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What is an example of a Hopf algebra $(H,\Delta,\epsilon)$ containing an invertible element $h$ which is not grouplike: An element $h \in H$ such that $$ \Delta(h) \neq h \otimes h\qquad\text{(not grouplike)} $$ and such that there exists a $h^{-1} \in H$ with $hh^{-1} = h^{-1}h = 1$ (invertible).

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    $\begingroup$ Hae you looked at the group ring of a (finite) group? $\endgroup$
    – Tyrone
    Commented Feb 22, 2021 at 23:30
  • $\begingroup$ Related (though clearly not the same question—this one seems much quicker to address, per @Tyrone's comment): mathoverflow.net/questions/86197/… . $\endgroup$
    – LSpice
    Commented Feb 22, 2021 at 23:45
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    $\begingroup$ Trivial example: scalar multiples of grouplikes are generally not grouplike. $\endgroup$
    – lambda
    Commented Feb 23, 2021 at 6:13
  • $\begingroup$ @Tyrone: But in the group algebra every element is grouplike? $\endgroup$ Commented Feb 23, 2021 at 13:23
  • $\begingroup$ You seem to have just changed your question... $\endgroup$
    – Tyrone
    Commented Feb 23, 2021 at 13:34

2 Answers 2

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Let $L$ be a finite-dimensional $p$-nilpotent restricted Lie algebra over a field of characteristic $p>0$ and consider its restricted enveloping algebra $u(L)$. Then the only group-like element of the Hopf algebra $u(L)$ is 1. On the other hand, every element of $u(L)$ that is not in the kernel of the counit is invertible.

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Let $G$ be any finite group. Then the group algebra $\mathbb{C}[G]$ is, as an algebra, isomorphic to $\bigoplus_V \mathrm{End}(V)$, where the direct sum is over irreducible representations of $V$ of $G$. So the group of units of $\mathbb{C}[G]$ is isomorphic to $\prod_V \mathrm{GL}(V)$. Only finitely many elements of this product come from the original group $G$. Any element of $\prod_V \mathrm{GL}(V)$ not coming from $G$ is invertible but not group like.

To be concrete, let $G$ be the cyclic group $C_2$, with nontrivial element $g$. Then $a+bg$ is invertible as long as $(a+b)(a-b) \neq 0$, but $a+bg$ is group like only for $(a,b) = (1,0)$ and $(a,b) = (0,1)$.

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