$\DeclareMathOperator\Gr{Gr}$Let $R$ be a local ring, let $A$ be a finite abelian group, and let $I$ be a Hopf ideal of the ring $R[A]$. The quotient $R[A]\twoheadrightarrow R[A]/I$ induces a map on group-like elements $f\colon \Gr(R[A])\to \Gr(R[A]/I)$.


**Is the map $f$ surjective?**


The answer is yes if the order of $A$ is invertible in $R$, so let's assume it's not.


(By $R[A]$ I just mean the usual ring with generators $T\_a$ for $a\in A$ and relations $T\_0=1$ and $T\_aT\_b=T\_{a+b}$).



Group-like elements in quotients of group rings
I found a counter-example, so the answer is no.


Let $B=\mathbb{F}\_2[a,b,c]/J$ where $J=(a,b,c)^3+(ab+ac-bc)$ and put $s=ab$ and $t=ac$. Let $A=(\mathbb{Z}/2\mathbb{Z})^2$ and define $R = B[x\_{10},x\_{01},x\_{11}]/J'$ where
$$
 J'=(x\_{10}^2-s, x\_{01}^2-t, x\_{11}^2-(s+t), x\_{10}x\_{01}-ax\_{11}, x\_{10}x\_{11}-bx\_{01}, x\_{01}x\_{11}-cx\_{10})\,.
$$
Then $R[A]\cong R[T\_{10},T\_{01}]/(T\_{10}^2-1, T\_{01}^2-1)$ with the usual Hopf-algebra structure. Define
$$
 I=(x\_{10}(T\_{10}-1), x\_{01}(T\_{01}-1), x\_{11}(T\_{10}T\_{01}-1))\,.
$$
I claim that element $g=1+t(T\_{10}-1)$ is not equal to 1 but group-like modulo $I$. To see that $g$ is group-like, we must show that $\varepsilon(g)=1$ and $\Delta(g)=g\otimes g$, where $\varepsilon$ denotes the counit and $\Delta$ denotes the coaction. It is clear that $\varepsilon(g)=1$ so we will show that $\Delta(g)=g\otimes g$. Since
$$
 T\_{\lambda}-1=(T\_{\lambda-\lambda'}-1)(T\_{\lambda'}-1)+(T\_{\lambda-\lambda'}-1)+(T\_{\lambda'}-1)\,,
$$
we have $x\_{\lambda'}(T\_\lambda-1)=x\_{\lambda'}(T\_{\lambda-\lambda'}-1)$ modulo $I$ for all $\lambda,\lambda'\in A$.
This implies that, modulo $I\otimes R[A]+R[A]\otimes I$, we have
$$
 \begin{split}
 \Delta(g)-g\otimes g 
 & = \Delta(1+t(T\_{10}-1))-(1+t(T\_{10}-1))\otimes (1+t(T\_{10}-1)) \\
 & = t(T\_{10}-1)\otimes (T\_{10}-1) \\
 & = (s+t)(T\_{10}-1)\otimes (T\_{10}-1) \\
 & = (s+t)(T\_{01}-1)\otimes (T\_{10}-1) \\
 & = t(T\_{01}-1)\otimes (T\_{10}-1)+(T\_{01}-1)\otimes s(T\_{10}-1) \\ 
 & = 0\,.
 \end{split} 
$$
Hence $g$ is a group-like element which is not the image of a group-like element in $R[A]$.


