TITLE: A question on the conjugacy class in infinite group
QUESTION [5 upvotes]: Construct an infinite group $G$ and a proper subgroup $G'$, such that the union of all the conjugate group of the subgroup can cover $G$, that is to say, $\forall g \in G$, $\exists h\in G,g'\in G'$, s.t. $h^{-1}g'h=g$.
Here is my thoughts: $G$ should be non-abelian, so $G$ can be constructed as the matrix multiplication or the infinite permutation group. However, I got stuck here because I can't get the subgroup and prove the covering.
Thanks for your help!

REPLY [4 votes]: Equivalently, you want $G$ to be the union of the conjugates of $H$ in $G$. This is equivalent to saying that, in the permutation action of $G$ by multiplication on the (left or right, whichever you prefer) cosets of $H$, no elements of $G$ act fixed-point-freely.
So we are looking for a transitive group $G$ of permutations on an infinite set with no fixed-point-free elements. We can then take $H$ to be the stabilizer of a point. How about taking all permutations of an infinite set that move only finitely many points?