Generators of Sylow $2$-subgroups of $ S_8 $ I know that $ S_8 $ has Sylow $2$-subgroups of order $ 2^7 $. I have no idea about their generators.
 A: The Sylow $2$-subgroups of $S_{2^{n}}$ are particularly easy to describe, because they are "simply" iterated wreath products of cyclic groups of order $2$.
In this case, this observation yields a somewhat natural minimal set of generators of the form
$$
(1,2), \quad (1,3)(2,4), \quad (1,5)(2,6)(3,7)(4,8).
$$
The pattern should be clear. 
The general form for a "wreath product" set of generators for a Sylow $2$-subgroup of  $S_{2^{n}}$ would be
$$
(1,2) = \prod_{i=1}^{1} (i, i+1), \quad
\prod_{i=1}^{2} (i, i+2), 
\quad
\prod_{i=1}^{4} (i, i+4), \quad
\dots
\quad,
\quad
\prod_{i=1}^{2^{n-1}} (i, i+2^{n-1}) 
$$
A: Let's build up the group, noting that because all the Sylow $2$ subgroups are conjugate, and because every 2-subgroup is a subgroup of some Sylow $2$ subgroup, any Sylow $2$ subgroup will contain elements from each conjugacy class of elements of order a power of $2$.
$S_0$ has the identity element, as does $S_1$ - we add that the Sylow $2$ subgroups of $S_{2n+1}$ are isomorphic to those of $S_{2n}$.
$S_2$ adds a transposition $(1,2)$ to give a subgroup of order 2.
$S_4$ has a Sylow subgroup of order $8$ with three types of element $(1,2,3,4); (1,3); (1,3)(2,4)$ it is generated by the $4$-cycle and the transposition (which is not in the subgroup generated by the $4$-cycle).
$S_6$ has a Sylow subgroup of order $16$ with new cycle-types $(1,3)(2,4)(5,6); (1,2,3,4)(5,6)$ - the transposition $(5,6)$ commutes with the elements of the group of order $8$ based on $S_4$.
$S_8$ adds several cycle-types. If we base it on $a=(1,2,3,4,5,6,7,8)$ we have $b=(1,3,5,7)(2,4,6,8)$; $c=(1,3,5,7)(2,6)(4,8)$; $d=(1,3,5,7)(2,6)$; $e=(1,3,5,7)$ as the types containing $4$-cycles. The group generated by $b,c,d,e$ is the direct product of the cyclic group generated by the $4$-cycle $(1,3,5,7)$ and a Sylow-subgroup of $S_4$ based on $(2,4,6,8)$ - hence a group $H$ of order $32$. The other conjugacy types are $(2,6)$, $(2,6)(4,8)$, $(1,5)(2,6)(3,7)(4,8)$, $(1,5)(3,7)(4,8)$ - with a bit of thought these can be derived from $a,b,c,d,e$.
Adding $a$ to $H$ gives the whole group (since conjugation by $a$ gets all the basic elements we want). Now we note that $b=a^2$. Also $de^{-1}=(2,6)$, and conjugating $(2,6)$ by appropriate powers of $a$ gives the transpositions $(1,5);(3,7);(4,8)$. The last of these gives us $c$. So the group is generated by the elements $a, d, e$. We could replace $d$ by $f=(2,6)$ since $d=ef$.
Then our generators are $a=(1,2,3,4,5,6,7,8); e=(1,3,5,7); f=(2,6)$ and it remains to show that each lies outside the group generated by the other two.
