6
$\begingroup$

Let $Syl_p(G)$ be the number of Sylow $p$-subgroups in a group $G$. Why does $|Syl_2(A_5)|=\binom{5}{4}$? Is this true in general, i.e., does $|Syl_2(A_n)|=\binom{n}{2^\alpha}$ where $2^\alpha$ is the maximal power of $2$ dividing $n!/2$?

Looking at the case of $A_5$, I know that Sylow subgroups are all conjugate. Taking $P=\langle (1234) \rangle$, it's easy to see that $P$ is a Sylow 2-subgroup, so $|Syl_2(A_5)|$ should be the size of the orbit of $P$ under the action of $A_5$ on its subgroups given by conjugation. Why does this orbit have size $\binom{5}{4}$? What can be said in general?

$\endgroup$
  • 1
    $\begingroup$ For $n>5$ the maximal power of $2$ dividing $\frac{n!}{2}$ is greater than $n$, so it also fails. It is true precisely for $n=1,4,5$. $\endgroup$ – Servaes Aug 13 '18 at 14:47
  • 1
    $\begingroup$ Nice observation. How did you prove the result for $n=4,5$? $\endgroup$ – applebees Aug 13 '18 at 15:03
  • 1
    $\begingroup$ In fact, for $n>5$, the number of Sylow 2-subgroups of $A_n$ is the same as the number for $S_n$: $n!/2^\alpha$. Here $2^\alpha$ is the maximal power of $2$ dividing $n!$. $\endgroup$ – Steve D Aug 13 '18 at 18:00
  • 1
    $\begingroup$ @Batominovski: see my proof here for the fact that Sylow 2-subgroups of $S_n$ are self-normalizing. So to finish my claim in the above comment, we need to show that when $n>5$, every Sylow 2-subgroup of $A_n$ is contained in exactly one Sylow 2-subgroup of $S_n$. Assume not, and let $S$ be a Sylow 2-subgroup of $A_n$ contained in $P$ and $Q$ in $S_n$. WE can conjugate to assume $(12)\in P$ and $(13)\in Q$. Now $P$ contains a transposition commuting with $(12)$: if it doesn't contain $3$, we're good... $\endgroup$ – Steve D Aug 14 '18 at 4:35
  • 1
    $\begingroup$ If it does -- say $(3a)$ -- then we use that $n>5$ to find another transposition $(bc)$ not using any of the already-seen numbers, that also commutes with $(12)$ and $(3a)$ and is thus contained in $P$. So no matter what there's a transposition in $P$ not moving $1$, $2$, or $3$. We can conjugate that to $(45)$. To wrap up, after several possible conjugations, $(12)$ and $(45)$ are in $P$ and $(13)$ is in $Q$. $S$ has index $2$ in $P$, so $(12)(45)\in S$; $S$ is normal in $Q$, so $[(13), (12)(45)]=(123)\in S$. This is absurd, and the final contradiction. $\endgroup$ – Steve D Aug 14 '18 at 4:38
5
$\begingroup$

Note that each $2$-Sylow subgroup of $A_5$ is of order $4$. We know a subgroup of order $4$ of $A_5$: $$V_1:=\big\{(),(2\,\,\,3)(4\,\,\,5),(2\,\,\,4)(3\,\,\,5),(2\,\,\,5)(3\,\,\,4)\big\}\,.$$ Thus, we have four more conjugates of $V_1$: $$V_2:=\big\{(),(1\,\,\,3)(4\,\,\,5),(1\,\,\,4)(3\,\,\,5),(1\,\,\,5)(3\,\,\,5)\big\}\,,$$ $$V_3:=\big\{(),(1\,\,\,2)(4\,\,\,5),(1\,\,\,4)(2\,\,\,5),(1\,\,\,5)(2\,\,\,4)\big\}\,,$$ $$V_4:=\big\{(),(1\,\,\,2)(3\,\,\,5),(1\,\,\,3)(2\,\,\,5),(1\,\,\,5)(2\,\,\,3)\big\}\,,$$ and $$V_5:=\big\{(),(1\,\,\,2)(3\,\,\,4),(1\,\,\,3)(2\,\,\,4),(1\,\,\,4)(2\,\,\,3)\big\}\,.$$ There cannot be more $2$-Sylow subgroups of $A_5$. You can easily check that these are the only conjugates of $V_1$. Thus, $A_5$ has five $2$-Sylow subgroups.


As for $A_4$, it has a normal subgroup of order $4$. Since $4$ is the largest power of $2$ that divides $|A_4|=\dfrac{4!}{2}=12$, we conclude that $A_4$ has only one $2$-Sylow subgroup $$V:=\big\{(),(1\,\,\,2)(3\,\,\,4),(1\,\,\,3)(2\,\,\,4),(1\,\,\,4)(2\,\,\,3)\big\}\,.$$

$\endgroup$
  • $\begingroup$ Ah, thanks. My mistake here was looking at the group generated by a 4-cycle, which although it has order 4, is not a subgroup of $A_5$... $\endgroup$ – applebees Aug 13 '18 at 16:03
  • 3
    $\begingroup$ Oh, the problem with $4$-cycles is that they do not belong in $A_5$. Cycles with even length is an odd permutation. $\endgroup$ – Batominovski Aug 13 '18 at 16:12

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.