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**NonCommAlg** __Question 3__: the center has only one element, i.e. the identity matrix. suppose G is the group of all $\displaystyle n \times n$ unit upper triangular matrices over a field or a unitary commutative ring. as usual

let $\displaystyle e_{ij}$ be the element of G with 1 in its (i,j)-entry and 0 elsewhere. let $\displaystyle g \in Z(G).$ we have $\displaystyle g=I + \sum_{1 \leq i < j \leq n} a_{ij}e_{ij},$ for some scalars $\displaystyle a_{ij}.$ now let $\displaystyle 1 \leq k < \ell \leq n.$ then $\displaystyle I + e_{k \ell} \in G$ and thus

we must have $\displaystyle (I+e_{k \ell})g=g(I+e_{k \ell}),$ which gives us: $\displaystyle \sum_{1 \leq i < j \leq n} a_{ij}e_{k \ell}e_{ij}=\sum_{1 \leq i < j \leq n} a_{ij}e_{ij}e_{k \ell}.$ using the fact that $\displaystyle e_{rs}e_{tu}=\delta_{st}e_{ru},$ we'll have: $\displaystyle \sum_{\ell < j \leq n} a_{\ell j}e_{kj}=\sum_{1 \leq i < k} a_{ik}e_{i \ell},$ and this gives us:

$\displaystyle a_{ik}=a_{\ell j} = 0, \ \ \forall 1 \leq i < k < \ell < j \leq n.$ therefore $\displaystyle g=I,$ and thus: $\displaystyle Z(G)=\{I \}. \ \ \Box$