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?
Best Answer
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 $$V:=\big\{(),(1\,\,\,2)(3\,\,\,4),(1\,\,\,3)(2\,\,\,4),(1\,\,\,4)(2\,\,\,3)\big\}\,.$$