About the closure of an orbit

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If $G$ is a Lie group, $H$ is a closed subgroup of $G$ and $I$ is a normal Lie subgroup of $G$. Consider the homogeneous space $X:= G/H$ and define $L:=cl_X(I\cdot x_0)$ and $K:=cl_G(I\cdot H)$ where $x_0$ is the base point of $X$.

How to prove that $K$ is a subgroup of $stab_G(L)$?

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Firstly, remark that if $x\in I.x_0, i\in I, h\in H,$ we can write $x=jx, j\in I$, $(ih)(x)=(ih)(jx_0)=i(hjh^{-1})h(x_0)$. Since $I$ is normal, $hjh^{-1}\in I$, we also have $hx_0=x_0$ so $(ih)(x)=i(hjh^{-1})(x_0)\in I.x_0$. Thus for $g\in I.H, cl_X(I.x_0)= cl_X(g.(I.x_0)=g.cl_X(I.x_0)$.

Let $g\in cl_G(I.H), g=lim_ng_n, g_n\in I.H$, for $x\in cl_X(I.x_0)$, $g(x)=lim_ng_n(x)$, since $g_n(x)\in cl_X(I.x_0), g(x)\in cl_X(I.x_0)$ since it is the limit of $g_n(x)$ which are elements of the closed subset $cl_X(I.x_0)$.