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?

  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.

    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.

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