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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?

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    $\begingroup$ I'm pretty sure $\dim M_0^{G_p} > \dim M_0^{G'_p}$ should not be strict inequality. $\endgroup$
    – LSpice
    Commented Jul 5, 2020 at 22:44

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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 $\nu_pG(p)$.

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