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Suppose we have two Hopf algebras, H and A and additionally A is (left) H-module algebra and H is (right) A comodule coalgebra. This means that A is left module over H and moreover that $h(ab)=h_{(1)}(a)\otimes h_{(2)}(b)$ (using the Sweedler notation) and $h(1_A)=\varepsilon(h)1_A$. Dualising these axioms (reversing the arrows on the suitable diagrams where the action of $H$ on $A$ is thought as $\alpha:H \otimes A \to A$) we obtain something which is call $A$-comodule coalgebra. One can define the suitable algebra and coalgebra structure on $A \otimes H$. When certain compatibility axioms are satisfied $A \otimes H$ becomes a Hopf algebra (after defining antipode in a appropriate way). This is done in S. Majid book on Quantum Groups, in chapter 6. I'm looking for the simplest example of such a construction: in "Basic Noncommutative Geometry" by M. Khalkhali i found the following: given a finite group $G$ suppose that it factories to $G_1G_2$ meaning that $G_1G_2=G, G_1 \cap G_2=\{e\}$ and $G_1,G_2$ are subgroups (nothing was said about the normality of them). Write each $g \in G$ as $g_1g_2$ with $g_i \in G_i$ (which are determined uniquely). Author claims that $g \cdot h:=(gh)_2$ defines left action of $G_1$ on $G_2$ and also conversely: $G_2$ acts from the right on $G_1$ by the formula $g*h:=(gh)_1$. It seems to me that this actions are trivial: for example, if $g\in G_1, h\in G_2$ then $gh$ is already written in such a way that $(gh)_2=h$. The same thing with the right action. I've stuck in this moment: is there really a problem, or I'm understanding something wrong? Further author claims that this action defines the structure of left $\mathbb{C}G_1$ algebra on the Hopf algebra of all functions on $G_2$. There is no formula for the action of $\mathbb{C}G_1$ in this book so I'm only making guess that is defined by the following: $(\alpha_1 g_1+...+\alpha_ng_n)f(h)=\alpha_1 f(g_1 \cdot h)+...+\alpha_nf(g_n \cdot h)$. I would be grateful if anybody who is familiar with this example could provide me a missing ingredients for this example to work. In particular: do we realy need the explicit form of the actions of $G_i$ on $G_j$ or it is enough to know that there is \textit{some} actions of $G_i$ on $G_j$?

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  • $\begingroup$ I think you have a left/right confusion. To avoid (maybe?) it, I will use different letters. Suppose the group $G$ factors as $G=HK$; I will denote the factorization as $g=g_Hg_K$ for any $g\in G$. Then define a left action of $K$ on $H$ by $k\triangleright h=(kh)_H$, and a right action of $H$ on $K$ by $k\triangleleft h =(kh)_K$. We have $kh=(kh)_H(kh)_K$, but not in the other order, so generically these are nontrivial actions. They are both trivial only for $G=H\times K$. One action is trivial and the other is not for (nontrivial) semidirect products; that example makes a good exercise. $\endgroup$ Jan 30, 2014 at 2:40
  • $\begingroup$ Thank You, probably it should look like You have described. But still I couldn't manage to prove that $A=\mathbb{C}^G_1$ and $H=\mathbb{C}G_2$ satisfy the compatibility conditions. Not going into details: it would be sufficent for me to show that these actions satisfy:$h'h \triangleleft g=(h' \triangleleft g)(h \triangleleft (h' \triangleright g))$ and $h \triangleright gg'=((h \triangleleft g') \triangleright g)(h \triangleright g')$. I was only able to show $h'h \triangleleft g=(h' \triangleleft (h \triangleright g))(h \triangleleft g)$ and "right handed" version of this. $\endgroup$
    – truebaran
    Feb 3, 2014 at 17:23
  • $\begingroup$ Well, that sure looks like another left/right error, either on you part or on the part of whichever author you're reading. I get the same equations you do when I try the exercise. $\endgroup$ Feb 4, 2014 at 3:31

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