Can $G$ have a Sylow -5 subgroups and Sylow -3 subgroups [CSIR-NET-DEC-2015] 
Let $G$ be a simple group of order $60$. Then
  
  
*
  
*$G$ has six Sylow -5 subgroups.
  
*$G$ has four Sylow -3 subgroups.
  
*$G$ has a cyclic subgroup of order 6.
  
*$G$ has a unique element of order $2$.
  

$60=2^2.3.5$ No. of Sylow -5 subgroups =$1+5k$  divides 12.So $1+5k=1,6\implies n_5=1,6\implies n_5=6$ as $G$ is a simple group.
Consider $n_3=1+3k$ divides $20\implies 1+3k=1,4,10\implies 1+3k=4,10$. If $n_3=4$ then we have $8 $ elements of order $3$ and $A_5$ has 20 elements of order $3$ which is a contradiction.Hence $n_3=10$.
Since  $A_5$ has no  element  of order $6$.So 3 is false.
$A_5$ has many elements of order $2$ viz. $(12)(34),(13)(24),,$. Hence $1$ is correct only .Please can someone check whether I am correct /not?
 A: As was remarked before, you do not have to assume that $G \cong A_5$.
(1) You did that one correctly! 
(2) If $|Syl_3(G)|=4$, and $S \in Syl_3(G)$, then $|G:N_G(S)|=4$. Now let $G$ act on the left cosets of $N_G(S)$ by left multiplication, then the kernel of this action is $core_G(N_G(S))=\bigcap_{g \in G}N_G(S)^g$, which is a normal subgroup. Hence, by the simplicity of $G$, it must be trivial and $G$ can embedded in $S_4$, a contradiction, since $60 \nmid 24$.
(3) We prove that if a non-abelian simple $G$, with $|G|=60$, has an abelian subgroup of order $6$, then $G \cong A_5$. This gives a contradiction, since it is easily seen that $A_5$ does not contain any elements of order $6$ (note that an abelian group of order $6$ must be cyclic). So assume $H \lt G$ is abelian and $|H|=6$. $H$ is not normal so, $N_G(H)$ is a proper subgroup (if not then $H$ would be normal) and since $|G:N_G(H)| \mid 10$, we must have $|G:N_G(H)|=5$ ($=2$ is not possible since subgroups of index $2$ are normal). Similarly as in (2), $G/core_G(N_G(H))$ embeds homomorphically in $S_5$ this time. Of course $core_G(N_G(H))=1$, so $G$ is isomorphic to a subgroup of $S_5$ and since it is simple it must be isomorphic to $A_5$ (if we write also $G$ for the image in $S_5$, consider $G \cap A_5 \lhd G$ and use $|S_5:A_5|=2$).
(4) In general: if $G$ is a group with a unique element $x$ of order $2$, then $x \in Z(G)$. Why? Because for every $g \in G$, $g^{-1}xg$ has also order $2$ and must be equal to $x$. In your case $G$ is non-abelian simple, so $Z(G)=1$. 
So only (1) is the true statement.

Edit For case (3) I forgot the case where $|G:N_G(H)|=10$. I have a proof that is quite sophisticated and maybe there is an easier way. Anyway, in this case $H=N_G(H)$. Consider the subgroup $P$ of order $3$ of $H$. This must be a Sylow $3$-subgroup of $G$, since $3$ is the higest power of $3$ dividing $|G|=60$. Observe that in fact $N_G(P)=H$. This follows from what we showed in (2): $|Syl_3(G)|=|G:N_G(P)|=10$ and of course $H \subseteq N_G(P)$. Trivially, $P \subset Z(N_G(P))$. Now $P$ satifies the criterion of Burnside's Normal $p$-Complement Theorem, see for example Theorem (5.13) here. But then $P$ has a normal complement $N$, such that $G=PN$ and $P \cap N=1$. Now $G$ is non-abelian simple, so $N=1$ or $N=G$, which both lead to a contradiction. 
