So I'm studying a simple group of order $|G|=60$, with $n_2=15,n_5=6,n_3=10$ and I'm trying to show that the centralizer $C_G(x)=\{g \in G | xg=gx\}$ of a $x \in P \cap Q$ where $P$ and $Q$ are 2-Sylow groups that intersect non-trivially such that $|P \cap Q|=2$. I would like to show that $|C_G(x)|= 12$ or $20$, and consequently show that it is isomorphic to a subgroup of $S_5$ using an action of $G$ on $G / H$. (for then show that the only subgroup of order $60$ in $S_5$ is $A_5$)
I was able to conclude this same thing with $n_2=5$ before, but I don't really know how to do it for this case.
Would you have any tips? Thanks.
edit: I found something related in https://coolnumbers.wordpress.com/2011/12/28/a_5-is-the-only-simple-group-of-order-60/ but is not quite the same.
edit: right now I'm trying to find $f$ such as $Ker(f)=C_G(x)$ et $Im(f)=5-Sylow$ ou $3-Sylow$ that would show that $G/C_G(x) \cong Im(f)$ therefore $[G:C_G(x)]=12$ or $20$.
Note that there is one element of order $1$, fifteen elements of order $2$, $24$ elements of order $5$ and $20$ elements of order $3$ within the (supposed) Sylow Subgroups accounting for all $60$ elements of $G$.
Since the Sylow subgroups of given order are permuted transitively by conjugation only the identity element can be central, so the centre must be trivial.