Consider the subgroup $H$ of $SO(3)$ generated by rotations of order $5$ (i.e., rotations by $\frac{2\pi}5$) about the $x$ and $y$ axes. This group certainly isn't finite or discrete (as it's not embedded in any of the small number of three-dimensional point groups), and it's obviously a subgroup of the group $\langle a,b\mid a^5, b^5\rangle$, with $a$ and $b$ mapping to the two generating rotations. But can it be shown isomorphic to this group, or is it possible that there's some additional relation in $a$ and $b$?
Relatedly, this group is obviously a subgroup of the group $K$ generated by a rotation of order 5 and one of order $4$ (using the previous notation and letting $c$ refer to a rotation of $\frac\pi2$ about the $z$-axis, we have $b=cac^{-1}$). I think (though without the confidence that I should have) that $H\lhd K$, since we have $c^2ac^{-2}=a^{-1}$ and $c^2bc^{-2}=b^{-1}$, and by using these along with $c^4=1$, it looks to me like the conjugate of any word in $b$ and $a$ by a member of $K$ can be 'unwound' to another word in $b$ and $a$. I strongly suspect that $H$ isn't the whole of $K$, but is the quotient finite? Can anything interesting be said about this quotient?
(Note that unlike $H$, $K$ clearly has a non-trivial relation, the aforementioned $c^2ac^{-2}a$, but I'm also curious whether that's the only one; i.e., whether $K\cong \langle a, c\mid a^5, c^4, c^2ac^{-2}a\rangle$. There's also an obvious generalization to groups generated by rotations of orders $m$ and $n$ about orthogonal axes, where there's a nontrivial relation if (at least?) one of $m$ and $n$ is even.)