The group morphism of tye ring

Question

Let $(G,+,\cdot)$ $(H,+,\cdot)$ be rings, we suppose that the unites $(G*,\cdot)$ and $(H*,\cdot)$ form groups respectively, for example, the matrix ring $M(n,\mathbb{R})$. There is a group morphism $\phi: (G,\cdot)\to (H,\cdot)$, then is it $\phi(-\textrm{id})=-\textrm{id}$?

Answer 1

If a ring homomorphism is required to send $1_G$ to $1_H$, then $\phi(-1_G)+1_H=\phi(-1_G)+\phi(1_G)=\phi(1_G-1_G)=\phi(0)=0$, so $\phi(-1_G)=-1_H$.

If $\phi$ is merely a group homomorphism between the groups of units $G^\ast$ and $H^\ast$, then the answer is trivially "no": you could map $\phi(g)=1_H$ for every $g\in G^\ast$ to provide a counterexample.