Question on definition: primitives in the enveloping algebra of a Lie algebra. 
Let $C$ be a coalgebra, and take $c\in C$. Then $c$ is group-like if $\Delta c=c\otimes c$ and $\epsilon(c)=1_k$, and the set of group-like elements is denoted $G(C)$.
For $g,h\in G(C)$, $c$ is $g,h$-primitive if $\Delta c=c\otimes g+h\otimes c$; the set of such elements is denoted $P_{g,h}(C)$.
If $C$ is a bialgebra and $g=h=1$, then elements of $P(C):=P_{1,1}(C)$ are simply called primitive elements of $C$.

So let $\mathfrak{g}$ be a Lie algebra over a field $k$ (with char$=0$) and let $B=U(\mathfrak{g})$ be its universal enveloping algebra. Then $B$ is a bialgebra via $\Delta x=x\otimes 1+1\otimes x$ and $\epsilon(x)=0_k$ for all $x\in\mathfrak{g}$.
The claim then is that $P(B)=\mathfrak{g}$. For this to be true, we need $1\in G(B)$, meaning we need $\Delta 1=1\otimes 1$ and $\epsilon(1)=1_k$. But neither of these is true!
What's going on here? What am I missing?
 A: The unit $1 \in U(\mathfrak g)$ is indeed grouplike.  Both $\Delta(1) = 1 \otimes 1$ and $\varepsilon(1) = 1$ hold because both $\Delta$ and $\varepsilon$ are homomorphisms of $k$-algebras.
Edit:
Here is an example.  Let $k$ be a field and let $\mathfrak g = kx \oplus ky$ be a two dimensional abelian lie algebra.  So $\{x, y\}$ is a basis of $\mathfrak g$ and $[x, y] = 0$.
Now the universal enveloping algebra of $\mathfrak g$ is the symmetric algebra $S^\ast(\mathfrak g)$ modulo the commutator relations.  Well, the symmetric algebra $S^\ast(\mathfrak g)$ can be identified with $k\langle x, y\rangle$, the ring of non-commuting polynomials in $x$ and $y$.  The commutator relations say exactly that $x$ and $y$ commute.  So when we take the quotient we get that the enveloping algebra is a polynomial ring:
$U(\mathfrak g) = k[x, y]$
The inclusion $\mathfrak g \subseteq k[x, y]$ is the obvious, it's the span of $x$ and $y$ in $k[x, y]$ so the map
$\Delta\colon k[x, y] \to k[x, y] \otimes_k k[x, y]$
is an algebra map defined by $x \mapsto x \otimes 1 + 1 \otimes x$ and $y \mapsto y \otimes 1 + 1 \otimes y$.
This does not mean that those formulas hold for all elements of $k[x, y]$.
For example $\Delta(xy) \neq xy \otimes 1 + 1 \otimes xy$.  Instead we compute $\Delta(xy)$ as:
$\begin{align*}
\Delta(xy) &= \Delta(x)\Delta(y) \\
&= (x \otimes 1 + 1 \otimes x)(y \otimes 1 + 1 \otimes y) \\
&= xy \otimes 1 + x \otimes y + y \otimes x + 1 \otimes xy
\end{align*}$
Now if you think about the degree's of the polynomials you'll see that if $\Delta$ is defined this way then a polynomial $f$ satisfies $\Delta(f) = f \otimes 1 + 1 \otimes f$ if and only if $f$ is homogeneous of degree $1$, i.e., if $f \in \mathfrak g \subseteq k[x, y]$.
