of groups. groups whose group structre is determined by $c^*$ algebraic structure of $c^*_{red} G$. $\endgroup$ Commented Aug 21, 2019 at 19:13
  • $\begingroup$ @KonstantinosKanakoglou but i really was considering the alternative coalgebra structure not the group like one. $\endgroup$ Commented Aug 21, 2019 at 19:18

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For a finite group algebra, $T_H(g)=g^2$ and $S_H(g)=g^{-1}$ for any group element $g$.

So for any two finite groups of exponent three and the same order, the quantities will be the same. For example, there's a non-abelian group of order 27 and exponent 3, and also an abelian one.

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  • $\begingroup$ Thank you very much for your answer. to be honest I was considering another comultiplcation with $\Delta f(x,y)=f(xy)$ but you are considering the group like structure. $\endgroup$ Commented Aug 21, 2019 at 11:14

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