Weyl group and Dynkin diagram

Question

Can somebody help me with following questions: 1)Prove that two simple roots in a Dynkin diagram that are connected by a single edge are in the same orbit under the Weyl group. and 2)For an irreducible root system, prove that all roots of a particular length form a single orbit under the Weyl group.

Answer 1

Let $\Phi$ be an irreducible root system, $W$ the Weyl group, and let $\alpha$ be a root. Are you familiar with/can you prove the following?

Lemma. The $W$-orbit of $\alpha$ spans the inner product space.

A direct result of this is that for a second root $\beta$, not every element of $\lbrace\sigma(\alpha):\sigma\in W\rbrace$ can be orthogonal to $\beta$.

So suppose that $\alpha$ and $\beta$ have the same length, then replacing one by a suitable $W$-conjugate, we can assume that they are not orthogonal. So $\langle\alpha,\beta\rangle=\pm 1$. Replacing $\beta$ by its negative $\sigma_\beta(\beta)$ if required, we may assume that $\langle\alpha,\beta\rangle=1$. Then: $$ (\sigma_\alpha\sigma_\beta\sigma_\alpha)(\beta)=\text{?} $$