Exotic group topologies on the affine group $ax+b$ Let $G = \{(x; y) : x \in \mathbb{R}, y > 0\}$. With $(x, y)(u, v) = (x + yu, yv)$, $G$
is a group. If we topologize $G$ as a subset of $\mathbb{R}^2$, it is known that $G$ is a locally compact group that is not unimodular (see (15.17) of Hewitt-Ross). Is there another
topological structure of the group $G$ such that $G$ is a locally compact group and the subgroup $K = \{(0, y) : y > 0\}$ is compact?
 A: If I understand the question, then the answer is no. Let $X = \{(x,1) : x \in \mathbb{R}\}$. Then the conjugation action of $K$ has only three orbits on $X$. All three orbits must be closed (since $K$ is compact). So the orbits are also open (in $X$), since there are only finitely many of them. One of these orbits is the singleton $\{(0,1)\}$, so the topology on $X$ must be the discrete topology. This contradicts the fact that the other two (infinite) orbits are compact.
A: [I realize I had misread the question, as I understood that the real subgroup required to be compact is the normal one. Since asking the question with the normal subgroup ($X$ in Dave's post) required to be compact seems much less trivial than the original question (answered by Dave), I'll include the full answer to the modified question.]
To avoid ambiguity, let me denote by $R$ the underlying additive group structure on the reals, without topology. So it's just a $\mathbf{Q}$-vector space of continuum dimension, so a group is isomorphic to $R$ iff it's abelian, torsion-free, and divisible, of continuum cardinal.
As mentioned in the comments, there exist compact group topologies on $R$. 
However, I claim that if $T$ is a compact topology on $R$, then the group of automorphisms of the topological group $(R,T)$ is not transitive on $R-\{0\}$. This implies that the underlying group of $\mathbf{R}\rtimes\mathbf{R}$ cannot be endowed with a group topology making the normal $\mathbf{R}$ compact.
To prove the claim, write $K=(R,T)$. This is a compact abelian group, with no continuous homomorphism onto any nontrivial finite group. Hence its Pontryagin dual is a torsion-free abelian group $A$. If $I$ is a maximal free family in $A$, then the embedding $\mathbf{Z}^{(I)}\subset A$ induces a (continuous) surjection $K\to (\mathbf{R}/\mathbf{Z})^I$. Since $K$ has continuum cardinal, this implies that $I$ has cardinal less than continuum. Hence, noting that $A$ has the same cardinal as $I$, the cardinal of $A$ is less than continuum. 
On the other hand, for each $x\in\mathbf{R}/\mathbf{Z}$, $x$ belongs to the image of some homomorphism $A\to \mathbf{R}/\mathbf{Z}$: this comes from basic Pontryagin duality: just find an infinite cyclic subgroup $\langle m\rangle$ in $A$, consider the homomorphism $m\mapsto x$, and extend it to $A$. As a consequence (using that $A$ has cardinal less than continuum), the number of possible images of homomorphisms $A\to\mathbf{R}/\mathbf{Z}$ is continuum. Now the group $\mathrm{Aut}(K)$ acts on this set of homomorphisms, and this action does not change the image, so the action of $\mathrm{Aut}(K)$ on $\mathrm{Hom}(A,\mathbf{R}/\mathbf{Z})$ has continuum many orbits. By Pontryagin duality, the action of $\mathrm{Aut}(K)$ on $\mathrm{Hom}(A,\mathbf{R}/\mathbf{Z})\simeq K$ has continuum many orbits, in particular it is not transitive on $K\smallsetminus\{0\}$.
