Remark 2

Let $x\in A$; then exp $x\in T_r$, so Z(exp $x$)$_0= T$. Further, exp $x$ is very regular (no. 1, Remark) if and only if $w(x)\not=x$ for all $w\in W'_a$ distinct from the identity. By Cor. 1, this also means that $h(x)\not=x$ for all $h\in H_A$ distinct from the identity. In particular, if G is simply-connected, then $Z_G(t) = T$ for all $t\in T_r$, and every regular element of G is very regular.