Show that a Sylow 2-subgroup of $A_5$ has exactly 5 conjugates

Question

The exact problem is as follows:

Show that a Sylow 2-subgroup of $A_5$ has exactly 5 conjugates.

My question is why are there not 15 conjugates. Seeing as every Sylow 2-subgroup of $A_5$ is the klein 4-subgroup, 15 of those subgroups, and every Sylow p-subgroup is conjugate to one another, why not conclude that there are 15 conjugates?

Answer 1

Hint: put $A_5=G$, if the number of conjugates of a Sylow $2$-subgroup $P$ would be $15$, then $|G:N_G(P)|=15$. But since $P$ is abelian, this gives $P \subseteq C_G(P) \subseteq N_G(P)$, and by comparing indices, we get $C_G(P)=N_G(P)$. Hence $P$ satisfies the conditions of Burnside's Normal Complement Theorem, and this gives the existence of a normal subgroup $N$ of $G$, such that $G=NP$ and $N \cap P=1$, which is absurd since $G$ is simple. I assume you have dismissed also the other cases where $|G:N_G(P)|=1, 3$.

Answer 2

Alternatively, observe that the only elements of $2$-power order in $A_5$ are permutations of the form $(a\,b)(c\,d)$. For each $x\in\{1,2,3,4,5\}$, the permutations of this type that fix $x$, fix no other element and form a subgroup isomorphic to Klein’s $4$-group. Since the order of $A_5$ is $60$, this must be a $2$-Sylow subgroup. All the other Sylow subgroups are conjugate and therefore fix a unique element of $\{1,2,3,4,5\}$. There are five elements and hence five Sylow subgroups