Suppose $G$ is a simple group of order $60$, show that $G$ can not have a subgroup isomorphic to $ \frac {\bf Z}{6 \bf Z}$.
Of course, one way to do this is to note that only simple group of order $60$ is $A_5$. So if $G$ has a cyclic subgroup of order $6$ then it must have a element $\sigma$ of order $6$, i.e. in (disjoint) cycle decomposition of $\sigma$ there must be a $3$ cycle and at least $2$ transposition, which is impossible in $A_5$.Hence,we are done.
I'm interested in solving this question without using the fact that $ G \cong A_5$. Here is what I tried:
Suppose $G$ has a subgroup say $H$ isomorphic to $ \frac {\bf Z}{6 \bf Z}$, then consider the natural transitive action $G \times \frac {G}{H} \to \frac {G}{H}$, which gives a homomorphism $\phi \colon G \to S_{10}$. Can some one help me to prove that $\ker \phi$ is nontrivial ?
Is there any other way to solve this question? Any hints/ideas?
Best Answer
(1) Suppose $H=C_6$ is a cyclic subgroup of order $6$ in $G$ (simple group of order $60$). Let $z$ be the element of order $2$ in $H$, so that $\langle z\rangle$ is subgroup of order $2$ in $H$.
(2) $C_G(z)=$centralizer of $z$ in $G$; it clearly contains $H$. Also, $\langle z\rangle$ is contained in Sylow-$2$ subgroup of $G$, say $P_2$, which has order $4$, hence it is abelian, and therefore, centralizer of $z$ contains this Sylow-$2$ subgroup.
(3) Thus, $C_G(z)$ contains a subgroup of order $3$ (of $H$) as well as Sylow-$2$ subgroup of order $4$; hence $|C_G(z)|\geq 12$.
(4) Then $[G\colon C_G(z)]\leq 5$, and since $G$ is simple group of order $60$, $G$ can not have a subgroup of index $<5$ [prove it]. Hence $|C_G(z)|$ must be $12$.
(5) Thus, $H\subseteq C_G(z)$ and $|C_G(z)|=12$. Then $H$ has index $2$ in $C_G(z)$, so $H\trianglelefteq C_G(z)$. If $P_3$ denotes the (unique, characteristic) subgroup of $H$ of order $3$ then it follows that $P_3$ is normal in $C_G(z)$. [Characteristic subgroup of a normal subgroup is normal].
(6) Thus, $P_3$ is a Sylow-$3$ subgroup of $G$, which is normal in a group of order $12$ namely $C_G(z)$, i.e. index of normalizer of a Sylow-$3$ subgroup is at most $5$; it must by $5$ (by similar reason as in (4)).
(7) So $P_3$ is a Sylow-$3$ subgroup of $G$ with index of its normalizer equal to $5$; this means the number of Sylow-$3$ subgroups must be $5$; this contradicts Sylow's theorem.