Find the number $n_p$ of Sylow $p$ subgroups of $Alt(5)$ for $p=5,3,2$.
So $Alt(5)=60=3*4*5$
Then for $p=5$, $n_5$ $=2,3,4,6,12$ but only $6=1$ mod $5$, hence there are $6$ Sylow $5$ subgroups in $Alt(5)$
Then for $p=3$, $n_3$ $=2,4,5,10,20$ but then both $4$ and $10$ are congruent to $1$ mod $3$. What do I do in this case?
Also, for $p=2$ exactly what do I do? We have been taught to compute the $n_i$ by writing the factors of the sum of the numbers not including i, if that makes sense, for example for $p=5$, $n_5$ had possibilities $2,3,4,6,12$ because they are the factors of $12$, and $60=3*4*5$, so discounting the $5$, we get $3*4=12$. So would I write $60=2*2*3*5$, but then do I discount both 2's, so $n_2$=factors of $15$, or just one so $n_2$=factors of $30$?
I am trying to work myself through questions using Sylow's Theorem but I am getting stuck when it comes to finding the orders and determining the structures of the subgroups/groups.
If someone could talk me through an example that would be amazing. I understand finding $n_p$ when there is only one value congruent to 1modp but if there are multiple I do not know where to go from there.
Many thanks