Transitive groups Someone told me the only transitive subgroup of $A_6$ that contains a 3-cycle and a 5-cycle is $A_6$ itself.
(1) What does it mean to be a "transitive subgroup?"  I know that a transitive group action is one where if you have a group $G$ and a set $X$, you can get from any element in $X$ to any other element in $X$ by multiplying it by an element in $G$.  Is a transitive subgroup just any group that acts transitively on a set?  And if so, does its transitiveness depend on the set it's acting on?
(2) Why is $A_6$ the only transitive subgroup of $A_6$ that contains a 3-cycle and a 5-cycle?
Thank you for your help :)

Let $H\leq A_6$ be transitive and generated by a 3-cycle and a 5-cycle.
Let if possible, $H\neq A_6$, and let us compute $|H|$.
$|H|$ is divisible by 15, and divides 360$=|A_6|$, so it is one of $\{15,30,45,60,90,120,180\}$.


*

*$|H|$ can not be $\{90,120,180\}$, otherwise we get a subgroup of $A_6$ of index less than $6$.

*$|H|$ can not be 15, since then $A_6$ will have an element of order 15, which is not possible,

*$|H|$ can not be 45, since a group of order 45 is abelian and so it contains an element of order 15.

*$|H|$ can not be 30, since a group of order 30 has normal Sylow-5 subgroup, and so it will contain a subgroup of order 15, hence an element of order 15.


Hence  $|H|$ should be $60$. 
Now in this subgroup of order 60, Sylow-5 subgroup can not be normal, since if it is normal, then it will also be normalized by an element of order 3, giving a subgroup of order 15, hence an element of order 15.
So $H$ is a group of order $60$, which has no normal Sylow-5 subgroup; $H$ must be isomorphic to $A_5$. There are 6 Sylow-5 subgroups of $H\cong A_5$, hence 24 elements of order 5; they will be 5 cycles, hence fixing an element in $\{1,2,...,6\}$. Let $(12345)\in H$. 
As $H$ is transitive subgroup, there will be an element $\sigma \in H$ such that $\sigma(6)=1$, so $\sigma (12345)\sigma^{-1}\in H$, will be a 5-cycle fixing 1; in this way all Sylow-5 subgroups of $A_6$, and hence all element of order 5 of $A_6$ will be in $H$, exceeding the size of $H$.
Hence we must have $H=A_6$.
