# How many subgroups of order 17 does $S_{17}$ have?

How many subgroups of order 17 does $S_{17}$ have ?

My attempt :

An order 17 group is of prime order, hence cyclic and each element in it is a generator and of order 17.
In $S_{17}$ group we can get an order 17 element only through a 17-cycle.

Number of elements of order 17 in $S_{17}$ is $\frac{17!}{17} = 16!$.

Now given that two sylow 17 subgroups have only a trivial intersection. We can conclude that 16 of these elements fall into each sylow 17 subgroup.

Hence the number of sylow 17 subgroups would be $\frac{16!}{16} = 15!$

• Looks ok to me. But surely you meant to write $17!/17=16!$ Bonus question: Can you think of another (=non-Sylow) way of proving that $(p-2)!\equiv1\pmod p$ for an odd prime $p$? – Jyrki Lahtonen Mar 19 '17 at 14:43
• @JyrkiLahtonen Isn't it wilsons theorem in case of odd prime?.. thanks for the insight ... – spaceman_spiff Mar 20 '17 at 4:09
• Correct (at least that's the idea I had). Goes nicely together with Sylow, doesn't it! – Jyrki Lahtonen Mar 20 '17 at 5:33

## 2 Answers

Your answer looks correct.

It is true that there are $16!$ elements of order $17$, but if this is a homework question a marker might want you to elaborate on why that is.

An alternative (but not necessarily better) proof is as follows:

Consider the set of subgroups of $S_{17}$ of order $17$. As you noted, these subgroups must be cyclic. In particular they must be transitive.

For each subgroup $G$ fix $g\in G$ with $g(1)=2$. $g$ is the only such element of $G$ and generates $G$ so uniquely defines $G$.

Write $g=(1,2,x_3,\ldots,x_{17})$. There are $15!$ choices for the $x_i$ so $15!$ such subgroups.

Prove that the number of $$p-$$Sylow subgroups in the symmetric group $$S_p$$ is $$(p − 2)!$$.

Proof : Any $$p-$$Sylow subgroup is cyclic of order $$p$$ and has precisely $$p − 1$$generators. Moreover, if two $$p-$$Sylow subgroups share a generator, they are identical. So, the elements of order p are partitioned according to which p-Sylow subgroup they belong to. We need to count the number of elements of order exactly p. This is precisely the number of distinct $$p-$$cycles, which is $$p!/p = (p−1)!$$. Grouping them into distinct $$p-$$Sylow subgroups (with $$p−1$$ in each clump), we see that the number of $$p-$$Sylow subgroups is $$(p − 1)!/(p − 1) = (p − 2)!.$$

Now take take $$p= 17$$ ,then

Number of $$p-$$Sylow subgroups in the symmetric group $$S_{17}$$ is $$(17 − 2)!=15!$$.