# grouplike elements in Hopf algebras

Recall, that if $k$ is a field and $G$ is a group, then the group algebra $kG$ can be turned into a Hopf algebra, by defining comultiplication $\Delta(g)=g\otimes g$, counit $\varepsilon(g)=1$ and antipode $S(g)=g^{-1}$.

Now let $H$ be a Hopf algebra over a field $k$, with identity $1$, comultiplication $\Delta$, counit $\varepsilon$ and antipode $S$. Recall that element $g\in H$ is called grouplike iff $g\neq 0$ and $\Delta(g)=g\otimes g$. The set of all grouplike elements $G(H)$ is nonempty, because $1\in G(H)$. Also, since comultiplication is an algebra morphism, then $G(H)$ is multiplicative, i.e. if $g,h\in G(H)$, then $gh\in G(H)$. Furthermore, it can be shown that for any $g\in G(H)$ we have $S(g)\in G(H)$ and $S(g)g=gS(g)=1$. Thus $G(H)$ is a group under multiplication inherited from $H$.

It is easy to see, that the vector subspace spanned by $G(H)$ is a Hopf subalgebra of $H$ isomorphic to $kG(H)$. It can be shown that $G(H)$ is always linearly independent, so if $H$ is finite dimensional, then $G(H)$ is a finite group. Also, if $H$ is finite dimensional, then it follows from the Nichols-Zoeller Theorem, that the order of $G(H)$ divides $\mathrm{dim}_{k}H$.

From these observations it follows that if $\mathrm{dim}_{k}H=p$ is a prime number, then $G(H)$ is either trivial or the order of $G(H)$ is equal to $p$ (i.e. $G(H)$ is cyclic of order $p$). The second case implies that $H$ is isomorphic to $k\mathbb{Z}_{p}$ and it can be shown that the first case cannot occur.

Title grouplike elements in Hopf algebras GrouplikeElementsInHopfAlgebras 2013-03-22 18:58:39 2013-03-22 18:58:39 joking (16130) joking (16130) 5 joking (16130) Definition msc 16W30