# Grouplike elements is a group

Let $(H, m, u, \Delta, \epsilon, S)$ be a $K$-Hopf Algebra. We call an element $x\in H$ 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 $\ \mu : G(H) \times G(H) \rightarrow G(H) : (x,y) \rightarrow m(x\otimes y)\\ e: \{*\} \rightarrow G(H):* \rightarrow u(1_K) \\ S':G(H) \rightarrow G(H):x \rightarrow S(x)$

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) $∆(m(x\otimes y)) = m(x\otimes y) \otimes m(x\otimes y)$ (EDIT: Solved)

2) $\epsilon(m(x\otimes y)) = 1_K$ (EDIT: Solved)

3) $∆(S(x)) = S(x) \otimes S(x)$

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

Any hints would be appreciated

• For 1) is is probably more obvious what is going on if you write $m(x\otimes y) = xy$ and remember that $\Delta$ is a homomorphism of algebras. – Tobias Kildetoft Jan 17 '17 at 9:59
• Ok. 1) and 2) are actually easily verified – tomak Jan 17 '17 at 10:18
• What is true is that the set of grouplike elements is a group. Please correct your title, so that it makes sense. – Mariano Suárez-Álvarez Jan 18 '17 at 3:06

Let $$g$$ and $$h$$ be grouplike elements. Then $$\Delta(gh)=\Delta(g)\Delta(h)=(g\otimes g)(h\otimes h)=(gh\otimes gh)$$. Indeed the multiplication on $$H\otimes H$$ uses the standard flip map $$\tau$$, thus $$m_{H\otimes H}=(m\otimes m)\circ (1\otimes \tau\otimes 1)$$. 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 $$\Delta(S(g))(g\otimes g)=1\otimes 1.$$ Hence $$\Delta(S(g))$$ is the inverse of $$g\otimes g$$ in the algebra $$H\otimes H$$. 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 $$1$$.

• Ok. 1) and 2) are actually easily verified – tomak Jan 17 '17 at 10:18
• I don't understand the final conclusion. Since $S(g)$ is the inverse of g, it follows... – tomak Jan 17 '17 at 10:54
• I'm going to edit it, it's not complete. – Mathematician 42 Jan 17 '17 at 11:04
• Ok. So we get $∆(S(g)) = ((S\otimes S) \circ ∆)(g) = (S\otimes S)(g\otimes g) = S(g) \otimes S(g)$. And for 4) $(\epsilon \circ S)(g) = \epsilon(g) = 1$ (since S is a anti-algebra morphism. PS: by anti-algebra morphism you mean coalgebra morphism? PPS: I didn't know that S was a coalgebra morphism. I'll try to show that. – tomak Jan 17 '17 at 11:34
• I edited it again, this is an easy argument. You can also use that $S$ is a bialgebra morphism from $H$ to $H^{\text{op cop}}$, but I believe the above argument is more streamlined. – Mathematician 42 Jan 17 '17 at 14:24

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.