Order of antipode in Sweedler's Hopf Algebra I'm learning the basics of Hopf Algebras, and I've found an example that I'm not able to follow, Sweedler's Hopf Algebra. 
In this Hopf Algebra, we have $g^2 = 1$, $x^2 = 0$, $xg = -gx$, so that in particular $g^{-1} = g$. Then the antipode $\sigma$ is defined by $g \mapsto g, x \mapsto -gx$. This is fine, except I have read (in Fundamentals of Hopf Algebras, by Robert G. Underwood) that in this example $\sigma$ has order $4$, which I'm struggling to follow, as it seems to me $\sigma$ has order $2$.
If we try:
$$(\sigma \circ \sigma)(x) = \sigma(\sigma(x)) = \sigma(-gx) = \sigma(xg)$$
Then (this is possibly where I'm going wrong) I assume we have that $\sigma(xg) = \sigma(x\cdot g) = \sigma(x) \cdot \sigma(g)$. In which case:
$$\sigma(x) \cdot \sigma(g) = -gx \cdot g = xg \cdot g = xg^2 = x$$
So $\sigma$ has order $2$. So my question is, why does my book claim that it has order $4$ (or: what have I done wrong)?
 A: Unlike the coproduct or the counit, the antipode is generally not a morphism of algebras. Often it's actually an antihomomorphism, i.e. $\sigma(ab) = \sigma(b) \sigma(a)$, but here we don't know that (yet).
The Wikipedia article doesn't say what $\sigma(xg)$ is, but we can find it. (Apparently you have a different convention from the Wikipedia article, since you wrote $\sigma(x) = -gx$ but Wikipedia has $\sigma(x) = -xg$. I use Wikipedia's convention but you can reverse everything if you want, you also need to change the coproduct if you want to keep your antipode.) First compute:
$$\Delta(xg) = \Delta(x) \cdot \Delta(g) = (1 \otimes x + x \otimes g) \cdot (g \otimes g) = g \otimes xg + xg \otimes g.$$
Then apply $\sigma \otimes 1$ to this, and multiply to get
$$\mu((\sigma \otimes 1))(\Delta(xg)) = \mu(g \otimes xg + \sigma(xg) \otimes 1) = gxg + \sigma(xg) = -g^2x + \sigma(xg) = -x + \sigma(xg).$$
But by definition of the antipode, this has to be equal to $\varepsilon(xg) \cdot 1 = 0$, hence $\sigma(xg) = x$. Now we get:
$$\sigma(\sigma(x)) = \sigma(-xg) = - x.$$
