# Why does $A_5$ have $\binom{5}{4}$ Sylow 2-subgroups?

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?

• 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 $n=1,4,5$. – Servaes Aug 13 '18 at 14:47
• Nice observation. How did you prove the result for $n=4,5$? – applebees Aug 13 '18 at 15:03
• 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 $n!$. – Steve D Aug 13 '18 at 18:00
• @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... – Steve D Aug 14 '18 at 4:35
• 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. – Steve D Aug 14 '18 at 4:38

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 $$V:=\big\{(),(1\,\,\,2)(3\,\,\,4),(1\,\,\,3)(2\,\,\,4),(1\,\,\,4)(2\,\,\,3)\big\}\,.$$
• 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 $A_5$... – applebees Aug 13 '18 at 16:03
• Oh, the problem with $4$-cycles is that they do not belong in $A_5$. Cycles with even length is an odd permutation. – Batominovski Aug 13 '18 at 16:12