# 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 $\mathrm{\Delta}(g)=g\otimes g$, counit $\epsilon (g)=1$ and antipode $S(g)={g}^{-1}$.

Now let $H$ be a Hopf algebra over a field $k$, with identity^{} $1$, comultiplication $\mathrm{\Delta}$, counit $\epsilon $ and antipode $S$. Recall that element $g\in H$ is called grouplike iff $g\ne 0$ and $\mathrm{\Delta}(g)=g\otimes g$. 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 |
---|---|

Canonical name | GrouplikeElementsInHopfAlgebras |

Date of creation | 2013-03-22 18:58:39 |

Last modified on | 2013-03-22 18:58:39 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 5 |

Author | joking (16130) |

Entry type | Definition |

Classification | msc 16W30 |