Problems about generators for Sylow p-subgroups

Question

There are several problems I met asking to find the generators for some different Sylow $p$-subgroup.

$(i)$ a Sylow 2-subgroup in $S_{8}$;

$(ii)$ a Sylow 3-subgroup in $S_{9}$;

$(iii)$ a Sylow 2-subgroup in $S_{6}$;

$(iv)$ a Sylow 3-subgroup in $S_{6}$;

$(v)$ a Sylow p-subgroup in $S_{p^{2}}$.

But here, I would like to ask what is "generators" here? (I found a similar idea in cyclic group, but seems not the same meaning here.)

There is a hint for me that using the De Polignac's formula ( also need to prove or show it that I don't know ).

Would you mind do me a favour, show some of $(i)-(v)$ as examples for me? Thanks :)

Answer 1

A set of elements $g_1, \ldots, g_n$ are called generators for a subgroup $S$ if $S$ is the smallest subgroup that contains $g_1, \ldots, g_n$. This is equivalent to saying that $S$ is the intersection of all the subgroups that contain $g_1, \ldots, g_n$. It's also equivalent to saying that $S$ is the subgroup consisting of all elements which can be written as a product of powers of the elements $g_1, \ldots, g_n$.

I'll do (ii) and (iii) as examples. For (ii) we're looking for a subgroup of $S_9$ that has order $3^4$. Think of the numbers $1$ - $9$ as being partitioned into $3$ groups, $\{1, 2, 3\}, \{4, 5, 6\}, \{7, 8, 9\}$. I wan't to take all the permutations where I'm allowed to cyclically permute within the three groups, and I'm allowed to cyclically permute the three groups themselves. Since permuting $3$ objects cyclically has order $3$ the order of this subgroup has a factor of $3$ for each of the groups and a factor of $3$ for permuting the groups themselves. That's $3^4$ as needed. For a set of generators, the generators for cyclically permuting the groups are $(1 \ 2 \ 3), (4 \ 5 \ 6)$, and $(7 \ 8 \ 9)$. And the generator that lets me permute the three groups themselves is $(1 \ 4 \ 7)(2 \ 5 \ 8)(3 \ 6 \ 9)$. So these four generators generate a Sylow $3$ subgroup of $S_9$.

For (iii) I'm gonna use basically the same trick. Now my groups are $\{1, 2\}, \{3, 4\}$, and $\{5, 6\}$. I need a group of order $2^4$ so I'm gonna allow permuting cyclically within the three groups and I'm gonna allow cyclically permuting the two groups $\{1, 2\}$ and $\{3, 4\}$ themselves. This gives me $4$ factors as needed. The generators are $(1 \ 2), (3 \ 4), (5 \ 6)$, and $(1 \ 3)(2 \ 4)$.

Answer 2

De Polignac's formula is a result in combinatorics, so I would think you're able to use it without proof on an algebra problem, because you don't have the tools to prove it right now. It's not as complicated as it looks. It just lets you count the multiplicity of a prime in $n!$, which is the order of $S_n$.

You can do this without any fancy formulas. In $S_8$, for instance, you have $8!$ elements. The sequence $1$ through $8$ has 4 even numbers. You can count the number of times $2$ appears as a factor easily: once in $2$, twice in $4$, once in $6$, and three times in $8$. So a Sylow 2-subgroup in $S_8$ has order $2^7$.

Every group has generators. These are a set of elements that you can use to get any element in the group by applying the group operation. For example, the dihedral group $D_8$ (symmetries of a square) has two generators: a rotation and a flip. A cyclic group is a group with only one generator.