the cardinal of $x^G$ factors the cardinal of $x^N$

Question

please give me hints to solve this problem: Let $G$ acts on $X$ and $N$ be a normal subgroup of $G$, show that for every $x\in X$ we have: the cardinality of $x^G$ factors the cardinality of $x^N$ .

Answer 1

Try showing every $N$-orbit of a $G$-orbit $Gx$ has the same size. Say $gx,hx\in Gx$ and exhibit a bijection $Ngx\cong Nhx$, or $\Leftrightarrow~gNx\cong hNx$. Can you think of a map that could serve as a bijection here? (Pick an element from $G$ to do the job, you want $g\mapsto h$. Show the map is invertible.)

If the $N$-orbits partition a $G$-orbit $Gx$ into parts of equal size, that size must divide the whole.

Moreover, stabilizer of $x$ in $N$ must be a subgroup of its stabilizer in $G$, so invoke Lagrange's theorem to obtain the relation $\#{\rm Stab}_N(x)\mid\#{\rm Stab}_G(x)$.

Answer 2

1) $|x^G|=|G:Stab_G(x)|$ and $|x^N|=|N:Stab_N(x)|$. 2) $Stab_N(x)=N \cap Stab_G(x)$. Now $|N:Stab_N(x)|=|NStab_G(x):Stab_G(x)| $ divides $|G:Stab_G(x)|$ for $N \lhd G$ and then $NStab_G(x)$ is a subgroup of $G$.