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Question is :

For $p=2,3$ and $5$ find $n_p(A_5)$ and $n_p(S_5)$. [Note that $A_4\leq A_5$]

What i have done so far is :

for $A_5$ we have $|A_5|=5.4.3$ possible number of sylow subgroups are

  • $n_2(A_5)=(1+2k)\mid 15\Rightarrow n_2(A_5)=1\text { or }3\text { or }5\text { or }15$
  • $n_3(A_5)=(1+3k)\mid 20\Rightarrow n_3(A_5)=1\text { or }4\text { or }10$
  • $n_5(A_5)=(1+5k)\mid 12\Rightarrow n_5(A_5)=1\text { or }6$

Now, by thanking the result $A_5$ is simple I would conclude that possibilities are :

  • $n_2(A_5)=(1+2k)\mid 15\Rightarrow n_2(A_5)=3\text { or }5\text { or }15$
  • $n_3(A_5)=(1+3k)\mid 20\Rightarrow n_3(A_5)=4\text { or }10$
  • $n_5(A_5)=(1+5k)\mid 12\Rightarrow n_5(A_5)=6$

With this it is concluded that there are exactly $6$ sylow $5$ subgroups.

Number of elements of order $3$ in $S_5$ (they are actually in $A_5$) are $\frac{5.4.3}{3}=20$

Now, each element of order $3$ has to be in some sylow $3$ subgroup.As there are $20$ non identity elements of order $3$ and each sylow subgroup affords only $2$ non identity elements there should be $10$ sylow $3$ subgroups ($10\times 2=20$)

So, now we have $n_5(A_5)=6,n_3(A_5)=10$

Suppose $n_2(A_5)=3$ we would have (assuming no two sylow subgroups intersect non trivially) $3\times 3=9$ non identity elements and adding up with above collection of non identity elements we would get $(3\times 3=9)+(10\times 2=20)+(6\times 4=24)=53$ which is a contradiction as there are $59$ non identity elements in $A_5$

So, Possibilities are $n_2(A_5)=5\text { or }15$ and there are two chances :

  • there are $5$ sylow $2$ subgroups with no two sylow subgroup intersecting non trivially adds upto $5\times 3=15$ non identity elements and with above calculation we have $(5\times 3=15)+(10\times 2=20)+(6\times 4=24)=59$ so we do not have any problem with this.
  • there are $15$ sylow $2$ subgroups such that for any two distinct sylow $2$ subgroups we have $|P_2\cap P_2'|=2$ then each sylow $2$ subgroup contribute only one non identity element which is not there in any other sylow $2$ subgroup adding upto $15\times 1=15$ non identity elements. In this case also we are not getting any problem as all non identity elements adds up to $59$ with $(15\times 1=15)+(10\times 2=20)+(6\times 4=24)=59$

But then I have seen that all sylow $2$ subgroups are of order $4$ and are of the form $\mathbb{Z_2}\times \mathbb{Z_2}$ and then $$\{e,(1 ~2)(3~ 4),(1~ 3)(2~ 4),(1~4)(2~3)\}\cap\{e,(1~2)(3~5),(1~5)(2~3),(1~3)(2~5)\}=\{e\}$$

So, I have two sylow $2$ subgroups that intersect trivially (I have actually listed out the other sylow subgroups and they also intersect trivially) which contradicts above possibility of $|P_2\cap P_2'|=2$ (I have also checked it and $2,2$ cycles got exhausted after writing $5$ sylow $2$ subgroups)

Thus i strongly believe there are $5$ sylow $2$ subgroups which are intersecting trivially... But then how do i write this with less labour work i mean I do not like the idea of writing down all sylow $2$ subgroups until the elements got exhausted so i would like to know if there is any better idea to say that $n_2(A_5)=5$.

Thus, we have :

  • $n_2(A_5)=5$
  • $n_3(A_5)=10$
  • $n_5(A_5)=6$

For $S_5$ we have $|S_5|=120=5.4.3.2=5.3.2^3$. Possible no.of sylow subgroups are

  • $n_2(S_5)=(1+2k)\mid 30 \Rightarrow n_2(A_5)=1\text { or }3\text { or }5\text { or }15$
  • $n_3(S_5)=(1+3k)\mid 40\Rightarrow n_3(A_5)=1\text { or }4\text { or }10\text { or }40$
  • $n_5(S_5)=(1+5k)\mid 24\Rightarrow n_5(A_5)=1\text { or }6$

As $S_5$ is not simple, I can not say for the same reason as above i would exclude the possibility of $n_p=1$ but then, I know that $H\leq G\Rightarrow n_p(H)\leq n_p(G)$. for this reason we have :

  • $n_2(S_5)=5\text { or }15$
  • $n_3(S_5)=10\text { or }40$
  • $n_5(S_5)=6$

So, It is confirmed thar number of sylow $5$ subgroups in $S_5$ are $6$.

