Lemma 6

**Let $L$ be a totally ordered finite set. For every $\lambda \in L$, let $V_\lambda$ be a free module of finite dimension over a commutative ring $k$; for $\lambda, \mu$ in $L$ such that $\lambda \leq \mu$, let $f_{\lambda \mu} \in \mathrm{Hom}_k(V_\mu, V_\lambda)$. Then the linear mapping

from $\prod_{\lambda \in L} V_\lambda$ into $\prod_{\lambda \in L} V_\lambda$, has determinant $\prod_{\lambda \in L} \det f_{\lambda \lambda}$.

One reduces immediately to the case that $L$ is an interval of integers, and the lemma then follows from A, III, §8, No. 6, formula (31).

If $Z \in T_J$, one then sees that there exists $Z' \in T_J$ such that $Z'Z = I_n$, whence $Z' = Z^{-1}$. Thus, $T_J$ is a closed subgroup of $\mathbf{GL}(n, K)$. On the other hand, Prop. 15 of §1, No. 10 shows that the measure

is a left Haar measure on $T_J$. By calculating $ZZ'$ one sees in the same way that this measure is a right Haar measure on $T_J$.

There is an analogous result if, in the definition of $T_J$, the roles of rows and columns are interchanged.

When $J$ is the set of pairs $(i, j)$ such that $i < j$, the group $T_J$ is called the upper strict triangular group of order $n$ over $K$, and is denoted $T_1(n, K)$. Its transpose is called the lower strict triangular group.