# Group generated by a conjugacy class

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.

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 $KH$.
• Very good! I'm not quite seeing why $x^{h_1k_1} \in H$, since we do not know that $x^{h_1} \not\in K$. But we do have $x^{kh} \in H$ for all $h \in H$, $k \in K$, so we could use the fact that $G=KH$ to deduce that $S \subset H$. – Derek Holt Sep 28 '13 at 9:47
• Why is $G=KH$? Shouldn't that be $G=\langle KH \rangle$? – Nicky Hekster Sep 28 '13 at 11:36
• I've added a proof that $G=KH$. – Derek Holt Sep 28 '13 at 11:42