Let $G$ be a group of order $1575 = 3^25^27$. Show that $G$ is solvable.
So far, my idea is to show that $n_7, n_5 $ or $n_3$ equals $1$, so that there exists a unique Sylow subgroup of order $7, 5^2$ or $3^2$ which is normal.
For example, suppose that $n_7 = 1$. Then, I can take the quotient group $H = G/P_7$, where $P_7$ is a $7$-Sylow subgroup. Then, $|H| = 3^25^2$. Now, in $H$, we have that $n_5=1$, so there exists a unique $5$-Sylow subgroup in $H$. Then, $H/P_5$ has order $3^2$, which is abelian. Hence, $P_5$ and $H/P_5$ are solvable groups, whence $H$ is solvable. Applying the same argument on $H = G/P_7$, since $P_7\leq G$ is solvable, then $G$ is solvable.
However, I can not make sure that $n_7=1$, since $n_7$ could be $n_7 = 15 $ or $n_7 = 225$. The old trick of assuming that $n_7>1$ would give me more elements of order $7$ than the number of elements of $G$ doesn't work here.
Is there any other idea I could use to rule out the cases $n_7 = 15$, $n_7 = 225?$ I am aware of this question here, but in this case it is assumed that we have a normal $3$-Sylow subgroup.
Undoubtedly there are many ways to settle this. The one outlined in the comments by Travis Willse and j.p. may be the simplest, and I am most certainly the wrong person to hunt for the simplest arguent (if we can agree on a metric :-). I proffer the following less general alternative that I cooked up.
As a starting point I also use the action of $G$ on the set $X$ of Sylow $5$-subgroups. In the interesting case $n_5=|X|=21$, so if we fix a Sylow $5$-subgroup $P$, we can assume that $N_G(P)$ has order $|G|/n_5=75$.
A useful observation about a group of order $75$ is that either it is abelian, or it is isomorphic to the semidirect product $\Bbb{F}_5^2\rtimes C_3$, with the conjugation action of an element or order three on the vector space $\Bbb{F}_5^2$ represented by a matrix of order three that has no eigenvalues in the prime field. See here for a proof of this fact.
A consequence of the italicized observation is that either A) $N_G(P)$ is abelian, or B) $N_G(P)$ has no subgroups of order fifteen. I shall deal with these two cases separately.
B: If $N_G(P)$ has no subgroups of index five, it follows that the conjugation action of $N_G(P)$ on $X$ cannot have orbits of size five. That leaves orbits of size $1,3,15$ as possibilities. Actually size $3$ is impossible as the Sylow $5$-subgroups in such an orbit would be normalized by $P$, and an elementary fact is that $P$ is the only Sylow $5$-subgroup it normalizes. Try as we may, we cannot build the set $X$ with $21$ elements out of such orbits unless there are several orbits of size one, a possibility excluded by the same reasoning. So case B is impossible.
A: And we can assume that $N_G(P)$ is abelian. Let $x\in N_G(P)$ be an element of order three. Consider its centralizer $C_G(x)$. We know that $P\le C_G(x)$. But all the Sylow $3$-subgroups have order $3^2$, and are thus known to be abelian. Therefore the order of $C_G(x)$ is also divisible by nine. So either $C_G(x)=G$, or $[G:C_G(x)]=7$. In the latter case we get a non-trivial homomorphism $\phi:G\to S_7$ from the transitive action of $G$ on the cosets of $C_G(x)$. As $5^2\nmid 7!$ this homomorphism has a kernel of order $m$ such that $5\mid m$ and $m\mid 3^2\cdot 5^2$.
So it follows that either $G$ has a normal subgroup $N$ of order divisible by five, or $3\mid |Z(G)|$. Leaving the rest to you. The most taxing case seems to be that of $|G/N|=315$, but there is another useful old thread. The case of $G/Z(G)$ of order $525$ has also been dealt with.