I was unable to solve the following question in abstract algebra quiz, so I am posting it here.
Let G be a simple group of order 60. Then prove that it doesn't have a cyclic subgroup of order 6.
Using sylow theorems and that group is simple , I got No. of sylow 5 subgroups are 6 and number of Sylow 3 subgroup are 4 or 10 and number if sylow 2 subgroups are 15 or 3 or 5.
But I noticed that in any combinations sum of elements in not 60.
Can you please tell what mistake I am making?
Another thought i had was that by cauchy theorem there exists an element og=f both 2 and 3 so they will generate a subgroup of order 2 and 3 respectively say A and B . SO, why doesnot there exists a cyclic subgroup of order 6 (A $\times$ B)?
Please illuminate on this.
Best Answer
A nonabelian simple group cannot contain a proper subgroup of index less than $5$. ($*$)
(See here, for example, for a proof.)
So the number of Sylow $3$-subgroups cannot be $4$, and hence it must be $10$.
Let $P \in {\rm Syl}_3(G)$. Then $|N_G(P)| = 6$, and to prove that $G$ has no cyclic subgroup of order $6$, we need to prove that $N_G(P)$ is not cyclic.
So suppose it is, and let $g \in N_G(P)$ have order $2$. Let $H = C_G(g)$. Since $g \in H$ and $H$must also contain a Sylow $2$-subgroup of $G$, we have $|H| \ge 12$, and then by $(*)$ we must have $|H| = 12$.
Since $N_G(P) \le H$, $H$ has at most two Sylow $3$-subgroups, and so by Sylow it has only one, but then $P \lhd H$, contradiction.