The compactness theorem for simple groups I would like to use the compactness theorem to prove the following:
Claim. If $H$ is a subgroup of countable index in a simple group $G$, then there are finitely many conjugates of $H$ with trivial intersection.
Let $L = \{\cdot, e, -^{-1}, P_H, x_n\}_{n \in \mathbb{N}}$ be the language of groups with a predicate for $H$ and a sequence of constants for a set of representatives in $G$ of left cosets of $H$.
Let $p_n(x)$ be $x \in H^{x_n}\ \&\ x \neq e$.
If the Claim is not true, then $\{p_n\}$ is finitely satisfiable and by the compacntess theorem it is satisfiable. This means there is $x \neq e$, $x \in \bigcap_{n \in \mathbb{N}}H^{x_n}$.
However, $\bigcap_{n \in \mathbb{N}}H^{x_n} = \bigcap_{g \in G}H^{g}$ that is a normal subgroup of $G$ simple, contradiction.
Is this proof a correct use of the compactness theorem?
 A: The claim is false. For example, let $G$ be the automorphism group of the random graph $R$. The simplicity of this group is due to Truss.
For a vertex $v$ in $R$, let $G_v$ be the subgroup of $G$ consisting of those automorphisms which fix $v$.
Let $\sigma\in G$, and let $w=\sigma(v)$. The left coset $\sigma G_v$ consists of all automorphisms which map $v$ to $w$. The conjugate $G_v^\sigma = \sigma G_v \sigma^{-1}$ consists of all automorphisms which fix $w$.
It follows that $G_v$ has countable index, since there is one left coset for each $w\in R$. But no finite intersection of conjugates of $G_v$ is trivial: such an intersection consists of those automorphisms which fix finitely many elements $w_1,w_2,\dots, w_n$. But the random graph has non-trivial automorphisms fixing these elements.
To see this, pick some $x\in R$ which is distinct from the $w_i$. Now use the extension property of $R$ to find some $y\in R$ which has an edge to exactly the same $w_i$ as $x$ does, and which also has an edge to $x$. Then $x\neq y$, but there is an automorphism of $R$ fixing the $w_i$ and moving $x$ to $y$.
