TITLE: Group generated by a conjugacy class
QUESTION [5 upvotes]: Let $G$ be a group, $x \in G$, and $S = \{ x^g \mid g \in G\}$. Suppose $\langle S \rangle = G$ and that $H$ and $K$ are 
subgroups of $G$ with $S \subseteq H \cup K$. Show that $H=G$ or $K=G$.
This is a problem from Kurzweil & Stellmacher's Theory of Finite Groups. 
If we let $\langle x \rangle = A$, then $G$ is the product of the distinct conjugates
of $A$ in any order, 
 $G=A_1 A_2 \cdots A_k$, because they generate $G$. Using this I've managed to show $G=HK$.
Also $k≥3$ because a group can't be the product of two proper conjugate subgroups.
The picture emerging 
is that since one of $H$, $K$ has to contain two distinct  $A_i$,
 somehow that forces 
it to be the whole group. Of course this only works if the cyclic subgroups generated by each individual generator of $G$ have finite index, which is not guaranteed in this introductory-level problem.
I suspect I'm overlooking something simple. Any help would be greatly appreciated.

REPLY [6 votes]: I corrected the answer thanks to  Derek Holt.
We can suppose that $x\in H\setminus K$. Then $x^k\in H$ for all  $k\in K$ and hence $x^{kh}\in H$ for all $h\in H$. Since $G=KH$ we get $S\subset H$.
Added by Derek Holt: Proof that $G=KH$. Every element of $G$ can be written as a product of conjugates of $x^{\pm 1}$, some of which lie in $H$ and some in $K$. Using $ba = ab^a$, we can move all of those in $K$ to the left, without increasing the length of the product, and we end up with a product in $KH$.