Let $G < Sym(\Omega)$ be a transitive group of degree $n$ and
$G_x$ a point-stabilizer. Is stabilizer subgroup $G_x$ Normal? The answer is in general no. Now read the following line (Source of the question):
If $G = G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \cdots$ be a composition chain of G. Then the subgroups $G_i \cap G_x$ form a subnormal chain $G_x \trianglerighteq G_1\cap G_x \trianglerighteq G_2\cap G_x \cdots$ (possibly with repetition).
I read in a text book by D. S. Malik, that if $H, K$ are normal subgroups of group, then so does the $H \cap K$. So if $G_x$ is not a normal subgroup, then how does $G_i\cap G_x$ become a normal subgroup?
The main question is, how does the subgroups $G_i \cap G_x$ form a subnormal chain $G_x \trianglerighteq G_1\cap G_x \trianglerighteq G_2\cap G_x \cdots$?
And what is the meaning of "possibly with repetition"?
Best Answer
The context of permutation groups is distracting. In general, if $H \leq G$ is a subgroup and $K \trianglelefteq G$ is normal then $H \cap K$ is normal in $H$. This is an easy exercise.
It follows that if $1 = K_0 \lhd K_1 \lhd \cdots \lhd K_n = G$ is a chain of subnormal subgroups in $G$ then $1 = K_0 \cap H \trianglelefteq \cdots \trianglelefteq K_n \cap H = H$ is a chain of subnormal subgroups in $H$. The only subtlety is that even if $K_i < K_{i+1}$ it may be that $K_i \cap H = K_{i+1} \cap H$ (for example $H$ might be trivial), so to get a chain of subgroups with each a proper subgroup of the next one may have to skip some indices: that is what is meant by the warning "with repetition".