Let $(R,+,\times)$ be a ring with an identity $1$. The set of all the multiplicative invertible elements is denoted by $U(R)$. Suppose $G$ is a non-cyclic subgroup of $U(R)$. If $|G|=4$, namely, $G=\{1,g_1,g_2,g_3\}$, then $g_i-1\notin U(R)$ for some $i$.
I am trying to (dis)prove the above result.
Since $|G|=4$ and $G$ is non-cyclic, then $G$ is Abelian and $G=\langle g_1,g_2 \rangle=\{1,g_1,g_2,g_1g_2\}$ where $g_3=g_1g_2$.
Note that $G\cong \langle g_1 \rangle\times \langle g_2 \rangle$. It means there is a group isomorphism $\varphi$ from $G$ to$\langle g_1 \rangle\times \langle g_2 \rangle$. Then $$\varphi(g_1)=(g_1,1).$$ It follows that $$\varphi(g_1-1)=(g_1,1)-(1,1)=(g_1-1,0).$$
To be continue...
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I think $g_1-1\notin U(R)$ if $g_1g_2-1\in U(R)$. But I do not know how to go on.
Since $G$ is non-cyclic, you have $g_{i}^{2} = 1$ for all $i$. Hence $$ 0 = g_{i}^{2} - 1 = (g_{i} - 1) (g_{i} + 1). $$ If $g_{i} - 1 \in U(R)$, then you can multiply by its inverse and obtain $g_{i} + 1 = 0$, that is, $g_{i} = -1$.