Does a subgroup have countable index in the group generated with a countable subgroup? Suppose $H$ is a subgroup of a group $G$ and both have the same cardinality of $\mathbb{R}$.
If $K$ is a countable subgroup of $G$, is it possible that $H$ has not countable index in the subgroup generated by $K$ and $H$?
Clearly $G$ must be not abelian and $HK \neq KH$.
 A: Yes, it's possible. Probably there are easier examples, but here's one. Consider the subgroups of $\mathrm{GL}_2(\mathbb{R})$:
\begin{align*} 
G &= \left\{\begin{pmatrix} a & b\\ 0 & c\end{pmatrix}\mid a,b,c\in \mathbb{R},a\neq 0,c\neq 0\right\}\\
H &= \left\{\begin{pmatrix} a & 0\\ 0 & c\end{pmatrix}\mid a,c\in \mathbb{R}, a\neq 0,c\neq 0\right\}\\
K &= \left\{\begin{pmatrix} 1 & n\\ 0 & 1\end{pmatrix}\mid n\in \mathbb{Z}\right\}.
\end{align*}
$G$ and $H$ have size continuum, and $K$ is countable. A coset of $H$, say $$\begin{pmatrix} d & e\\ 0 & f\end{pmatrix}H = \left\{\begin{pmatrix} ad & ec\\ 0 & fc\end{pmatrix}\mid a,c\in \mathbb{R},a\neq 0,c\neq 0\right\}$$
consists of all matrices in $G$ with a particular ratio $\frac{e}{f}$ between the top-right and bottom-right elements. Since every real number can occur as this ratio, $H$ has index continuum in $G$.
Also, the subgroup of $G$ generated by $H$ and $K$ is $G$ itself, since when $b\neq 0$, $$\begin{pmatrix} a & b\\ 0 & c\end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & \frac{c}{b}\end{pmatrix}\begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix}\begin{pmatrix} a & 0\\ 0 & b\end{pmatrix}.$$
So $H$ has index continuum in the subgroup generated by $K$ and $H$.
A: Let $K$ be a nontrivial countable group and $H$ any infinite group, say of cardinal $\alpha$. Then the index of $H$ in the free product $K\ast H$ is $\alpha$.
Indeed, fix a nontrivial element $k$ of $K$. Then the left cosets $hkH$ for $h$ ranging over $H$ are pairwise distinct. Indeed otherwise there are $h\neq h'$ in $H$ such that $hk\in h'kH$. So $hk=h'kh''$ for some $h''\in H$. Since $h\neq h'$ we have $h''\neq 1$. So the expression $k^{-1}(h^{-1}h')kh''$ is reduced, but it the represents the trivial element, contradiction. So the index is indeed $\alpha$. In particular, it is uncountable as soon as $H$ is uncountable.
