Number 3

It seems like we need this map $\epsilon (x) = 1$ for $x \in$ the coalgebra to do this problem. Or am I not understanding this correctly?

1. From my understanding, you need to show that if you have any ${x} \in \mathbb{F}{G}$ such that $\Delta ({x}) = {x} \otimes {x}$, then ${x}$ has to be strictly only in ${G}$, ie. ${x}$ is not of the form $\sum_{i=1}^{n}{\lambda_{i}}{g_{i}}$.

2. katrina888

And you don’t think we need the counit map to show this?

3. What if all of the lamdas are 0 except one which is the 1 of F. Doesn’t that give us what we want?

• Yes it does, but you have to derive to that conclusion via the condition of the coproduct.

4. I addressed this problem without having to use the coproduct. Just pick an arbitrary element of F(G), and carefully expand it out under the coproduct. Set it equal to x$\otimes$x, and see what conditions are forced on you.

5. Sebastián O.

https://hopfcombinatorics.wordpress.com/2012/02/07/hw-3rd-exercise/

Maybe that could help, a little observation on the definition of $grouplike$. I think it could help.

6. Brian Cruz

I did what Zach mentions here. I had to be careful though in drawing conclusions from setting the co-product of the arbitrary element (expanded out) equal to $x \otimes x$. The concern is that a linear combination of linearly independent pure tensors can add together to be a pure tensors also. We of course remember $$(1+g)\otimes(1+g) = 1\otimes 1 + 1 \otimes g + g \otimes 1 + g \otimes g$$.
I found it useful to solve the equation for zero and then try to apply the “useful lemma” we saw in class.