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The following is an excerpt of Truss's 1989 paper Infinite Permutation Groups II.: Subgroups of Small Index. (The transcript will appear later.)

Lemma 3.6

Here, $2^\omega$ the Cantor space, automorphisms mean autohomeomorphisms, and I take "fr" to mean "the frontier" or the boundary.

His argument is convincing, but I have a small issue here: What if the complement of $X$ is finite? Then we wouldn't be able to pick distinct $y_n$ in $2^\omega \setminus X$. I tried to argue that $H$ is too big (so to speak) to result in such a situation by computing the the order or the index of the stabilizer of $2^\omega \setminus H$ but to no avail.

Pteromys
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It also appears to me that the case in which $X$ or $2^\omega \setminus X$ is finite, needs to be dealt with separately. We may suppose without loss of generality that $X$ is finite, and $H$ fixes $X$ (since $H$ fixes $X$ if and only if $H$ fixes $2^\omega \setminus X$). Let us show that the index of $H$ is at least $|\text{Aut}(2^\omega):H|\ge 2^{\aleph_0}$.
Denote $\text{Stab}(X)$ and $\text{Orbit}(X)$ the stabiliser resp. orbit of $X$ under the action of $\text{Aut}(2^\omega)$. By assumption we have $H\subset \text{Stab}(X)$. By the orbit-stabiliser theorem we have $\text{Aut}(2^\omega)/\text{Stab}(X) \cong \text{Orbit}(X).$ Since $\text{Aut}(2^\omega)$ acts transitively on the set of subsets of $2^\omega$ having the same (finite) number of elements as $X$, and latter set has cardinality $2^{\aleph_0}$, it follows that $$|\text{Aut}(2^\omega):H|\ge |\text{Aut}(2^\omega):\text{Stab}(X)|=|\text{Orbit}(X)|=2^{\aleph_0},$$ as desired.

Samuel
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    I agree, the original argument doesn't make sense if $X$ is finite or cofinite. This is the correct one in this case. Although I think OP got it right: by hypothesis, $H$ is too big to fix a finite set, which is what your argument shows. – tomasz Jan 16 '22 at 18:08
  • The trick is to consider the setwise stabilizer. That makes sense. – Pteromys Jan 16 '22 at 18:27