1
$\begingroup$

I am preparing for an algebra prelim at my university, and I came across this question which I cannot finish. Suppose that a finite group $G$ acts transitively on a set $S=\{ a_1,a_2,a_3,a_4,a_5\}$ satisfying the following: $$ Stab(a_1)\cap Stab(a_i) = {e} \textrm{ for all i } \neq 1.$$ Let $H=Stab(a_1)$. Assume that H contains an element of order 4.

(1) Prove that H is cyclic of order 4

(2) Prove that $|G|=20$

Here is my attempt

(1) Let $\psi: G\rightarrow S_5$ be the permutation representation of $G$ on $S$. Then $$Ker(\psi)=\{g\in G|g\cdot a_i = a_i \hspace{3mm} 1\leq i \leq 5\}$$ But $Stab(a_1)\cap Stab(a_i) = {e}$, so this forces $Ker(\psi)=1$. By the first isomorphism theorem and the fact that $Ker(\psi)=\{1\}$, we have $$ G\cong Im(\psi)\leq S_5$$ Now identify $H$ with its isomorphic subgroup of $S_5$ and represent $a_i$ by $i$. $H$ has an element of order 4 that stabilizes $a_1$, so possible generators of $H$ are $\{(2 \hspace{1mm} 3 \hspace{1mm} 4\hspace{1mm} 5),(2\hspace{1mm} 5\hspace{1mm} 4\hspace{1mm} 3), (2\hspace{1mm} 4\hspace{1mm} 3\hspace{1mm} 5), (2\hspace{1mm} 3\hspace{1mm} 5\hspace{1mm} 4),(2\hspace{1mm} 5\hspace{1mm} 3\hspace{1mm} 4), (2\hspace{1mm} 4\hspace{1mm} 5\hspace{1mm} 3) \}$. We could view this set isomorphically as the 6 cycles of length 4 of $S_4$. Now, there are 3 distinct groups that are generated by these 4 cycles. My question is, how can I determine with one $H$ is isomorphic to? $G$ acts transitively on $S$, but I fail to see how this helps since with determining $H$.

(2) Since $G$ acts transitively on $G$, we must have that there is only one orbit. One possibility for $H$ is $\{(2 \hspace{1mm} 3 \hspace{1mm} 4\hspace{1mm} 5), (2\hspace{1mm} 4)(3\hspace{1mm} 5),(2\hspace{1mm} 5\hspace{1mm} 4\hspace{1mm} 3), e \}$. Notice that for any $i$ and $j$ such that $2\leq i, j \leq 5$, there is a element of $H$ that sends $i$ to $j$. So it would seem to me that $H$ acts transitively on $S$. How can I show that $|G|=20$?

$\endgroup$
  • $\begingroup$ hint: you can use the orbit-stabiliser theorem to answer both your questions $\endgroup$ – Robert Chamberlain Aug 13 at 21:19
  • $\begingroup$ For (b) we can use the orbit-stabilizer theorem ($|orb(a_1)|\cdot |stab(a_1)| = |G|$), but since $G$ acts on $S$ transitively and $|stab(a_1)|=4$ by part (a), we see that $|G|=5\cdot 4 = 20$. But how can I use the orbit-stabilizer theorem for part a? How does the equation $|H|\cdot 4 = |G|$ help? $\endgroup$ – MEG Aug 13 at 23:32
  • $\begingroup$ Sorry, I meant $|H|\cdot 5 = |G|$ in the last equation. $\endgroup$ – MEG Aug 13 at 23:52
  • 1
    $\begingroup$ Well done on getting (b), I think (a) is harder. Let $H_1$ be the stabiliser in $H$ of $a_2$. That is, $H_1$ is the set of elements of $G$ that fix $a_1$ and $a_2$, so $H_1=Stab(a_1)\cap Stab(a_2)=e$. Hence by orbit-stabiliser applied to $H$ you have $|H|=4\cdot |H_1|=4$. $H$ has an element of order $4$ so is cyclic $\endgroup$ – Robert Chamberlain Aug 14 at 19:56
  • $\begingroup$ That works, thanks! A clever argument indeed. $\endgroup$ – MEG Aug 15 at 0:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.