9
$\begingroup$

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 $\frac{16!}{16} = 15!$

$\endgroup$
  • 5
    $\begingroup$ 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 $p$? $\endgroup$ – Jyrki Lahtonen Mar 19 '17 at 14:43
  • 2
    $\begingroup$ @JyrkiLahtonen Isn't it wilsons theorem in case of odd prime?.. thanks for the insight ... $\endgroup$ – spaceman_spiff Mar 20 '17 at 4:09
  • 2
    $\begingroup$ Correct (at least that's the idea I had). Goes nicely together with Sylow, doesn't it! $\endgroup$ – Jyrki Lahtonen Mar 20 '17 at 5:33
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $(17 − 2)!=15!$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.