If $R(G)$ is a group ring constructed from a group $G$ and a ring $R$, and $H\leq R(G)$ is a subring, then must there be a surjective homomorphism from $R(G)$ to $H$ in general?
What if $G$ is a finite group, so the group ring $R(G)$ is a free module with finite basis?
Specifically, I am asked to find a surjective homomorphism from $\mathbb{R}[Q_8]$ to $H$, the ring of quaternions over the reals. My attempts to find an explicit homomorphism is not successful, so I was wondering perhaps the projection of $\mathbb{R}[Q_8]$ to $H$ or something similar will yield the desired surjective homomorphism.
There need not be such a surjection in general. Recall the universal property of group rings: $$ \operatorname{Hom}_{Ring}(R[G],S) = \operatorname{Hom}_{Ring}(R,S)\times\operatorname{Hom}_{Grp}(G,S^\times). $$ That is, giving a map $R[G]\to S$ of rings is the same as giving a map $G\to S^\times$ of groups and a map $R\to S$ of rings. But a subring $S\subseteq R[G]$ need not admit any maps from $R$ (Consider $R = \Bbb Q$ and $S = \Bbb Z$; a map $f : \Bbb Q\to\Bbb Z$ must send $\Bbb Z\to\Bbb Z$ isomorphically, but then where can it send a general element of $\Bbb Q$? Take $1/2$ for instance. We have $1/2\cdot 2 = 1$ in $\Bbb Q$, so $$ 1 = f(1) = f(1/2\cdot 2) = f(1/2)\cdot f(2) = 2\cdot f(1/2), $$ implying that $f(1/2)$ is an inverse of $2$ in $\Bbb Z$, which is absurd. So even if $G$ is trivial (!) there are no maps $\Bbb Q[G]\to\Bbb Z$, let alone surjections.
As I mentioned in my comment, the real quaternion division algebra is not naturally a sub of $\Bbb R[Q_8]$, it is a quotient, which does give you a surjective map $\Bbb R[Q_8]\to H$. If $Q_8 = \{\pm e,\pm i,\pm j,\pm k\}$ (think of $-e$ as $-1$, so $(-e)(-e) = e$; I'm using this notation to distinguish the group element $-1$ from the ring element $-1$), then $-1\cdot e\neq 1\cdot (-e)$ in $\Bbb R[Q_8]$. But in $H$, these two things should be the same. In $H$, $1\cdot e$ becomes $1$ and $1\cdot(-e)$ becomes $1\cdot(-1)$, and similarly for $\pm i$, $\pm j$, and $\pm k$. So you have a map $$ \Bbb R[Q_8]\to H $$ sending $\Bbb R\cdot e$ to $\Bbb R$ isomorphically and $a\cdot(\pm e)\mapsto \pm a$, $a\cdot(\pm i)\mapsto \pm ai$, $a\cdot(\pm j)\mapsto\pm aj$, $a\cdot(\pm k)\mapsto\pm ak$, and extending by linearity. This is the same as the natural quotient map $$ \Bbb R[Q_8]\to\Bbb R[Q_8]/(1\cdot e - (-1)(-e), 1\cdot i - (-1)(-i), 1\cdot j - (-1)(-j), 1\cdot k - (-1)(-k)), $$ which is surjective as all quotient maps are.