How many $2$-Sylow subgroups in $G$ with $|G| = 2^2\cdot 3$? I have a group $G$ with $|G| = 2^2\cdot 3$. I also know it has $4$ Sylow-$3$ subgroups.
I need to show that there is $1$ Sylow-$2$ subgroup. 
(This is all I have left from the full question.)
Any help will be appriciated.
 A: This follows from a straightforward counting argument. Each pair of Sylow-3 subgroups can intersect only trivially (if two of these intersected nontrivially, they would be the same subgroup). This accounts for nine group elements, out of the twelve total. There are only three extant group elements. Thus there can not be more than one Sylow-2 subgroup. 
A: The group $G$ has $2^2\cdot 3=12$ elements. Given two distinct subgroups $H_1,H_2\subset G$ of order $3$, then their only element in common is the identity element of $G$ (because in any group of order $3$, both non-identity elements generate the entire group). Therefore, since you have concluded that there are $4$ distinct Sylow $3$-subgroups of $G$, they together comprise
$$\underbrace{(3-1)}_{\substack{\text{non-identity elements}\\\text{per Sylow 3-subgroup}}}\cdot \underbrace{4}_{\substack{\text{number of}\\\text{Sylow 3-subgroups}}}+\underbrace{1}_{\substack{\text{identity}\\ \text{element}}}=9\text{ elements}$$
Now remember that any given Sylow $2$-subgroup of $G$ has, by definition, $2^2=4$ elements, one of which will be the identity element (which we already counted).
