I need to calculate the $p$-Sylow subgroups of a Galois group with order $5^3 \cdot 29^2$, i.e. $|\mathrm{Gal}(K/F)|=5^3 \cdot 29^2$.
I've already established that there is only one 29-Sylow-subgroup (with $|G_{29}|=29^2$) by the following conditions: $n_p \equiv 1 \pmod p$ and $n_p \mid m$ where $|G|=m\cdot p^r$ and $p\nmid m$.
(Indeed $5,25,125 \not{\!\equiv}\; 1 \pmod {29}$ and they are the only $\ne1$ divisors of $5^3$, so $n_{29} = 1$.)
I want to apply it to find 5-Sylow subgroups. $n_5 \equiv 1\pmod 5$ and $n_5\mid 29^2$. We have that $29 \not{\!\equiv}\; 1 \pmod 5$ but $481 = 29^2 \equiv 1 \pmod 5$ so $n_5 = 1$ and $n_5 = 29^2$ are both valid options.
I want to show that $n_5 = 1$.
Edit:
I will explain the rational behind my question: we have $F \subset K$ a Galois extension of degree $5^3 \cdot 29^2 = | \mathrm{Gal}(K/F)|$ and I want to find two subfields $F \subset K_1 , K_2 \subset K$ which are Galois extensions and $K = K_1 K_2$.
My idea was to find the corresponding p-Sylow groups of $\mathrm{Gal}(K/F)$ and use the fundamental theorem of Galois theory and deduce the extensions. That's why I looked for a normal subgroup of order $5^3 = 125$.