that for any $U\in\U$ there is $f\in G$ such that $U$ and $f(U)$ are disjoint. Take a cover $\V$ which is starwise finer than $\U$. This is possible since $M$ is metrizable, so paracompact. We may assume that $\H_c(M)$ is factorizable with respect to $\V$ (see Def. 5.1(1) and Prop. 5.2 below). In view of the commutator equalities $$[fg,h]=f[g,h]f^{-1}[f,h],\quad [f,gh]=[f,g]g[f,h]g^{-1}$$ and Theorem 1.1, it follows that $$\H_c(M)=[\H_c(M),\H_c(M)]=\prod_{U\in\U}[\H_U(M),\H_U(M)].$$ Let $[h_1,h_2]\in[\H_U(M),\H_U(M)]$ with $U\in\U$ and $f\in G$ such that $U\cap f(U)=\emptyset$. Then $[h_1,h_2]=[[h_1,f],h_2]\in G$. Thus $\H_c(M)\s G$ as required.

*Proof of Corol. 1.3.* $(\Rightarrow)$ It follows from a theorem of Ling [@li] since $\H_c(M)$ is factorizable (Prop. 5.2 below) and *transitively inclusive*. The latter means that for any $U,V\in\B$ there is $h\in\H_c(M)$ such that $h(U)\s V$.

$(\Leftarrow)$ $\H_c(M^o)$ is a normal subgroup of $\H_c(M)$. $\square$

If $\pm\neq\emptyset$, $\H^{\p}_c(M)$ is not simple

In fact, $\H_c(M^o)$ is a proper normal subgroup of $\H^{\p}_c(M)$.

Conjugation-invariant norms 
============================

The notion of the conjugation-invariant norm is a basic tool in studies on the structure of groups. Let $G$ be a group. A (or *norm* for short) on $G$ is a function $\nu:G\r[0,\infty)$ which satisfies the following conditions. For any $g,h\in G$

1.  $\nu(g)>0$ if and only if $g\neq 