Quotient Group of Stabilizer and Larger (probably) Normal Subgroup

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Let $G < Sym(\Omega)$ be a transitive group of degree $n$ and $G_x$ a point-stabilizer. If $G = G_0 \triangleright G_1 \triangleright G_2 \triangleright \cdots$ be a composition chain of G. Then the subgroups $G_i \cap G_x$ form a subnormal chain $G_x \trianglerighteq G_1\cap G_x \trianglerighteq G_2\cap G_x \cdots$.Now I would like to prove below 3 relations (Source of the question, page 2, 3):

i) $G/G_1 \geq G_1 \cdot G_x/G_1$

ii) $G_i/G_{i+1} \geq G_{i+1} \cdot (G_i \cap G_x)/G_{i+1} $

iii) $ G_{i+1} \cdot (G_i \cap G_x)/G_{i+1} \cong (G_i \cap G_x)/(G_{i+1} \cap G_x)$

It appears to me that $G_1$ could be larger normal subgroup than $G_x$, then how will we get the quotient group $G_x/G_1$, Similarly, $/G_{i+1}$ could be larger normal subgroup than $G_x$, in general I don't know hoe to begin with i) and ii).

In the case of iii), second isomorphism theorem could be used where $H/(H\cap K) \cong (HK)/K$, to get,

$ (G_i \cap G_x)/(G_{i+1} \cap G_x) \cong (G_i \cap G_x) \cdot G_{i+1}/G_{i+1} $

if we set,

$H= (G_i \cap G_x),$

$K=G_{i+1},$

$(H\cap K)= (G_i \cap G_x) \cap G_{i+1}= (G_i \cap G_x)$, since, $G_i \geq G_{i+1}$.

Is it correct? Also how do I prove i) and ii)?