$p$-Sylow subgroups of a group of order $5^3\cdot 29^2$ 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 $5^3 = 125$.
 A: Here is a survey of the groups of order 105125.
There are basically two kinds of groups of order $5^3 \cdot 29^2$: the nilpotent and the non-nilpotent.
Nilpotent: These are the groups with both $n_5=1$ and $n_{29}=1$.  Such a group is the direct product of its Sylow 5-subgroup and its Sylow 29-subgroup.  Since there are 5 isomorphism classes of groups of order $5^3$ and 2 different isomorphism classes of groups of order $29^2$, there are (5)(2) = 10 isomorphism classes of nilpotent groups of order $5^3 29^2$.  They are: 125×481, 5×25×481, 5×5×5×481, (5⋉(5×5))×481, (5⋉25)×481, 125×29×29, 5×25×29×29, 5×5×5×29×29, (5⋉(5×5))×29×29, and (5⋉25)×29×29.
Every such group has normal subgroups of indices $5^i 29^j$ for 0 ≤ i ≤ 3, 0 ≤ j ≤ 2.  In particular, any field with such a Galois group has a tower of Galois extensions of length 5 (but not 6).
Every such group has a direct product factorization $G_5 \times G_{29}$ as the product of two proper normal subgroups (in fact Sylow p-subgroups).
Non-nilpotent: These are the groups with $n_5=29^2$ and $n_{29}=1$.  Such a group is a semi-direct product of a normal Sylow 29-subgroup of structure  29×29 with a Sylow 5-subgroup acting with a kernel of size $25$.  There are 5 such Sylow 5-subgroups, and most of them have only a single action (up to iso): 125, 5×5×5, and 5⋉(5×5) all have a single action, so we get 3 groups 125⋉(29×29), 5×5×5⋉(29×29), and (5⋉(5×5))⋉(29×29). There are two actions for 25×5 (one where the kernel is 5×5 and one where it is 25×1), and 3 actions for 5⋉25.  This gives a grand total of 8 non-nilpotent groups.
Every such group has normal subgroups of indices 1 and $5^i 29^j$ for 1 ≤ i ≤ 3 and j ∈ { 0, 2 }. In particular, any field with such a Galois group has a tower of Galois extensions of length 4 (but not 5).
No such group has a factorization $G=MN$ where M and N are normal, proper subgroups since every normal proper subgroup has index divisible by 5.
A: It is possible to have $n_5 = 29^2$ as in the following group:
Let $F$ be a finite field of order $29^2$, and let $\zeta \in F$ have multiplicative order $5$. In other words, let $F$ be the splitting field of $\zeta^2 + 6\zeta + 1$ over $\mathbb{Z}/29\mathbb{Z}$ and let $\zeta$ be a root in $F$.
Then consider the matrix group:
$\displaystyle G = \left\{ ~~\begin{bmatrix} \zeta^i & 0 & 0 & \beta \\ 0 & \zeta^j & 0 & 0 \\ 0 & 0 & \zeta^k & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} : 0 ≤ i,j,k < 5, \beta \in F \right\} \leq \operatorname{GL}(4,29^2)$
which is a group of order $5^3 29^2$ with the following $481$ Sylow 5-subgroups indexed by the elements $\gamma$ of $F$:
$\displaystyle P_\gamma = \left\{ ~~\begin{bmatrix} \zeta^i & 0 & 0 & \frac{1-\zeta^i}{1-\zeta} \gamma \\  0 & \zeta^j & 0 & 0 \\ 0 & 0 & \zeta^k & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} : 0 ≤ i,j,k < 5 \right\} \qquad (\beta \text{ has a special form })$
In terms of Steve's answer,
$\displaystyle H = \left\{ ~~\begin{bmatrix} \zeta^0 & 0 & 0 & \beta \\ 0 & \zeta^0 & 0 & 0 \\ 0 & 0 & \zeta^0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} : \beta \in F \right\} \qquad (i=j=k=0)$
$\displaystyle K = \left\{ ~~\begin{bmatrix} \zeta^0 & 0 & 0 & 0 \\ 0 & \zeta^j & 0 & 0 \\ 0 & 0 & \zeta^k & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} : 0 ≤ j,k < 5 \right\} \qquad (i=0, \beta = 0)$
$\displaystyle \alpha =~~~~~ \begin{bmatrix} \zeta & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \qquad (i=1, j=k=0, \beta=0)$
In terms of my favorite groups, $G \leq \operatorname{AGL}(1,29^2) \times K$, and in fact is a Hall {5,29}-subgroup of this group.
