It is known that almost every pair of elements in a connected compact Lie group (topologically) generates the group.


Obviously this isn't true for non-connected groups but



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> Given a compact Lie group $G$, is it true that almost every pair of elements of $G$ generates a subgroup containing the connected component $G^\circ$ of 1?
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Does almost every pair of elements in a compact Lie group generates the connected component?
The closed subgroups of $G$ not containing the identity component lie in countably many conjugacy classes of subgroups. So it is sufficient to show that for each closed subgroup $H$ not containing the identity component, the probability that Haar-random $g\_1$ and $g\_2$ both lie in some conjugate of $H$ vanishes.


Such pairs are parameterized by the manifold of triples $x \in G/H$, $g\_1 \in x H x^{-1}, g\_2 \in x H x^{-1}$, which is a manifold of dimension $(\dim G - \dim H) + 2 \dim H = \dim G + \dim H$. 


The image of this manifold in $G \times G$ under the projection $(x,g\_1,g\_2)\mapsto (g\_1,g\_2)$ must have measure $0$, as it is the image of a smaller-dimensional manifold (as $H$ does not contain the identity we have $\dim H < \dim G$) under a smooth map (by Sard's theorem).


For $H$ a subgroup, the same argument just barely fails to show that the probability that a single $g$ is contained in a conjugate of $H$ vanishes. Indeed in this case both manifolds have dimension $\dim(G)$. This is convenient as that statement is false, because we could take $H$ to be a maximal torus, or alternately the subgroup generated by a reflection in an infinite dihedral group, as in Noam's example. 


