Conjugates of a sylow 2-subgroup of $A_5$ 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?
 A: 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
A: 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 $|G:N_G(P)|=1, 3$.
A: I'm just elaborating Rene Schoof's answer a little here.
$$V\mathrel{\mathop:}=\{1,(12)(34),(13)(24),(14)(23)\}$$
is a Sylow 2-subgroup of $A_5$, and there are four more subgroups of this form, so $n_2\geq5$.
On the other hand, any conjugate of $V$ is clearly one of these five subgroups. Since $\operatorname{Syl}_2(A_5)$ is a conjugacy class, it follows that these five subgroups are the only Sylow 2-subgroups.
