Let $P$ be a Sylow $5$-subgroup of the symmetric group $S_{25}$ on $25$ letters. Show that there are two elements of $P$ that generate $P$ as a group.

Question

Let $P$ be a Sylow $5$-subgroup of the symmetric group $S_{25}$ on $25$ letters. Show that there are two elements of $P$ that generate $P$ as a group.

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We know that the order of $P$ is $5^{\left\lfloor\frac{25}{5}\right\rfloor+\left\lfloor\frac{25}{5^2}\right\rfloor}=5^6$ and $P$ consists of elements whose orders are of the form $5^n$ for some $n=0,1,...,6$. Meanwhile, we know that elements of $S_{25}$ can be written as products of disjoint cycles. Note that $5$ is a prime, so elements in $P$ are

\begin{align} \prod_{\text{distinct}\ a_{i_k}}(a_{i_1}a_{i_2}a_{i_3}a_{i_4}a_{i_5}) \end{align}

But I am not sure how to choose two elements generating $P$. Any help? Thank you.

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Edit: Now I am pretty sure my deduction below the question is incorrect. Since we know nothing about the number of Sylow $5$-subgroup of $S_{25}$ and we should have discussed things generally. Clearly, all the products of $5$-cycles do not form $P$. I am sorry for the misleading idea.

As a result, the two elements as suggested in the comment: $(1\ 6\ 11\ 16\ 21)$ and $(1\ 2\ 3\ 4\ 5)(6\ 7\ 8\ 9\ 10)(11\ 12\ 13\ 14\ 15)(16\ 17\ 18\ 19\ 20)(21\ 22\ 23\ 24\ 25)$ do not generate $P$ since we have never specified a Sylow $5$-subgroup but the question needs us to prove for every Sylow $5$-subgroup. Simply claiming this is meaningless. Meanwhile, there is no proof for that, so this question remains unsolved.

Answer 1

Here is a very brief proof that the two elements suggested in the comment by Lord Shark The Unknown do indeed generate a Sylow $5$-subgroup of $S_{25}$. Note that it is enough to prove that they generate a subgroup of order $5^6$.

Let $a = (1,6,11,16,21)$ and $b=(1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25)$, and let $H = \langle a,b \rangle$.

You can calculate the conjugates of $a$ under the powers $b^i$ for $0 \le i \le 4$, and these are $(1,6,11,16,21)$, $(2,7,12,17,22)$, $(3,8,13,18,23)$, $(4,9,14,19,24)$, and $(5,10,15,20,25)$.

These five elements clearly commute with each other, and generate an abelian subgroup $N$ of $H$ of order $5^5$. Furthermore, $N$ is normalized by $b$, so $N$ is normal in $H$, and $H = \langle N, b \rangle$ has order $5^6$ as claimed.

Answer 2

Here is another, possibly more conceptual, explanation of for the two generators suggested by Lord the Shark and proved by Derek Holt:

The 5-Sylow subgroup $P$ of $S_5$ can be described as follows: Consider five small pentagons which themselves are arranged on the vertices of a larger pentagon. $P$ is the group of symmetries where we are allowed to rotate each of the 5 small pentagons independently as well as rotate the larger one on which they sit.

Clearly this group is generated by a single rotation of one of the small pentagons along with a rotation of the large one. These are the two generators suggested by Lord the Shark.