Antipode of Cocommutative Hopf Algebra I’m reading about affine group schemes by Waterhouse and in the proof of showing the (Jordan) decomposition of Abelian affine group scheme (equivalently cocommutative Hopf algebra), I came across the argument, saying:
“Since the Hopf algebra $A$ is cocommutative, the antipode $S$ is a coalgebra morphism and therefore sends $ A’ $ (a sub-bialgebra of $A$) into $A’$“
I know that the antipode is an algebra automorphism of $A$, and in this case a bialgebra automorphism of A, but I still don’t understand why does the restriction of $S$ to $A’$ would send $A’$ back to itself.
P/S: we are assuming that Hopf Algebras are commutative here.
 A: I do not have access to the book you are citing, but if i understand correctly, you are speaking about a commutative, cocommutative hopf algebra.
If your assumptions are complemented by: finite dimensional hopf algebra over an algebraically closed field of characteristic zero, then the author's claim:  

Since the Hopf algebra $A$ is cocommutative, the antipode $S$ is a coalgebra morphism and therefore sends $ A’ $ (a sub-bialgebra of $A$) into $A’$

seems to be correct.
Here is my argument:    
Under these assumptions (i.e.: f.d., commutative, cocommutative over alg. closed field $k$ with $chark=0$), it is relatively easy to show that the hopf algebra $H$ is actually isomorphic to the group algebra of the group $G(H)$  formed by  its grouplike elements:
$$
H\cong kG(H)
$$
(This isomorphism can also be viewed as a direct consequence of the Cartier-Konstant-Milnor-Moore theorem applied in the finite dimensional setting.)
Thus, by the same argument,  the sub-bialgebras $A'$ of $H$ -which are again f.d., commutative,  cocommutative, over an algebr. closed field of $chark=0$- are group algebras $A'=kN$, for some subgroup $N$ of $G(H)$. But then, since for any grouplike $S(g)=g^{-1}$ we get that:
$$
S(\sum_{g\in N}a_g g)=\sum_{g\in N}a_g S(g)= \sum_{g\in N}a_g g^{-1}\in kN
$$ 
thus $S(kN)\subseteq kN \ \ $ i.e.: $ \ S(A')\subseteq A'$.  
