Let $R_{1}$, $R_{2}$, $\cdots$, $R_{m}$ be rings with identity. I need to prove that the following group isomorphism holds:
$U(R_{1} \oplus R_{2} \oplus \cdots \oplus R_{n}) \simeq U(R_{1}) \oplus U(R_{2}) \oplus \cdots \oplus U(R_{n})$.
I surmise that induction is going to be necessary here, but I'm having trouble even just getting started to prove it for just the base case, where $n = 2$: $U(R_{1} \oplus R_{2}) \simeq U(R_{1}) \oplus U(R_{2})$.
I have absolutely no idea where to begin, so any kind of point in the right direction would be appreciated. Just be willing to answer lots of follow-up questions, please.
Thank you in advance.
Keep in mind that for a finite number of rings, the finite direct sum is isomorphic to the finite direct product. It is easier, then, to think of $R_1 \times R_2$ rather than of $R_1 \oplus R_2$. In particular, the elements of $R_1 \times R_2$ are pairs of the form $(r_1, r_2)$, the multiplication is $(r_1, r_2) (s_1, s_2) = (r_1 s_1, r_2 s_2)$ and the multiplicative neutral element is $(1,1)$.
It then follows easily that $(r_1, r_2)$ is invertible in $R_1 \times R_2$ if and only if there exist $(s_1, s_2) \in R_1 \times R_2$ such that $(r_1, r_2) (s_1, s_2) = (s_1, s_2) (r_1, r_2) = (1,1)$, which is equivalent to saying that $(r_1 s_1, r_2 s_2) = (s_1 r_1, s_2 r_2) = (1,1)$, which in turn is equivalent to saying that $r_1 s_1 = s_1 r_1 = 1 \in R_1$ and $r_2 s_2 = s_2 r_2 = 1 \in R_2$, which finally is equivalent to saying that $r_1$ is invertible in $R_1$ and $r_2$ is invertible in $R_2$.