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Let $(H, m, u, Delta, epsilon, S)$ be a $K$-Hopf Algebra. We call an element $xin H$ grouplike if $Delta(x) = x otimes x$ and $epsilon(x) = 1_K$. The set of all grouplike elements is a group (denoted $G(H)$).

How do we show this?

I thought I might have the group $(G(H), mu, e, S')$ with $ mu : G(H) times G(H) rightarrow G(H) : (x,y) rightarrow m(xotimes y)\ e: {*} rightarrow G(H):* rightarrow u(1_K) \ S':G(H) rightarrow G(H):x rightarrow S(x)$

I can show that this structure gives us a group. But what I can't show is that the images of $mu$ and $S'$ are indeed in $G(H)$.

We have to show that

1) $∆(m(xotimes y)) = m(xotimes y) otimes m(xotimes y)$ (EDIT: Solved)

2) $epsilon(m(xotimes y)) = 1_K$ (EDIT: Solved)

3) $∆(S(x)) = S(x) otimes S(x)$

4) $epsilon(S(x)) = 1_K$

Any hints would be appreciated

Let $g$ and $h$ be grouplike elements. Then $Delta(gh)=Delta(g)Delta(h)=(gotimes g)(hotimes h)=(ghotimes gh)$. Indeed the multiplication on $Hotimes H$ uses the standard flip map $tau$, thus $m_{Hotimes H}=(motimes m)circ (1otimes tauotimes 1)$. Thus the multiplication of grouplike elements is inner.

Now since $H$ is a Hopf algebra, the antipode axiom tells us that $S(g)g=varepsilon(g)1$. Since $varepsilon(g)=1$ (because $g$ is grouplike), this becomes $S(g)g=1$ and thus $S(g)$ is a left inverse for $g$. Similarly, $S(g)$ is a right inverse for $g$.

Now we show that $S(g)in G(H)$. By applying $Delta$ to $S(g)g=1$, we get that $$Delta(S(g))(gotimes g)=1otimes 1.$$ Hence $Delta(S(g))$ is the inverse of $gotimes g$ in the algebra $Hotimes H$. By uniqueness of inverses, $Delta(S(g))=S(g)otimes S(g)$ as desired.

Questions 2 and 4 follows immediately from the fact that the counit applied to grouplike elements is $1$.

First, note that the set of grouplikes $G(H)$ is closed wrt the multiplication of $Η$. This is a consequence of the fact that the comultiplication is an algebra homomorphism:
$$
Delta(gh) =
Delta(g) Delta(h) = (g otimes g)(h otimes h) = gh otimes gh
$$
i.e. $g,h in G(H)$ $Rightarrow$ $gh in G(H)$.

On the other hand, $1_{Η} in G(H)$, since in any Hopf algebra we have: $Delta(1_{Η}) = 1_{Η} otimes 1_{Η}$.

Next, using that $S$ is a coalgebra antimorphism, i.e. if $Delta(h)=sum h_{(1)}otimes h_{(2)}$ then $Delta(S(h))=sum S(h_{(2)})otimes S(h_{(1)})$, implies that (for any $g in G(Η)$):
$$
Delta(S(g)) = S(g) otimes S(g)
$$
thus: $g in G(H)$ $Rightarrow$ $S(g) in G(H)$.

Finally, by the definition of the antipode, we get (for all $g in G(Η)$):
$$
S(g)g = gS(g) = 1_{H}
$$
thus:
for all $g in G(Η)$ we have $S(g) = g^{-1}$ completing the proof.

P.S.1: If $Η$ is only a bialgebra then the above imply that $G(H)$ is a monoid. The presence of the antipode differentiates the situation, and makes $G(H)$ a group.

P.S.2: The above proof of the relation $Delta(S(g)) = S(g) otimes S(g)$, actually implements user's Mathematician_42 comment (posted in his answer above) that $S$ is a bialgebra morphism from $H$ to $H^{op cop}$ or equivalently an algebra antihomomorphism and a coalgebra antimorphism.