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?
Let $A$ be a $k$-algebra. If $I$ is a two-sided ideal in $A$ then the orthogonal $$ I^\perp = \{ \alpha \in A^* \mid \text{$\alpha(x) = 0$ for every $x \in I$} \} $$ is a subcoalgebra of $A$, and if $C$ is a subcoalgebra of $A$ then its orthogonal $$ C^\perp = \{ x \in A \mid \text{$\alpha(x) = 0$ for every $\alpha \in C$} \} $$ is a two-sided ideal in $A$. If $A$ is finite dimensional then these constructions are mutually inverse and therefore give a one-to-one correspondence between the two-sided ideals in $A$ and the subcoalgebras of $A^*$.
The quaternions are a skew-field and thus admit only the two trivial ideals $0$ and $\mathbb{H}$. Hence $\mathbb{H}^*$ admits only $0^\perp = \mathbb{H}^*$ and $\mathbb{H}^\perp = 0$ as subcoalgebras. Thus every nonzero element of $\mathbb{H}^*$ generates the whole coalgebra.
(This is taken from Sweedler’s Hopf Algebras, Proposition 1.43 and pages 63-64.)
You’re right that the cocommutativity of an element $f \in \mathbb{H}^*$ means that $f(xy) = f(yx)$ for all $x, y \in \mathbb{H}$. Indeed, if we identify $\mathbb{H}^* \otimes \mathbb{H}^*$ with $(\mathbb{H} \otimes \mathbb{H})^*$ then the cocommutativity of $f$ means $$ \Delta(f)(x \otimes y) \overset{!}{=} \Delta(f)(y \otimes x) \,. $$ Then also $\Delta = m^*$ for the multiplication $m \colon \mathbb{H} \otimes \mathbb{H} \to \mathbb{H}$, and we find that $$ f(xy) \overset{!}{=} f(yx) \,. \tag{$\ast$} $$
You’re also right that the only possible $f$ is up to scalar multiple the one you give (with $f(1) = 1$ and $f(i) = f(j) = f(k) = 0$). One can make your argument a bit less ad-hoc by observing that $(\ast)$ is equivalent to $$ f(xy - yx) \overset{!}{=} 0 \,, $$ which means that $f$ must vanish on the subspace $$ \langle xy - yx \mid x, y \in \mathbb{H} \rangle \,. $$ One can now check that this subspace is generated by $i$,$j$,$k$, giving again your result. (It sufficies to consider for $x$ and $y$ basis vectors, and the basis vector $1$ can be ignored. It now sufficies to consider the cases $(x,y) = (i,j), (i,k), (j,k)$, which give the vector space generators $\pm 2k$, $\pm 2j$, $\pm 2i$.)
There is also a canonical way to construct for any finite dimensional $k$-algebra $A$ with $\operatorname{char}(k) \nmid \dim(A)$ some nonzero cocommutative element $f \in A^*$: Every element $a \in A$ gives rise to an endomorphism $L_a \colon A \to A$ given by $L_a(x) = ax$. Then $f(a) := \operatorname{tr}(L_a)$ satisfies $$ f(ab) = \operatorname{tr}(L_{ab}) = \operatorname{tr}(L_a L_b) = \operatorname{tr}(L_b L_a) = \operatorname{tr}(L_{ba}) = f(ba) \,, $$ and $f$ is nonzero because $f(1) = \operatorname{tr}(L_1) = \operatorname{tr}(\mathrm{id}_A) = \dim(A) \neq 0$. For $A = \mathbb{H}$ this gives $f(i) = f(j) = f(k) = 0$ and $f(1) = 4$.
(This is again taken from Sweedler’s Hopf Algebras, pages 63-64.)