Find the $5$-sylow subgroup of the symmetric group $S_{24}$

Question

Show that the $5$-sylow subgroup of $S_{24}$ is isomorphic to $\mathbb{Z}_5\ \times \mathbb{Z}_5 \times \mathbb{Z} _5 \times \mathbb{Z}_5$.

So as I understand this, I need to prove that every element in the $5$-sylow subgroup is a multiplication of $4$ cycles of length $5$, and the rest are cycle of length $1$.

Which makes sense, as the elements should be of order $5$, and cycles of length $5$ are of order $5$, so I can see why it should be isomorphic to $\mathbb{Z}_5\ \times \mathbb{Z}_5 \times \mathbb{Z} _5 \times \mathbb{Z}_5$, as $4$ cycles of length $5$, and another $4$ elements of order $1$ (since it is $S_{24}$).

How can I prove it formally?(Assuming all I said is correct... Otherwise, any hints?)

Answer 1

A $5$-Sylow subgroup of $S_{24}$ has order $5^4$ because $24!=1 \cdots 5 \cdots 10\cdots 15 \cdots 20 \cdots 24$ has four $5$ factors.

The $5$-cycles $(1,2,3,4,5)$, $(6,7,8,9,10)$, $(11,12,13,14,15)$, $(16,17, 18,19,20)$ commute and so generate a subgroup of order $5^4$ isomorphic to $\mathbb{Z}_5 \times \mathbb{Z}_5 \times \mathbb{Z} _5 \times \mathbb{Z}_5$.

This subgroup must be a $5$-Sylow subgroup of $S_{24}$.