groupring morphisms and bialgebra

Question

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}] $ ?

Answer 1

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.

Answer 2

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.