Let $\bar{\mathbb{Q}}$ be a (fixed) algebraic closure of $\mathbb{Q}$ and $\tau\in\bar{\mathbb{Q}},\tau\notin\mathbb{Q}.$ Let $E$ be a subfield of $\bar{\mathbb{Q}}$ maximal with respect to the condition $\tau\notin E.$ Show that every finite dimensional extension of $E$ is cyclic.
Attempt
Let $K = \bar{\mathbb{Q}}$.
Edit
Since $\tau \notin E$ we can define $E=\left\{ \tau\in K \vert\alpha\left(\tau\right)\neq\tau\,\,\forall\:\alpha\in H\right\} $ be the fixed field of $H,$ a minimal closed subgroup of $\mathrm{Aut}(K/\mathbb{Q})$.
But then, such a subgroup would be generated by a single element. So $H$ is cyclic. Thus $\mathrm{Aut}(K/E)$ is a cyclic extension. Hence every finite extension, say, $M$ of $E$ is cyclic.
I am not really convinced by my argument. Hints and suggestions are very much welcomed. Thanks.