Let $G$ be a finite group acting transitively and faithfully on $A$ and $B$ (both finite). Futhermore $hcf(|A|,|B|)=1$. Show: $G$ is acting transitively on $A\times B$.
My ideas:
Since $G$ acts transitively there only exists one orbit. Orbit-stabilizer-theroem yields $|A|=|\alpha^G|$ for $\alpha \in A$ and $|B|=|\beta^G|$ for $\beta \in B$. Since $G$ acts faithfully $\alpha^g=\alpha$ for every $\alpha\in A,B$ implies $g=e$. How can I use this to show the proposition? Is there a possibilty to show that for every $a,a'\in A$ and $b,b'\in B$ there exists $g\in G$ so that $(a^g,b^g)=(a',b')$? Or is it easier to show this by contraposition?
Any help is greatly appreciated!
Edit:
By $hcf$ I intend hcf or $\gcd$.
Best Answer
I denote the stabilizers of (some fixed)$a\in A, b\in B$ by $H,K$ respectively,by the Orbit-stabilizer theorem, we ahve $$\lvert H \rvert \lvert A \rvert=\lvert G \rvert =\lvert K\rvert\lvert B \rvert$$ ,Since $Gcd(\lvert A\rvert;\lvert B\rvert)=1$,then there is positive integer $d=gcd(\lvert H\rvert,\lvert K \rvert)$ such that $$\lvert K \rvert=d\lvert A \rvert$$ and $$\lvert H \rvert=d\lvert B \rvert$$ and $$\lvert G \rvert=d\lvert A \rvert \lvert B \rvert$$.Now use the fact that if $G$ is a group acting transitively on a set $X$, then a subgroup $H$ acts transitively on $B$ iff $G=H stab_{G}(x)$,so this in our case we need to prove that $G=HK$,but we know that $\lvert HK \rvert=\frac{\lvert H \rvert\lvert K \rvert}{\lvert H\cap K \rvert}=\frac{d}{\lvert H\cap K \rvert}d\lvert A \rvert \lvert B \rvert$,since $\frac{d}{\lvert H\cap K \rvert}$ is an integer,we deduce that $HK=G$ and $d=\lvert H\cap K \rvert$ ; Now u can verify that $\lvert Hb \rvert=\frac{\lvert H \rvert}{\lvert H\cap K \rvert}=\lvert B \rvert$, So $H$ acts transitively on $B$,so starting from any $(a',b)$ u can achieve any $(a,b)$,then applying $$H u can achieve any $(a,b')$ . or u can see directly that $G$ acts transitively on $A\times B$ by applying the orbit stablizer theorem and remarking that the stablizer of $(a,b)$ is $H\cap K$,