Show that any group of order 3025 is solvable 
Show that any group of order $3025$ is solvable.

Prime factorize the order of the group such that $3025 = 5^{2}11^{2}$. Let $n_{11}$ be the number of Sylow $11$-groups. Then, by Sylow's Third Theorem we have that
$$n_{11} \equiv 1 \bmod{11} \quad \text{and} \quad n_{11}|5^{2}11^{2}.$$
By the second condition we have $n_{11} = 1, 5, 25$ (I always wondered here why we never consider for instance $n_{11} = 11, 11^{2}, 5 \cdot 11, \ldots$ as potential divisors? In the examples given in class we always just looked at the divisors of the other prime factor. How come?).
But only $n_{11} = 1$ satisfies $n_{11} \equiv 1 \bmod{11}$. Thus, there is precisely one Sylow $11$-group, call it $N$. This means that $N$ is a normal subgroup. Also every finite $p$-group is solvable and hence $N$ is solvable.
It remains to show that $G / N$ is solvable. But $|G / N| = \frac{|G|}{|N|} = \frac{5^{2}11^{2}}{11^{2}} = 5^{2}$. Hence, $G / N$ is a finite $5$-group and hence solvable.
Since $N$ is a normal subgroup of $G$ and $N$ and $G / N$ are solvable, this implies that $G$ is solvable.
 A: If $|G|=p^km$ where $(m,p)=1$ then $n_p|m$ because it is coprime with $p$ ($n_p \equiv 1 \mod p)$ and divides $p^km$.
A: Sylow theorem says:

Let $G$ be a finite group, and let $|G|=p^rm$ with $m$ not divisible by $p$. Then $n_p\equiv 1\pmod{p}$ and $n_p|m$

In your case it is also $11, 11\cdot 5,...\equiv 0\pmod{11}$
A: If you list all the factors of 3025 as you suggest, the ones divisible by 11 would get filtered out at the next step anyway: the proof would go

By the second condition, $n_{11} = 1, 5, 25, 11, 55, 275, 121, 605, 3025$.  But of these, only $n_{11} = 1$ is congruent to $1$ modulo $11$.

and from there, would be as in the original.
But this always happens — when trying to find $n_p$, you’re looking for factors of $|G|$ that are 1 mod $p$, and factors divisible by $p$ itself certainly won’t be 1 mod $p$.  So when listing the candidates for $n_p$, there’s no point in including those — you only need to consider the factors of $|G|$ that don’t include $p$.  And since this always happens when determining Sylow numbers, it’s not usually explicitly explained.
