Number of Sylow bases of a certain group of order 60 We have $G=\langle a,b \rangle$ where $a=(1 2 3)(4 5 6 7 8)$, $b=(2 3)(5 6 8 7)$.  $G$ is soluble of order 60 so has a Hall $\pi$ subgroup for all possible $\pi\subset\{2,3,5\}$. $\langle a\rangle$ is a normal subgroup.
A Hall $\{2,3\}$ subgroup is $\langle a^5,b \rangle$, a Hall $\{2,5\}$ subgroup is $<a^3,b>$ and a Hall $\{3,5\}$ subgroup is $\langle a\rangle$.  This means a Sylow basis is $\langle a^3 \rangle$, $\langle a^5\rangle$ and $\langle b\rangle$.
I am trying to determine the number of Sylow bases.  Here is my attempt:
$\langle a^3\rangle$ and $\langle a^5 \rangle$ are normal as characteristic subgroups of the normal subgroup $\langle a\rangle$, and so as all Sylow bases are conjugate, their number is equal to the number of Sylow 2 subgroups, which is 1 or 5 by Sylow's theorem and $\langle b\rangle$ is not normal so it is $5$.
The answer we were given is 15, so I wondered if anyone else could spot where I'm going wrong.  Thanks.
 A: The number of Sylow bases is exactly equal to the number of Sylow complement bases. A Sylow basis has a complicated definition involving permuting subgroups, but a Sylow complement basis has a very simple definition: a choice of one Sylow $p$-complement for each prime $p$. Thus we just need to count the number of Sylow $p$-complements. This is just finding the index of their normalizer, and since the Sylow 2-complement is normal, but the other complements are not and have prime index, we can easily count them.


*

*$p=2$: The Sylow $p$-complements are the Hall $\{3,5\}$-subgroups, the conjugates of the normal subgroup $\langle a \rangle$. So just one.

*$p=3$: The Sylow $p$-complements are the Hall $\{2,5\}$-subgroups, $\langle a^3,b \rangle$ and its 3 conjugates (not normal, but has index 3)

*$p=5$: The Sylow $p$-complements are the Hall $\{2,3\}$-subgroups, $\langle a^5, b \rangle$ and its 5 conjugates (nor normal, but has index 5)


All told that is $1 \times 3 \times 5 = 15$ possibilities.
Another method is to find a system normalizer. In this case it is a Sylow 2-subgroup, which can be seen from Carter's work on abnormal subgroups (given the large number of normal subgroups).
