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This is my original problem:

Find the number of elements of order $7$ in $S_7$. Find the number of Sylow $7$-subgroups of $S_7$.

Since $7$ is prime, we know that there must be a $7$-cycle in $S_7$ (since no $P_1 \cdot P_2 = 7$).

My next thought is to calculate the total number of combinations a number can have in $(A B C D E F G)$, which is simply: $7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 7!$.

But then you have to consider the fact that $(1 2 3 4 5 6 7) = (2 3 4 5 6 7 1)$, so divide by $7$, which gives $6!$.

So there should be $6!$ elements of order $7$.

(**) Now because $7$ is a unique prime when considering the order of $S_7$, I believe that means that each element should exist in only one subgroup.

That means there is a way to partition this into a set of $6$ elements, then consider the union of each set with the identity $\{ e \}$, and you get all the subgroups of order $7$.

So calculating the number of such subgroups may be done by: $\dfrac {6!} 6 \cdot 7 = 7 \cdot 5! = 840$.

If this method does indeed work, it should follow that this method would work for solving similar questions whenever one solves for elements and subgroups of order $p$ in the group $S_p$ (just use arbitrary constants in place of numbers).

In fact, it should give you a formula. The number of elements is $(p-1)!$ and the number of subgroups of that order are $\dfrac {p!} {p-1}$.

Expanding this further, I also think this method should work for $|S_n| = p^m$, where $p∤m$, though that is not my concern here.

Am I making a mistake or is this correct? I believe there is a theorem for my statement at (**), but I do not know what it is called.

Alex M.
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  • I think the theorem is probably one of the Sylow's theorems, I am not sure though. I am struggling to learn these concepts. –  Dec 03 '15 at 18:36
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    Every subgroup of order $7$ contains exactly $6$ elements of order $7$ which are not in any other subgroup of order $7$. Therefore, if there are $6!$ elements of order $7$, then there are $6! / 6 = 5!$ subgroups of order $7$. –  Dec 03 '15 at 18:49
  • Oh, yes, that is so obvious. I don't know how I mess that up. You should write an answer fixing my mistake! –  Dec 03 '15 at 19:05
  • Wow, I just realized I thought the number of groups was higher than the number of elements of that order! I don't know how I missed that mistake. Thank you, @Bungo –  Dec 03 '15 at 19:15
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    Ha, I hadn't noticed that. Nice sanity check :-) –  Dec 03 '15 at 19:20

1 Answers1

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Every subgroup of order $7$ contains exactly $6$ elements of order $7$ which are not in any other subgroup of order $7$. Therefore, if there are $6!$ elements of order $7$, then there are $6!/6 = 5! = 120$ subgroups of order $7$.

As you indicated, we can generalize this to $S_p$ for any prime number $p$. There are $p! / p = (p-1)!$ elements of order $p$. Each subgroup of order $p$ contains exactly $p-1$ of these elements, and these subgroups intersect trivially, so there are exactly $(p-1)! / (p-1) = (p-2)!$ subgroups of order $p$.

You might be able to relate this to Sylow theory, but the Sylow theorems are quite general, not specific to the symmetric groups, so they don't really give us enough information to get an exact count. If $n_p$ is the number of subgroups of order $p$, the Sylow theorems tell us that $n_p$ divides $(p-1)!$ and that $n_p \equiv 1$ (mod $p$), but that still leaves a lot of possibilities.

For example, if $p = 5$, the Sylow theorems tell us that $n_5$ divides $4! = 24$ and $n_5 \equiv 1$ (mod $5$). These two facts imply that $n_5$ is either $1$ or $6$. Of course the latter is equal to $(p-2)! = (5-2)! = 3!$, but without knowing that we are working with the symmetric group, we could not rule out $n_5 = 1$.

The situation gets worse as $p$ increases. For example, if $p = 7$, then Sylow tells us that $n_7$ divides $6! = 720$ and $n_7 \equiv 1$ (mod $7$). This allows many possibilities: $n_7$ can be $1$, $8$, $15$, $36$, $120$, and maybe some others I was too lazy to check. The correct answer, $5! = 120$, appears in this list, but so do many other numbers.

  • Awesome answer. +1 Could you please explain how the following is true?
    Each subgroup of order $p$ contains exactly $p−1$ of these elements     – Panglossian Oporopolist Jul 02 '18 at 04:23
  • @Kugelblitz If $H$ and $K$ are two subgroups of order $p$, where $p$ is prime, then $H \cap K$ has order dividing $p$, so is either $1$ or $p$. In the latter case, we have $H = K$, so if $H$ and $K$ are distinct, we must have $H \cap K = 1$. Thus $H$ and $K$ don't have any elements in common except the identity. Every element of order $p$ generates a subgroup of order $p$, hence every such element must be in exactly one of the subgroups of this order. –  Jul 02 '18 at 21:58
  • Thanks!! @bungo – Panglossian Oporopolist Jul 03 '18 at 02:49