Addition and Rudin-Keisler ordering in $\beta \omega$

Question

$\DeclareMathOperator{\RK}{\mathrm{RK}}$Let $\beta\omega$ be the Stone-Cech compactification of $\omega$ with the discrete topology. We can endow $\beta\omega$ with an addition operation that extends the addition on $\omega$ in a natural way. (The same statement holds for multiplication.)

The Rudin-Keisler preorder on $\beta\omega$ is defined by $${\bf p} \leq_{\RK} {\bf q} :\;\Leftrightarrow\; (\exists f:\omega\to\omega)\,(\forall U\in{\bf p})\, f^{-1}(U)\in {\bf q}$$ for ${\bf p},{\bf q}\in\beta\omega$.

Question. Given ${\bf a},{\bf b}, {\bf c}\in\beta\omega$, do we have

1. ${\bf a} \leq_{\RK} {\bf a+b}$, and
2. if ${\bf a}\leq_{\RK}{\bf b}$, then $({\bf a}+{\bf c}) \leq_{\RK} ({\bf b}+{\bf c})$?

Answer 1

For Question 1: First, there are idempotent ultrafilters, so some ultrafilters satisfy 1 in a very strong form. But 1 does not hold in general. The reason is that the semigroup $\beta\omega-\omega$ has subsemigroups $G$ that are groups of cardinality $2^{\mathfrak c}$ where $\mathfrak c$ is the cardinal of the continuum. (See Hindman and Strauss, "Algebra in the Stone-Cech Compactification", Theorems 2.7(d) and 2,25, and remember that all infinite closed subsets of $\beta\omega-\omega$ have cardinality $2^{\mathfrak c}$.) If the answer to Question 1 were always affirmative, then all elements of $G$ would be RK-below each other (and therefore isomorphic, but I don't need that). That's impossible as any ultrafilter is RK-above only $\mathfrak c$ others.

I expect the answer to Question 2 to be negative also, but I'll need to think some more to prove it. I also expect that there's an easier proof for Question 1. Finally, the answer to Question 1 becomes positive if either a or b is assumed to be a P-point.

Edit: A negative answer to Question 2 follows from the continuum hypothesis. Under CH (or certain weaker hypotheses, but not in ZFC alone), there exist P-point ultrafilters. Let a be one of these, and let b be an idempotent ultrafilter Rudin-Keisler above a. (It's provable in ZFC that any non-principal ultrafilter on $\omega$ is below an idempotent one.) Because a is a P-point, any sum $\mathbf a+\mathbf c$ (with c noprincipal) is isomorphic to the tensor product (also called Fubini product) $$ \mathbf a\otimes\mathbf c=\{S\subseteq\omega\times\omega:\{x:\{y:(x,y)\in S\}\in\mathbf c\}\in\mathbf a\}, $$ and is therefore strictly above both a and c in the Rudin-Keisler order. In particular, $\mathbf a+\mathbf b\not\leq_{\text{RK}}\mathbf b$. But since b is idempotent, this gives $\mathbf a+\mathbf b\not\leq_{\text{RK}}\mathbf b+\mathbf b$ even though $\mathbf a\leq_{\text{RK}}\mathbf b$.