This is the full question:
Let G be a finite group, $|G| = 1575$ and suppose $n_3 =1$,then $n_5=1$ and $n_7=1$. Prove $G$ is Abelian. $n_p$ denotes the number of sylow $p$-subgroups in $G$
$1575=3^2\cdot5^2\cdot7$, I was able to prove that if $n_3=1$ then $n_5=1$ and $n_7=1$. But how do i show from here that $G$ is Abelian ? I was considering using the thm : $ G/Z(G)$ is cylic iff $G$ is abelian. But then there are many cases to consider. Any hints ?
There is an extra property of Sylow subgroups, namely: if $H$ and $K$ are Sylow-subgroups (not necessarily of the same order and not with respect to the same prime number) that are normal (and thus unique), then $$\forall x \in H \text{ and } \forall y \in K, xy = yx. (1)$$ In particular, the above relation holds also for arbitrary normal subgroups whose intersection is only the neutral element.
In your question, the Sylow-subgroups are of order $3^2, 5^2$ and $7$. It is know that a group of order $p$ or $p^2$ where $p$ is a prime is abelian.
Equation $(1)$ also gives you an isomorphism between groups. If we let $S_3, S_5$ and $S_7$ be the respective Sylow subgruops from your question, then each of them is abelian, and the function $$g:S_3 \times S_5 \times S_7 \to G, g(x,y,z) = xyz $$ is an isomorphism, and since a direct product of abelian groups is abelian, then $G$ is abelian.