Let $G_{1}$ and $G_{2}$ be two groups. Suppose that we have a morphism $\mathbb{Z}[G_{1}]\rightarrow \mathbb{Z}[G_{2}] $ of bialgebras is it true that this morphism comes from a morphism of groups $G_{1}\rightarrow G_{2}$ ? In case when the answer is "no", is it true that if $\mathbb{Z}[G_{1}]\rightarrow \mathbb{Z}[G_{2}] $ is an isomorphism of rings then there exists an isomorphism of bialgebras $\mathbb{Z}[G_{1}]\rightarrow \mathbb{Z}[G_{2}] $ ?

  • 4
    $\begingroup$ The answer I believe is yes because your morphism will take grouplike elements to grouplike elements and the grouplike elements in this case are the elements of the groups. $\endgroup$ – Benjamin Steinberg Mar 7 '17 at 18:21
  • 1
    $\begingroup$ The answer to you second question is no. There are non-isomorphic finite groups with isomorphic group rings. By my comment above, it follows that the isomorphism is not a bi-algebra isomorphism (although it can be chosen to be augmentation preserving). $\endgroup$ – Benjamin Steinberg Mar 7 '17 at 22:13
  • $\begingroup$ @BenjaminSteinberg so what you are saying in your first comment is that a map morphism of bialgebras is actually a morphism of hopf algebras ?! Is it that clear? $\endgroup$ – Ofra Mar 8 '17 at 13:19
  • $\begingroup$ I am not saying this in general. Just for group algebras. The result OP wants is true for monoids and since a monoid homomorphism preserves the inverse we are done. $\endgroup$ – Benjamin Steinberg Mar 8 '17 at 13:23
  • $\begingroup$ By group like element I mean $\Delta(g)=g\otimes g$ and the counit gives $1$. Then in a monoid algebra the only group like elements are the elements of the monoid. A bialgebra map preserves group like elements. $\endgroup$ – Benjamin Steinberg Mar 8 '17 at 13:30

I will show that any bialgebra homomorphism $\mathbb QM_1\to \mathbb QM_2$ of monoid algebras is induced by a monoid homomorphism $M_1\to M_2$. This will imply what the OP wants.

An element $g$ of a bialgebra is called group-like if $\Delta(g)=g\otimes g$ and $\eta(g)=1$ where $\eta$ is the counit. It is well known that the group-like elements are linearly independent and form a monoid (cf. Lemma 2.1 http://www.math.wisc.edu/~passman/balgebra.pdf).

If $M$ is a monoid then the elements of $M$ are group-like in the monoid algebra and from the above linear independence they are the only group-like elements.

Since bialgebra morphisms preserve grouplikes it follows any bialgebra morphism of monoid algebras is inducted by a monoid homomorphism. Hence any bialgebra morphism of group algebras is induced by a group homomorphism.


Your first claim is true even if you substitute $\mathbb{Z}$ with any integral domain $\Bbbk$. Actually what is true is that we have a bijection $$\text{Bialg}_\Bbbk(\Bbbk[G],B)\cong\text{Mon}(G,\mathcal{G}(B))$$ where $\mathcal{G}(B)$ denotes the monoid of group-like elements in $B$ and $B$ is a $\Bbbk$-algebra via a ring homomorphism $\gamma:\Bbbk\to B$.

Notice that if $f:\Bbbk[G]\to B$ is a morphism of bialgebras then the relations \begin{gather} \Delta_B(f(g))=(f\otimes_\Bbbk f)(\Delta_{\Bbbk[G]}(g))=f(g)\otimes_\Bbbk f(g), \\ \varepsilon_B(f(g))=\varepsilon_{\Bbbk[G]}(g)=1_{\Bbbk}, \end{gather} imply that $f(g)$ is group-like in $B$ for every $g\in G$. Thus we may (co)restrict $f$ to $f':G\to \mathcal{G}(B)$, which gives the assignment from left to right.

Conversely, every morphism of monoids $f:G\to \mathcal{G}(B)$ can be extended in a unique way to a morphism of $\Bbbk$-algebras $F:\Bbbk[G]\to B$ by letting $$F\left(\sum_{g\in G}k_gg\right)=\sum_{g\in G}\gamma(k_g)f(g)$$ and this turns out to be a morphism of bialgebras.

If you take $B=\Bbbk[H]$ for $H$ another group, then you may check that $\mathcal{G}(\Bbbk[H])=H$ (here you should need the integral domain hypothesis).

About the question you asked in the comments, every $\Bbbk$-bialgebra morphism $ f:A\to B$ between Hopf algebras preserves the antipodes as both $fS_A$ and $S_Bf$ are convolution inverses of $f$ in $\text{Hom}_\Bbbk(A,B)$ (see also Sweedler, Hopf algebras, Lemma 4.0.4).

To complete Benjamin answer to your last question, in this paper you may find an example of two non-isomorphic groups whose group algebras are instead isomorphic.

  • 1
    $\begingroup$ This is essentially the same as my answer since you can replace $\mathbb Z$ and $\mathbb Q$ by any integral domain and its field of fractions. I was unaware you could ignore the antipode even for hopf algebras that are not group algebras. +1 $\endgroup$ – Benjamin Steinberg Mar 8 '17 at 14:27
  • $\begingroup$ @BenjaminSteinberg: you are right and I beg your pardon as I was writing and I didn't notice you already answered the question. About the antipode condition, it should work for Hopf algebras in any symmetric monoidal category by essentially the same trick. $\endgroup$ – Ender Wiggins Mar 8 '17 at 14:46
  • 1
    $\begingroup$ It's fine. Your answer is more detailed and adds the antipode. $\endgroup$ – Benjamin Steinberg Mar 8 '17 at 16:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.