group of order $p\cdot q^{2}\cdot r$, where $p,q,r$ are primes with $p<q<r$.

Question

Is there an any reference for classifying the group of order $p\cdot q^{2}\cdot r$, where $p,q,r$ are primes with $p<q<r$?

I'm interesting in case of $p=3$, $q=5$, and $r=7$.

In M.S.E., I found that the case of order $p^{2}\cdot q^{2}\cdot r$ with $p=3$, $q=5$, and $r=7$.

But, I could hardly find anything about the case of order $p\cdot q^{2}\cdot r$.

In someone M.S. thesis, all the number of sylow $p,q,r$-subgroups are $1$, i.e. $n_{3}=n_{5}=n_{7}=1$ if a group $G$ of order $3\cdot5^{2}\cdot7$. (He or she said just 'by calculation'..)

How do i derive the result that $n_{3}=n_{5}=n_{7}=1$?

Give some advice. Thank you!

Answer 1

For groups of order $525=3\cdot 5^2\cdot 7$ see here:

Proof that no group of order $525$ is simple

Show that $|N_G(\langle x\rangle)|$ is a multiple of $5^2*7$, where $|G|=3*5^2*7$ and $x\in C$, a cyclic subgroup of order 35.

We have $n_5=n_7=1$, see the duplicates, but not necessarily $n_3=1$.