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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!

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2 Answers

up vote 2 down vote accepted

Sorry about my previous answer, didn't quite think it through. Here's an explicit example:

Let $G = S_\infty = \bigcup_nS_n$. This means $G$ is the set of permutations $\sigma\colon\mathbb{N \to N}$ with the property that there exists an $N_\sigma$ such that $\sigma(n) = n$ for all $n \geq N_\sigma$.

Let $G' \subseteq G$ be the set of $\sigma$ satisfying $\sigma(1) = 1$. This is clearly proper.

Now let $\sigma \in G$ be an arbitrary element. Let $h = (1 \ N_\sigma)$ be the permutation that swaps $1$ and $N_\sigma$ (which is a fixed point of $\sigma$). Then $h\sigma h^{-1}$ fixes $1$ and therefore is contained in $G'$.

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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?

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