I was reading up on the Shannon capacity of graphs and came across this article on the Shannon capacity of Cayley graphs. Though the article deals primarily with the linear Shannon capacity, defined differently from the regular Shannon capacity, it makes the statement that $\Theta(G) \geq \sqrt{n}$ where $\Theta(G)$ denotes the regular Shannon capacity and $G$ is a Cayley graph.
How can one prove that the Shannon capacity of a Cayley graph is lower bounded by $\sqrt{n}$?
In general, it is of course not lower bounded by $\sqrt{n}$: the complete graph $K_n$ is a Cayley graph (every nonzero generator of $\mathbb{F}_n$), and it has a max independent set of size 1, so its Shannon capacity is just 1.
That paper is largely discussing self-complementary Cayley graphs, those where G is isomorphic to its own complement. In that case they show that a linear code achieves a capacity of at least $\sqrt{n}$, so the Shannon capacity is definitely at least $\sqrt{n}$.