Let $R$ be a commutative ring with identity and put $U=U(n,R)$, the group of $n\times n$ (upper) unitriangular matrices over R. Define $U_i$ to be the be the subgroup of $U$ having (at least) $i-1$ zero superdiagonals. Prove that $1=U_n<\ldots<U_1=U$ is both the upper and lower central series of $U$. (From an exercise in Robinson's A Course in the Theory of Groups).
One can easily show by calculation that $1=U_n<\ldots<U_1=U$ is a central series for $U$. It is hence well-known that $U_i\leq\zeta_{n-i+1}(U)$ and that $\gamma_{i}(U)\leq U_i$ for each $i\leq n+1$, where $\zeta_{n-i+1}(U)$ and $\gamma_{i}(U)$ are the ${n-i+1}$-th term of the upper central series and the $i$-th term of the lower central series of $U$, respectively. How can we prove that these series are really the same one?