Let $H$ be a finite dimensional Hopf algebra over a field $k$. Let $G(H)$ be the set of group-like elements of $H$. Is $G(H)$ finite?
1 Answers
Yes. Distinct grouplike elements are linearly independent in any coalgebra (we don't need the antipode or even the multiplication), so there are at most $\dim H$ grouplike elements.
To see this let $\sum_{i=1}^n c_i g_i = 0$ be a linear combination of $n$ distinct grouplike elements $g_i$; our goal will be to show that each $c_i = 0$. By definition we have
$$\Delta \left( \sum_i c_i g_i \right) = \sum_i c_i g_i \otimes g_i = 0$$
and more generally we have $\Delta^k \left( \sum c_i g_i \right) = \sum c_i g^{\otimes (k+1)}_i = 0$. It follows that the Vandermonde matrix with entries $g_i^{\otimes j} \in S(H)$ (the symmetric algebra on $H$) has $(c_1, \dots c_n)$ as a null vector (not sure if it's on the left or the right but it doesn't matter for this argument). But since we assumed that the $g_i$ are distinct, the determinant of this Vandermonde matrix (as an element of $S(H)$) does not vanish; this is a contradiction unless $(c_1, \dots c_n) = 0$ as desired.
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