I'm working on the following problem:
Let $A$ and $B$ be subgroups of a group $G$, for which $G=AB$. Further, let $g,h\in G$ be arbitrary.
Show that $(A^g)(B^h)=G$.
Additionally, show that there is an element $k\in G$ so that $A^g=A^k$ and $B^h=B^k$.
(To clarify, the notation $A^g$ means $\{g^{-1}ag:a\in A\}$.)
I believe I can prove Part 1:
Proof. Let $g,h\in G$ be given. Clearly $AB=BA$. Express $g=a_1b_1$ and $h=b_2a_2$ with $a_1,a_2\in A$ and $b_1,b_2\in B$. Then $A^gB^h=A^{b_1}B^{a_2}$. Rewrite $b_1a_2^{-1}$ as $a_3b_3$ with $a_3\in A,b_3\in B$. Then,
\begin{align*} A^{b_1}B^{a_2}&=b_1^{-1}A(a_3b_3)Ba_2\\ &=b_1^{-1}ABa_2\\ &=G.\qquad\blacksquare \end{align*}
The problem I'm having is, I don't know how to show that there is a $k\in G$ for which $A^g=A^k$ and $B^h=B^k$. I've tried defining $k$ as a few combinations of $g$ and $h$, but didn't get anywhere.
Most likely, I feel like I'll need to use the fact that $G=AB$ somewhere. I know also that conjugacy can be used as an equivalence relation, although I'm not sure if that will be relevant to this problem or not. Any help will be appreciated!
$A^gB^h=G \Rightarrow A B^{hg^{-1}} = G$.
Since $G=AB=BA$ we can write $hg^{-1}=ba$ with $a \in A$, $b \in B$.
Let $k=ag$. Then $A^k = A^g$, and $B^k = B^{ag} = B^{bag} = B^h$.