Are there always nontrivial primitive elements in a Hopf algebra?

The answer is no, even when the algebra is finite dimensional. Consider $k = \mathbb C$, and let $H = \mathbb C$ with the usual algebra structure and the comultiplication given by $\Delta(1) = 1 \otimes 1$, hence $\Delta(z) = z (1 \otimes 1)$. The counit is the identity. Then $\mathbb C$ has no nontrivial primitive elements, because $2z = z \implies z = 0$.

For a less trivial example, consider the group algebra $\mathbb{C}[\mathbb{Z}/2\mathbb{Z}]$ with basis $\langle 1, x \rangle$ ($x$ corresponds to the generator of the cyclic group, $x^2 = 1$). Its coalgebra structure is given by $$\begin{align} \Delta(1) & = 1 \otimes 1 & \Delta(x) & = x \otimes x \\ \epsilon(1) &= 1 & \epsilon(x) &= 1 \end{align}$$

Suppose $u = a + bx \in H$ is primitive, where $a,b \in \mathbb C$. Then $\Delta(u) = a (1 \otimes 1) + b (x \otimes x)$ should equal $1 \otimes (a+bx) + (a+bx) \otimes 1$, which is impossible unless $a=b=0$. More generally, I'm almost sure that the group algebra of a finite group has no primitive elements, but I haven't checked all the details yet.

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However you can still say something in restricted cases. The Milnor–Moore theorem tells you that a locally conilpotent^* cocommutative Hopf algebra $H$ is isomorphic to $\mathbb{U}(\mathbb{P}H)$, that is the enveloping algebra of the Lie algebra of primitive elements. So if $\mathbb{P} H = 0$, then $H = 0$.

^*: Let $\pi : H \to H$ be the projector onto the augmentation ideal $\overline{H}$ of $H$. A Hopf algebra $H$ is said to be locally conilpotent if it can be written as a direct colimit $H = \operatorname{colim} K^m$ such that:

• $\pi^{(n)} \Delta^{(n)}$ is zero when restricted to $K^m$, $m > n$ (this is the iterated coproduct followed by the $n$th tensor power of $\pi$);
• the coproduct restricted to $K^m$ factors through $\operatorname{colim}_{p+q \le m} K^p \otimes K^q \to H \otimes H$.

This is in particular the case when $H$ is a connected Hopf algebra. This is not the case for a group algebra $\mathbb{C}[G]$ with nontrivial $G$: no matter how many times you iterate the coproduct on $g \in G$, none of the factors will ever fall into the augmentation ideal.

For more details about this notion of locally conilpotent and a proof of the Milnor–Moore theorem, I refer you to Benoit Fresse, Homotopy of Operads and Grothendieck-Teichmüller Groups, chapter I.7.

In any group hopf algebra $\mathbb{k}G$ (where $\mathbb{k}$ is a field of characteristic zero), we have a trivial set of primitives: $P(\mathbb{k}G)=0$, because: $$ \Delta(\Sigma_{i} k_ig_i)=\Sigma k_i\Delta(g_i)=\Sigma k_ig_i\otimes g_i \neq 1\otimes \Sigma_{i} k_ig_i + \Sigma_{i} k_ig_i \otimes 1 = \Sigma_{i} k_i(1\otimes g_i + g_i \otimes 1) $$ Note that $\Sigma k_ig_i\otimes g_i$ cannot equal $\Sigma_{i} k_i(1\otimes g_i + g_i \otimes 1)$, due to the fact that $g_i \otimes 1$ and $g_i\otimes g_i$ (for $g_i\neq 1$) are linearly independent in $\mathbb{k}G\otimes\mathbb{k}G$.

And if you want a finite dimensional example, you can always use a finite group $G$.

More generally, it can be shown that:

"If $H$ is a $\mathbb{k}$-Hopf algebra of finite $\mathbb{k}$-dimension and $\mathbb{k}$ is a field of characteristic zero, then $P(H)=0$"

The proof of the above fact, amounts to showing -inductively- that, if there is $0\neq x \in P(H)$ then $x^{n}$ (for all positive integers $n$) are linearly independent, which leads us to an obvious contradiction to the finite dimensionality of $H$. Details on the proof can be found for example at Hopf algebras - an introduction (Dascalescu-Nastasescu-Raianu), exercise 4.2.16, page 174.