Let $G$ group of order $5^2q^m,\ m>1$, with $5\neq q$ and $q$ prime. We want to show that $G$ is solvable without using Burnside theorem. If $P\in \mathrm{Syl}_{q}(G)$ and $n_{q}=[G:N_{G}(P)]$ we have the following cases :
(i) If $n_{q}=1$ we have that $P\triangleleft G$, so the claim is true.
(ii) If $n_{q}=5$ or $25$ we have that $[G:N_{G}(P)]=5$ or $25$ we know that exist $N\triangleleft G$ with $N\subseteq N_{G}(P)$. If we prove that $N$ is not trivial, it's easy to prove the claim, but how can we do this?
If we can't solve the problem with this way, can you suggest me another to prove this claim?