Let $(A, \Delta, \epsilon)$ be a coalgebra and $f\in A$. What is the subcoalgebra generated by $f$ like?
For example, if $A$ is the dual of the quaternions $\mathbb{H}$ (which is $\mathbb{R}$-algebra with the usual product and $u(1)=1$) and $f$ is the linear functional that takes $1$ to $1$ and $i, j, k$ to $0$.
My target is actually to find an element of $A=\mathbb{H}^*$ that is cocommutative but such that the subcoalgebra generated by it isn't cocommutative. But I don't know where to begin to come up with an $f$, since I don't know how the generated subcoalgebra looks like. I know that the comultiplication is
$$\Delta f(x \otimes y) = f(xy)$$
and to be cocommutative means $f(xy) = f(yx)$, right?
I think the only cocommutative $f$ is the one I mentioned (and its multiples), since $f(ij) = f(ji) \implies f(k)=f(-k) \implies f(k)=0$ and similarly for the others.
EDIT: So, $A = \mathbb{H}^*$ has the basis $1^*, i^*, j^*, k^*$ and "my $f$" is $1^*$. Now
$$\Delta 1^* (x \otimes y) = 1^*(x\otimes y) = \text{ the component of 1 in the product }xy \\= (1^*\otimes 1^* - i^*\otimes i^* - j^*\otimes j^* - k^*\otimes k^* )(x\otimes y)$$.
And I now realize that of course the subcoalgra generated by a set $S$ must be the smallest subcoalgebra containing that set(?). So let $D\subset A$ be a subcoalgebra with $1^* \in D$. It's a subcoalgebra so $\Delta(D) \subset D\otimes D$. Hence
$$1^*\otimes 1^* - i^*\otimes i^* - j^*\otimes j^* - k^*\otimes k^* \in D\otimes D$$
But where to go from here? Do we see that all $i, j, k$ must belong to $D$? How?