TITLE: Group cannot be the union of conjugates
QUESTION [6 upvotes]: I have seen that if $G$ is a finite group and $H$ is a proper subgroup of $G$ with finite index then $ G \neq \bigcup\limits_{g \in G} gHg^{-1}$. Does this remain true for the infinite case also?

REPLY [10 votes]: In your original question you require $H$ to have finite index in $G$. Most of the other answers are treating this as unintended. If you actually did want to require this, then the result is true for infinite groups as well. 
Proof: Let $[G:H]=n$. The action of $G$ on $G/H$ gives a map $G \to S_n$; let $K$ be the kernel of this map. Then $H/K$ and $G/K$ are finite groups, so we know $G/K \neq \bigcup g (H/K) g^{-1}$. In particular, there is some coset $fK$ which is not in any $g H g^{-1}$. So $f$ is not in any $g H g^{-1}$.