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.