The number of elements not in conjugate of a subgroup There is a proposition about the number of elements not lying in any conjugate of a subgroup when the group is finite. It states: 

If $G$ is a finite group and $H$ is a proper subgroup, then the number of elements not in any conjugate of $H$ is at least $|H|$.

(See, for example, Isaac's Finite Group Theory Page 7.
However, I wonder whether the statement still holds when $G$ is an infinite group. I suppose it's not so hard to prove the statement for the infinite case, but I failed to find a way. Though it's clear that the statement holds when $G$ has an element with infinite order, I couldn’t find a way if I already get $G$ has a finite subgroup contains $H$ and the existence of a such subgroup is something I can't work out as the bounded Burnside problem shows there exists finitely generated infinite group with every element having a bounded order.
In fact, if there's any estimation about this number of elements not in conjugate of a subgroup, it'll be helpful! 
Hope for an answer, thanks in advance!
 A: For an infinite group, it is in fact possible that every element is contained in some conjugate of a proper subgroup containing just two elements.  That is, it is possible for an infinite group to consist of just two conjugacy classes:  the identity and everything else.
The construction of such a group is not elementary.  A very nice outline of the process has been given by in Arturo Magidin's answer to Infinite group with only two conjugacy classes.
A: Here's an explicit example:
let $G$ be the group of increasing self-homeomorphisms of $\mathbf{R}$ with bounded support (i.e., identity outside a compact subset), and $H$ the subgroup of those increasing self-homeomorphisms with support in $[-1,1]$. 
Indeed, if $g\in G$, there exists $u\in G$ such that $u(\mathrm{Supp}(g))\subset [-1,1]$, and hence $ugu^{-1}\in H$.
Also, if we restrict to those elements in $G$ (and in $H$) that are piecewise affine, for which slopes are integral powers of $2$ and breakpoints are dyadic numbers, we get a countable example for $G$.
