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 '19 at 20:32
  • $\begingroup$ $(1,2,3)$ should be $(123)$, and perhaps $H=\{(12)(34),(13)(24),(14)(23)\}$. $\endgroup$ – ashpool Jun 19 at 3:23

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 $H$.

| cite | improve this answer | |
  • $\begingroup$ 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

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.