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

We have $|G|=60=2^2\cdot 3\cdot 5$

Let $n_p$ be the number of Sylow $p$-subgroups of $G$. By Sylow's third theorem, we have $n_3\in\{1,4,10\}$. But $G$ contains $20$ elements of order $3$, each of which generates a Sylow $3$-subgroup, so $n_3=10$

Similarly $n_5\in\{1,6\}$ and contains $30$ elements each generating a Sylow $5$-subgroup so $n_5=6$

Consider $n_2$. By Sylow's third theorem, we have $n_2=\{1,3,5,15\}$

(1) Clearly, $n_2\in\{5,15\}$ since $G$ contains $15$ elements of order $2$

(2) If $n_2=15$ then $|G:N_G(H)|=15$ thus $H=N_G(H)$

(3) but this is false since $(1,2,3)\in N_G(H)\backslash H$

Therefore $n_2=5$

Can someone clarrify any of the three lines in bold (lines (1),(2),(3))

Line (1), why is $3$ excluded from the list for $n_2$

Line (2), $N_G(H)$ is the normalizer of $H$ in $G$

Please comment if you can clarify any of the lines, any help would be greatly appreciated.

  • $\begingroup$ I'm pretty sure there are only 24 elements of order 5 in $A_5$. $\endgroup$ – Junglemath Apr 19 at 20:32

A Sylow 2-subgroup contains at most 3 elements of order 2 because it is of order 4. If there were only 3 such subgroups, there would be at most 9 elements of order 2. Since there are 15, 3 is excluded.

$N_G(H)$ is the normalizer of $H$ in $G$, meaning the set of all elements $g$ such that $gHg^{-1}=H$. $G$ acts on the set of 2 Sylow subgroups transitively by conjugation, hence if there are 15 then the stabilizer of $H$, which is $N_G(H)$, must have order 4 (hence index 15), forcing it to be equal to $H$. The author shows that this cannot be by producing an element that normalizes $H$ that is not in $H$.

  • $\begingroup$ thanks for the answer, but why is a Sylow $2$ subgroup of order $4$ $\endgroup$ – Sam Houston Jan 20 '15 at 6:57
  • 4
    $\begingroup$ @Dansmith 4 is the largest power of 2 dividing 60. $\endgroup$ – Matt Samuel Jan 20 '15 at 6:58

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