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Given a finite group $G$ and a field $k$, we can form the group algebra $kG$ with basis the elements of $G$. There is a natural augmentation $\varepsilon\colon G\to k$ that sends an element to the sum of its coefficients in this basis. This is an algebra map, and is the counit in Hopf algebra terms. It also makes $kG$ into a symmetric frobenius algebra.

It is well known that $kG$ has algebra morphisms (either to itself or another group algebra) that do not come from group homomorphisms. Indeed, non-isomorphic groups can have algebra isomorphisms between their group algebras. But do any of these also preserve the augmentation map?

Specifically, if $f\colon kG\to kH$ is an algebra map, with $G,H$ finite groups, such that $\varepsilon(f(a))=\varepsilon(a)$ for all $a\in kG$, then is $f$ a group homomorphism?

I can't find any that aren't, but have little intuition on how to construct such maps or prove they don't exist.

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Let $k=\mathbb{C}$ and let $G$ be a cyclic group of order $4$. Then $\mathbb{C}G\cong\mathbb{C}^4$ as algebras, with the augmentation map projection onto one of the four factors. Any of the six permutations of the other three factors is an algebra isomorphism that commutes with the augmentation map, but only two of them are induced by group isomorphisms.

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  • $\begingroup$ Let me see if I understand well enough to generalize. The projection is onto the trivial representation. If I have any 'repeated' representation, I can permute them around. Such as linear ones of same order, or any higher dim rep. $\endgroup$ – zibadawa timmy Jun 22 '14 at 17:49
  • $\begingroup$ @zibadawatimmy: Yes. In fact, if you have higher dimensional irreducible representations, they correspond to matrix ring factors in the group algebra, which have non-trivial automorphisms, so as well as permuting you can act on individual factors by algebra automorphisms. $\endgroup$ – Jeremy Rickard Jun 23 '14 at 6:57

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