Let $G$ be a group of isometries acting effectively by isometries on a connected Riemannian manifold. And let $G'\triangleleft G$ be a normal subgroup. I am trying to prove that $\dim \operatorname{St}\_G(p)\leq \dim \operatorname{St}\_{G'}(p)$ for every $p\in M$; where $\operatorname{St}\_X(p)$ stands for the stratum of the action of $X$ on $M$ through $p$.


We have the formula
$$ \dim\operatorname{St}\_G(p)=\dim G+\dim M\_0^{G\_p}-\dim N\_G(G\_p); $$
where $M\_0^{G\_p}$ is the connected component of the fixed-point set $M^{G\_p}$ through $p$.


It is obvious that $\dim M\_0^{G\_p}\geq\dim M\_0^{G'\_p}$, since $G'\_p\triangleleft G\_p$. Thus, the result is true if we show that $\dim N\_G(G\_p)\geq \dim N\_G'(G'\_p)$, which I'm not sure if is true. If it is not true, is there any other way to prove my claim?



Let $G'\triangleleft G<\operatorname{Iso}(M)$ be a normal subgroup. A $G'$-stratum is the union of $G$-strata of lesser dimension
We have to assume that the index $[G':G]$ is finite. In this case:


Let $G'$ be a normal subgroup of $G$ such that the quotient $\Gamma=G/G'$ is finite and acts by isometries in $X'=M/G'$, and $X=X'/\Gamma$. Thus, $(G')^0=G^0$ and, therefore the orbits $G'(p)$ and $G(p)$ have same connected components through $p$. This way, $\nu\_pG(p)=\nu\_pG'(p)$. Also, as $G\_p^0=(G\_p')^0$, the orbits associated to their respective slice representations, i.e., their infinitesimal actions on the normal space to the tangent space to their respective orbits, have same dimension. And $(\nu\_pG(p))^{G\_p}\subset (\nu\_pG(p))^{G'\_p}$. As the orbits of their associated slice representations have same dimension, the cohomogeneity of the action $(G'\_p,(\nu\_pG(p))^\dagger)$ can be no greater then the cohomogeneity then the action of $(G\_p,(\nu\_pG(p))^\dagger)$. Thus, $\dim\operatorname{St}\_X(\pi(p))\leq \dim\operatorname{St}\_{X'}(\pi'(p))$.


Here, $(\nu\_pG(p)^\dagger)$ is the orthogonal complement of $(\nu\_pG(p))^{G\_p}$ in $\nu\_pG(p)$.


