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?