Calculate the number of Sylow p-subgroups of $A_5$

We have $|G|=60=2^2⋅3⋅5$

Let $n_p$ be the number of Sylow p-subgroups of G.

By Sylow's third theorem, we have $n_3∈{1,4,10}$. But G contains 20 elements of order 3, each of which generates a Sylow 3-subgroup.

I have just been looking at this online and I understand why $n_3$ has to be 1, 4 or 10, however why does G contain 20 elements of order 3?

  • 1
    $\begingroup$ I think you've to just count the three cycles. $\endgroup$ – snulty Apr 19 '17 at 11:15
  • $\begingroup$ How? Because it also says that G contains 30 elements of order 5 and I just don't understand how they've got that $\endgroup$ – Alex Carter Apr 19 '17 at 11:18
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    $\begingroup$ From a naive perspective, you can eliminate $n_3 = 1$ right away since the existence of only one $p$-subgroup implies that subgroup is normal. But $A_5$ is simple. The right way to look at this though is by counting the number of elements of size $p$. If you think about decomposing each element of $A_5$ into products of disjoint cycles, what is the order of a decomposed element if you know the order of the disjoint cycle lengths? You'll find that the elements of order $5$ are the $5$-cycles, and likewise for $3$. $\endgroup$ – Kaj Hansen Apr 19 '17 at 11:19
  • $\begingroup$ A three cycle is of the form $(a\, b\, c)$ with $a,b,c\in\{1,2,3,4,5\}$ all distinct. There are $5\times 4\times 3$ ways of picking these numbers, but that overcounts because cyclic permutations are the same, so divide by $3$. This gives $5\times 4=20$. $\endgroup$ – snulty Apr 19 '17 at 11:21
  • $\begingroup$ How come we are dividing by $3$ though? Because if we are dividing it by it because there are $3$ elements in the cycle $(a b c)$ then surely it would mean there are $24$ elements of order $5$ $\endgroup$ – Alex Carter Apr 19 '17 at 11:25

Sylow's third theorem says that if $|G|=p^na$ where $p$ and $a$ are coprime and $p$ is prime, then the number of Sylow $p$-subgroups (those are subgroups of order $p^n$) denoted by $n_p$ satisfy $n_p\equiv1(\textrm{mod} \hspace{3pt}p)$ and $n_p$ divides $a$. So in your case, $n_3$ divides $20$ and $n_3\equiv 1(\textrm{mod} \hspace{3pt}3)$

We have $20$ elements of order $3$ and not $20$ subgroups of order $3$. Since each Sylow 3-subgroup will have two non-identity elements we have $\displaystyle\frac{20}{2}=10$ Sylow-3 subgroups in $A_5$.


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