Hopf Algebra - Adjoint Representation I've been asked to prove the following;
$$ a \circ (bc) = \sum_{(a)} (a_{(1)} \circ b)(a_{(2)} \circ c)$$
Using the fact that the adjoint representation is as follows;
$$ a \circ b = \sum_{(a)} a_{(1)} b S(a_{(2)})$$
I've tried the expansion of the LHS as follows;
$$ a \circ (bc) = \sum_{(a)} a_{(1)} (bc) S(a_{(2)})$$
$$ = \sum_{(a)} a_{(1)} \epsilon(bc) S(a_{(2)})$$
$$ = \sum_{(a)} a_{(1)} \epsilon(b) \epsilon(c) S(a_{(2)})$$
$$ = \sum_{(a)} a_{(1)} b_{(1)} S(b_{(2)}) c_{(1)} S(c_{(2)}) S(a_{(2)})$$
However, from here, I'm not sure where to go. Or if I've even done the correc tthing (in using the antipode property).
Any help would be great!!
EDIT: Is the following appropriate??
$$ a \circ (bc) = \sum_{(a)} a_{(1)} \epsilon(b) \epsilon(c) S(a_{(2)})$$
$$ = \sum_{(a)} \epsilon(a_{(1)}) \epsilon(b) \epsilon(c) S(a_{(2)})$$
$$ = \sum_{(a)} a_{(1)} S(a_{(2)}) \epsilon(b) \epsilon(c) S(\epsilon(a_{(3)}) a_{(4)})$$
$$ = \sum_{(a)} a_{(1)} S(a_{(2)}) \epsilon(b) \epsilon(c) \epsilon(a_{(3)} S( a_{(4)})$$
$$ = \sum_{(a)} a_{(1)} S(a_{(2)}) (b) (c) a_{(3)} S( a_{(4)})$$
$$ = \sum_{(a)} a_{(1)} bS(a_{(2)}) a_{(3)}c S( a_{(4)})$$
$$ = \sum_{(a)} (a_{(1)} \circ b)(a_{(2)} \circ c)$$
I know I can pretty much drop the $\epsilon$ functions, but, is it okay to move the $b$ and $c$ terms around?? If they're just numbers, it seems to me as if it wouldn't matter where they go??
 A: By definition, we have
$$
\sum ( a_{(1)}  \circ b ) ( a_{(2)} \circ c)  = \sum a_{(11)} b  S(a_{(12)}) \ a_{(21)}   c S(a_{(22)}).
$$
By coassociativity, this is equal to
$$
\sum a_{(1)} b S(a_{(211)}) a_{(212)} c S(a_{(22)}).
$$
As $\sum S(a_{(211)}) a_{(212)} = \varepsilon( a_{(21)})$, this yields
$$
\sum a_{(1)} b\varepsilon( a_{(21)}) c S(a_{(22)}).
$$
Using the fact that $\varepsilon( S(a_{(21)}) ) = \varepsilon( a_{(21)})$, we rewrite this as
$$
\sum a_{(1)} bc \ \varepsilon( S(a_{(21)}))  S(a_{(22)}).
$$
But this is equal to 
$$
\sum a_{(1)} bc S(a_{(2)}),
$$
as desired.
Your second approach has basically the right idea: leave $b$ and $c$ alone and use coassociativity and the properties of the antipode and counit to move $a_{(1)}$ and $a_{(2)}$ around, but I think the details are incorrect. You cannot just replace $bc$ and $a_{(1)}$ by $\varepsilon(bc)$ and $\varepsilon (a_{(1)})$ as you do. Also, in general a Hopf algebra is not commutative and you cannot move $b$ and $c$ around. In my approach, $\varepsilon( S(a_{(21)}) )$ is a constant and thus does commute with the other elements of the algebra.
To clarify my notation: 
$$
\Delta(a) = \sum a_{(1)} \otimes  a_{(2)}, \quad (\iota \otimes \Delta)\circ \Delta (a)= \sum a_{(1)} \otimes  a_{(21)} \otimes a_{(22)},
$$
and so on.
