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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 $5^3 = 125$.

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    It's not true. It is true there is a subgroup of index $5$ which is normal, though.2012-06-08
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    In particular, let $H=C_{29}\times C_{29}$, let $K=C_5\times C_5$, and let $\alpha\in GL(2,29)$ be an element of order $5$. Then $HK\rtimes\langle\alpha\rangle$ has more than one Sylow 5-group.2012-06-08
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    $K \times (H \rtimes \langle \alpha \rangle)$ even.2012-06-08
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    It might help if I add that I do this calculation to find a normal subgroup of order $5^3$ and index $29^2$ in order to prove a claim about (at least 2) intermediate Galois extensions between $F$ and $K$. $G_{29}$ gives one Galois extension $[ K_{29} : F ] = 5^3$.2012-06-08
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    Ah, if you just want 2 intermediate Galois extensions (so in group theory he is asking about the chief length of this group, right?) then you mod out by the normal G29 and use the isomorphism theorems to see that there are normal subgroups of index 1, 5, 25, 125, and 125*481 (but there are not any normal subgroups of index $5^i*29$ unless the galois group is nilpotent). You also get normal subgroups of index 5*481 and 25*481, but these require more than I was taught in my first semester algebra class to find.2012-06-08
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    @JackSchmidt I already found the normal sugroup of index $5^3$, namely $G_{29}$ of order $29^2$ and then $[G : G_{29}] = |G|/|G_{29}|=5^3$ which gives one Galois extension $F \subset K_1 \subset K$ of $[K_1 : F ] = 5^3$ (since $G/G_{29} \cong G_{{K_1}/F}$ is its Galois group). I believe the second extension (which is not isomorphic to $K_1$) should be of degree $29^2$ and thus imply on a normal subgroup of order $5^3$.2012-06-08
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    Zachi, I was suggesting having $F \subset K_2 \subset K_1 \subset K$ with $[K_2:F] = 25$, $[K_1:K_2] = 5$ and $[K:K_1] = 481$. Notice that $G/G_{29} = G_{K_1/F}$ is a $p$-group, and so it has a normal subgroup $Z/G_{29}$ of order $p=5$, then $G/Z$ has order 25 and is (isomorphic to) the Galois group of $K_2/F$, where $F$ is the fixed field of $G$ and $K_2$ is the fixed field of $Z$.2012-06-08
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    Oh, and now I see your edit. You don't want a tower. Let me think for a bit. I think you want express $G=MN$ where both $M$ and $N$ are normal (and proper) subgroups. I think that might be impossible in my group, at least if you choose $N=G_{29}$, since $M$ must then have order a multiple of $5^3$, but there is no proper normal subgroup of my group with index a divisor of $29^2$.2012-06-08
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    And I just checked, sometimes there is no M for any choice of N. So the edited task of finding two such Galois subfields is impossible.2012-06-08
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    $29^2=841{}{}{}$2012-06-08

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