groupring morphisms and bialgebra

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 $\mathbb{Z}[G_{1}]\rightarrow \mathbb{Z}[G_{2}]$ ?

• 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. – Benjamin Steinberg Mar 7 '17 at 18:21
• 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). – Benjamin Steinberg Mar 7 '17 at 22:13
• @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? – Ofra Mar 8 '17 at 13:19
• 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. – Benjamin Steinberg Mar 8 '17 at 13:23
• 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. – 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.

• 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 – Benjamin Steinberg Mar 8 '17 at 14:27
• @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. – Ender Wiggins Mar 8 '17 at 14:46
• It's fine. Your answer is more detailed and adds the antipode. – Benjamin Steinberg Mar 8 '17 at 16:13