Let $\mathbf{S}=\left\{\left[\begin{array}{ll}\mathbf{a} & \mathbf{b} \\ 0 & \mathbf{a}\end{array}\right]: \mathbf{a}, \mathbf{b} \in \mathbb{Z}_{2}\right\}$, $\mathbf{R}=\mathbb{Z}_{2} \times \mathbb{Z}_{2}$.
Prove or disprove :
i) $(\mathbf{S},+,.)\cong (\mathbf{R},+,.)$ (as rings)
ii) $(\mathbf{S},+)\cong (\mathbf{R},+)$ (as groups)
I think we have $|\mathbf{S}|=2^3$ but $|\mathbf{R}|=2^2$ so we have $|\mathbf{R}| \neq |\mathbf{S}|$ so $(\mathbf{S},+,.)\ncong (\mathbf{R},+,.)$ but $(\mathbf{R},+)$ is Klein four-group, and every element of $(\mathbf{S},+)$ has order $2$, then $(\mathbf{S},+)\cong (\mathbf{R},+)$ (as groups). Is this true ?
$S$ is not isomorphic to $R$ as rings since $S$ has a non-zero nilpotent, while $R$ hasn't. More precisely, there is $x\in S$, $x\ne0$ such that $x^2=0$. Can you find it?