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Calculate the number of $n_5$ of Sylow 5-subgroups in $S_{10}$ (a product of several integers is an acceptable answer) and check that $n_5 \equiv 1 mod 5$. Prove that every element of order 5, $\sigma \in S_{10}$, lies in some Sylow 5-subgroup. Calculate the number of Sylow 5-subgroups containing $(12345) \in S_{10}$.

if $G = S_{10}$, for the first bit, I got $|G| = 10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$. So my $n_5$'s will be the numbers that satisfy

$$n_5 \equiv 1 mod 5 \hspace{1cm} n_5 | 2^8 \cdot 3^4 \cdot 7 = 518400$$

Is there a quick way of working this out rather than writing out all the numbers? Doing this will automatically answer the "check" part of question.

I'm unsure how to do the other two bits. I was hoping that they would lead on from part one so once I can do that, I could do the rest.

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What do you mean by "my $n_5$'s"? There can be only one $n_5$, and the congruence/divisibility properties you list aren't enough to find it in and of themselves; you should find the value in a different way and then confirm that it is $\equiv 1\pmod 5$. – Steven Stadnicki Jan 7 '13 at 19:56
Do you know what the Sylow 5-subgroups look like in terms of the cycle decomposition of their elements (there are two types of element of order 5 - how many of each type in each Sylow subgroup?). The 5-cycles (eg (12345)) can be in more than one Sylow 5-subgroup - what about the other type, and can you count them? – Mark Bennet Jan 7 '13 at 20:04
@MarkBennet No, this is the trouble I'm having with this module, I can't like picture all these groups and stuff in my head and so its making it harder for me to understand. I get that the cycle decomposition comes from the $S_{10}$ but I don't get how to do it. – Kaish Jan 7 '13 at 20:36
@StevenStadnicki Oh yeah, I see how I wrote it wrong. What other ways are there? I've only done example with small orders so its easy to write everything out – Kaish Jan 7 '13 at 20:37

So you need to understand what is going on in a symmetric group - that's fine.

Each element of $S_{10}$ can be thought of as permuting elements of a ten element set, say $\{1,2,3,4,5,6,7,8,9,X\}$.

Now there are different conventions for writing these elements, but let's take the cycle $(1 2 3 4 5)$ in your question. This maps $1\to 2,2\to 3, 3\to4,4\to5,5\to1, 6\to6,7\to7,8\to8,9\to9,X\to X$. Performing the same operation five times leaves us where we started, so this is an element of order 5.

We can always split an element into disjoint cycles (normally the fixed elements are assumed rather than stated). The order of an element is the least common multiple of the lengths of its (disjoint) cycles, so:

$(1 2)(3 4 5)$ has order 6; $(1 2 3)(4 5 6 7 8)$ has order 15; $(1 2)(3 4 5 6)$ has order 4; $(1 2 3 4)(5 6 7 8 9 X)$ has order 12.

You can only get an element of order 5 if the only cycles are of length (1 or) 5. So the shapes are $(1 2 3 4 5)$ and $(1 2 3 4 5)(6 7 8 9 X)$.

To get a Sylow 5-subgroup - since we don't have any elements of order 25, and we know that a group of order 25 is abelian, we must find two elements of order 5 which commute with each other. For example, we might choose (1 2 3 4 5) and (6 7 8 9 X) - these are disjoint, they affect different elements of the underlying set, so it does not matter which order we do them in. They form a group with elements:

1 (identity)

(1 2 3 4 5); (1 3 5 2 4); (1 4 2 5 3); (1 5 4 3 2)

(6 7 8 9 X); (6 8 X 7 9); (6 9 7 X 8); (6 X 9 8 7)

(so we take the simple powers of the generators)

And then we find the remaining 16 elements by combining each element from the top row with each element from the bottom - like:

(1 2 3 4 5)(6 7 8 9 X); (1 3 5 2 4)(6 9 7 X 8); (1 5 4 3 2)(6 X 9 8 7)

That gives us the 25 elements we need.

