# Determining structure of group based on its permutation representation.

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 $$|G|=20$$?

• hint: you can use the orbit-stabiliser theorem to answer both your questions – Robert Chamberlain Aug 13 at 21:19
• 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? – MEG Aug 13 at 23:32
• Sorry, I meant $|H|\cdot 5 = |G|$ in the last equation. – MEG Aug 13 at 23:52
• 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 – Robert Chamberlain Aug 14 at 19:56
• That works, thanks! A clever argument indeed. – MEG Aug 15 at 0:18