any element in $3$ sylow subgroup is a $3$ cycle so it has to be in $A_5$ so there is no possibility of having another sylow $3$ subgroup outside $A_5$ thus $n_3(S_5)=10$ for this reason we have :

  • $n_2(S_5)=5\text { or }15$
  • $n_3(S_5)=10$
  • $n_5(S_5)=6$

We would count the number of non identity elements in more worst cases :

  • Suppose $n_2(S_5)=5$ with trivial intersection between any two sylow $2$ subgroups gives $5\times 7=35$ non identity elements adding up non identity elements of sylow $3$ subgroups and sylow $5$ subgroups adds upto $(5\times 7=35)+(10\times 2=20)+(6\times 4=24)=79$ but then there are $119$ non identity elements in $S_5$.

So, I should get (for no good reason) that there are $15$ sylow $2$ subgroups of $S_5$. Conclusion is that :

  • $n_2(A_5)=5;n_2(S_5)=15$
  • $n_3(A_5)=n_3(S_5)=10$
  • $n_5(A_5)=n_5(S_5)=6$

I would be thankful if some one verify this calculation and i would be much more thankful if someone suggest me a better/well written/less labored approach.

Thank you :)

P.S: I do not even understand the special mention he have in the question saying : $\textbf{[Note that $A_4\leq A_5$]}$. please help me to know if i am missing anything.

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  • $\begingroup$ Since Gerry checked $S_5$, i say $A_5$ looks good too. $\endgroup$ – quis23 Feb 8 '14 at 7:46
  • $\begingroup$ @quis23 : Thank you :) $\endgroup$ – user87543 Feb 8 '14 at 10:58
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I think the $S_5$ work is correct (sorry, haven't looked at the $A_5$ work). I would do it a somewhat different way:

Since 5, but not 25, divides 120 (the size of $S_5$), the Sylow-5 subgroups of $S_5$ must be cyclic of order 5. There are 24 5-cycles in $S_5$, 4 of them in each of these subgroups, so, 6 Sylow-5 subgroups.

Similarly, the Sylow-3 subgroups must be cyclic of order 3. There are 20 3-cycles in $S_5$, 2 to a subgroup, so 10 Sylow-3 subgroups.

Since 8, but not 16, divides 120, the Sylow-2 subgroups must have order 8. Now, $S_4$ contains three copies of the dihedral group of order 8, and $S_5$ contains 5 copies of $S_4$, so I get 15 Sylow-2 subgroups.

Something like this ought to work for $A_5$.

EDIT: I think OP wants me to elaborate on the dihedral-group part of the argument.

Take a square, label its vertices, cyclically, with 1, 2, 3, 4. Then the element $(1234)$ of $S_4$ has a natural interpretation as the rotation, one-fourth of the way around, of the square, and the element $(13)$ is the flip in the diagonal through 2 and 4, so these two elements of $S_4$ generate a subgroup isomorphic to the dihedral group of order 8.

The same is true for the elements $(1342)$ and $(14)$, and also for the elements $(1423)$ and $(12)$, and those are the three copies of the dihedral group in $S_4$.

Now if you pick any one of the numbers 1, 2, 3, 4, 5, and consider all the elements of $S_5$ that fix that number, you get a subgroup of $S_5$ isomorphic to $S_4$. Those are your five copies of $S_4$ in $S_5$.

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  • $\begingroup$ I do not understand your idea when you say $5$ but not $25$ divide $120$ and $8$ but not $16$ divide $120$.. could you please explain your idea $\endgroup$ – user87543 Feb 8 '14 at 7:44
  • $\begingroup$ If $p$ is a prime dividing the order $n$ of a group $G$, suppose $p^r$ divides $n$ but $p^{r+1}$ doesn't. Then the Sylow-p subgroups of $G$ have order $p^r$. $\endgroup$ – Gerry Myerson Feb 8 '14 at 9:43
  • $\begingroup$ Yes I know that.... I was expecting that you want to say something more... I do not understand "Now, $S_4$ contains three copies of the dihedral group of order $8$, and $S_5$ contains $5$ copies of $S_4$, so I get $15$ Sylow-$2$ subgroups. .. could you please extend that a bit more.. Thank you :) $\endgroup$ – user87543 Feb 8 '14 at 9:59
  • $\begingroup$ Sorry, I don't know what more to say. What can I say about 5 dividing 120, and 25 not dividing 120? $\endgroup$ – Gerry Myerson Feb 8 '14 at 10:01
  • $\begingroup$ What i meant is i know what is sylow $p$ subgroup but when you specially mentioned it i got confused and i was thinking that there may be something more... $\endgroup$ – user87543 Feb 8 '14 at 10:03

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