You need to get to grips with the fact that if we have two 5-cycles which are not disjoint, either one is a power of the other, in which case we have a group of order 5, or they fail to commute with each other and don't make the kind of group we want. But any pair of disjoint 5-cycles will generate a Sylow 5-subgroup.

On the other hand, with the second kind of element, which is already a product of two disjoint 5-cycles, we can take those 5-cycles separately as generators on the unique Sylow 5-subgroup containing the original element.

These statements really need to be established rather than assumed ...

... and that ought to help with the first part of your question.

To count the subgroups note that each "type 2" element belongs to just one subgroup, and there are 16 such elements in each subgroup, so you just need to count properly and divide by 16.

If you are given a 5-cycle, you make a Sylow 5-subgroup by finding a disjoint 5-cycle. Look at the little table of the elements of the Sylow 5-subgroup and you will see that within each of the relevant Sylow subgroups there are four such disjoint cycles to pair with the given cycle. So count the number of possible disjoint cycles and divide by 4.

There are other ways of doing this - and the whole thing could be done more slickly and presented more efficiently - and there is a bit of work still left to do. But I put the hands on stuff in in the hope that it will help you to understand a bit what is going on, so that you can tackle other, similar, problems in future.

Note that the general topic of Sylow subgroups of symmetric groups gets a whole lot more involved - this example should not be taken as typical - it has been carefully chosen to illustrate some things about the Sylow theorems, and seems to assume that you are confident within the symmetric group.

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A sequence of hints. The idea is to count the order of the normalizer of a given Sylow 5-subgroup.

  1. Show that the two 5-cycles $a=(12345)$ and $b=(6789A)$ ($A$ stands for ten here) generate a Sylow 5-subgroup $P$. Check that there are altogether eight 5-cycles in $P$, namely $a^i, b^i$ for $i=1,2,3,4$. The other 16 elements of $P$ (in addition to these and the identity) are then products of two disjoint 5-cycles $a^ib^j$.
  2. Show that $g\in S_{10}$ satisfies both of the equations $gag^{-1}=a$ and $gbg^{-1}=b$ if and only if $g\in P$. To see this you have to recall how conjugation works on a cycle, and when two cycles are the same permutation.
  3. Show that there are altogether 32 ordered pairs of elements $(c_1,c_2)$ such that they are both 5-cycles in $P$ and that $P=\langle c_1,c_2\rangle$.
  4. Let $N=N(P)$ be the normalizer of $P$. Show that $g\in N$ if and only if $(gag^{-1},gbg^{-1})$ is one of the pairs $(c_1,c_2)$ in step 3.
  5. Conclude that $|N|=32\cdot|P|=800.$
  6. Conclude that $$n_5=\frac{10!}{|N|}=\cdots \equiv 1\pmod5.$$

As an independent verification you can also count the number $n_5$ differently. This approach also gives an answer to the other part of the question. Notice that above we got a Sylow 5-subgroup $P$ of $S_{10}$ as a direct product of Sylow 5-subgroups of $Sym(\{1,2,3,4,5\})$ and $Sym(\{6,7,8,9,A\})$. Clearly these two groups are both isomorphic to $S_5$. Because all the Sylow 5-subgroups are conjugates of $P$, they are all of this form for some two subsets of size 5 of the set $\{1,2,3,4,5,6,7,8,9,A\}$. In the same vein all the Sylow 5-subgroups share with $P$ the properties in step 1 above, as those properties are unchanged by conjugation. Another approach is then to:

  1. Recall that $S_5$ has six Sylow 5-subgroups.
  2. Conclude that the number of Sylow 5-subgroups of $S_{10}$ is $$n_5=6^2\cdot M,$$ where $M$ is the number of ways of splitting a set of ten distinguishable elements into a disjoint union of two subsets of five elements.

Have fun!!!

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Does the permutation $gag^{-1}$ is really just "swallowing"? :-) – Asaf Karagila Jan 7 '13 at 22:25
@Asaf: I don't know. May be it can suppress the gag reflex and swallow? – Jyrki Lahtonen Jan 7 '13 at 22:30

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