# $p$-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 $5^3 = 125$.

• It's not true. It is true there is a subgroup of index $5$ which is normal, though. – user641 Jun 8 '12 at 16:56
• 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. – user641 Jun 8 '12 at 16:57
• $K \times (H \rtimes \langle \alpha \rangle)$ even. – Jack Schmidt Jun 8 '12 at 17:00
• 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 $[ K_{29} : F ] = 5^3$. – Zachi Evenor Jun 8 '12 at 17:46
• $29^2=841{}{}{}$ – Chris Eagle Jun 8 '12 at 23:53

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.

• Thank you very much, if $\mathrm{Gal}(K/F)$ is indeed nilpotent then the direct product of its Sylow subgroups (which are normal) yields two Galois extension as desired. It remains to show that this Galois group is nilpotent (which is stronger property than solvability). – Zachi Evenor Jun 8 '12 at 21:23
• Zachi: exactly, and need not be true given what you've said so far. – Jack Schmidt Jun 8 '12 at 21:24
• The edit to the question says the OP is looking for two intermediate normal extension such that their compositum is all of $K$. So he is looking for two normal subgroups whose intersection is trivial. These always exist. – user641 Jun 9 '12 at 14:31

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 $F$:

$\displaystyle P_\gamma = \left\{ ~~\begin{bmatrix} \zeta^i & 0 & 0 & \frac{1-\zeta^i}{1-\zeta} \gamma \\ 0 & \zeta^j & 0 & 0 \\ 0 & 0 & \zeta^k & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} : 0 ≤ i,j,k < 5 \right\} \qquad (\beta \text{ has a special form })$

$\displaystyle H = \left\{ ~~\begin{bmatrix} \zeta^0 & 0 & 0 & \beta \\ 0 & \zeta^0 & 0 & 0 \\ 0 & 0 & \zeta^0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} : \beta \in F \right\} \qquad (i=j=k=0)$

$\displaystyle K = \left\{ ~~\begin{bmatrix} \zeta^0 & 0 & 0 & 0 \\ 0 & \zeta^j & 0 & 0 \\ 0 & 0 & \zeta^k & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} : 0 ≤ j,k < 5 \right\} \qquad (i=0, \beta = 0)$

$\displaystyle \alpha =~~~~~ \begin{bmatrix} \zeta & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \qquad (i=1, j=k=0, \beta=0)$

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.

• Thank you for the example. So my claim is not true in general, but is the fact that the "mother" group of order $5^3 \cdot 29^2$ is Galois can give additional information that may imply that $n_5=1$? – Zachi Evenor Jun 8 '12 at 17:42
• @Zachi: I don't think so. I believe this group is in fact the galois group of a polynomial over the rationals, but it is a little large for me to give a specific polynomial (I think such a polynomial has degree 481). – Jack Schmidt Jun 8 '12 at 18:31
• It is a conjecture that any finite group can be obtained as the Galois group of some polynomial over the rationals, so being a Galois group does not really give you any extra information, unless you know something about the polynomial. – Tobias Kildetoft Jun 8 '12 at 18:55
• In this specific case, we only know that $[ K : F ] = 5^3 \cdot 29^2$. We are not even told to assume that $\mathrm{char}(F)=0$. So I don't think using a specific polynomial helps. – Zachi Evenor Jun 8 '12 at 19:12
• @ZachiEvenor: To add to what Tobias said, it is a theorem (of Shafarevich) that every finite solvable group is the Galois group of a polynomial over the rationals, and it is is a theorem of Burnside that every finite group whose order is divisible by only two primes is solvable. So you definitely gain no new information from just knowing your group is a Galois group. – Arturo Magidin Jun 8 '12 at 19:45