# What is wrong with that counting of $S_{3}\times S_{3}$ subgroups?

I want to find all 2-sylow subgroups of $S_{3}\times S_{3}$. I know that there are nine such subgroups, but I tried to count them in the following way - I know that every 2-sylow subgroup isomorphic to $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ and there are 15 elements of order 2 in $S_{3}\times S_{3}$. Every subgroup that isomorphic to $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ is generated by 2 different elements of order 2 - and I have $15\cdot\left(15-1\right)$ ways to select 2 generators. but every such subgroup is generated by $3\cdot\left(3-1\right)$ couples of generators, thus I have $\frac{15\cdot\left(15-1\right)}{3\cdot\left(3-1\right)}=35$ subgroups which are isomorphic to $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. What is wrong ?

Not every pair of order $2$ elements generates a $2$-Sylow subgroup. For example $((1 \ 2), 1)$ and $((2 \ 3), 1)$ generate $S_3 \times 1$.
every desired subgroup will have exactly one element of the form $(a,e)$ and one element of the form $(e,b)$. There are three of each kind.
Not every pair of elements of order 2 generates a Sylow subgroup. Indeed, this happens if and only if the pairs are of the form $(x,e),(e,y)$ where $e$ is the identity. If the pairs have the form $(x,e),(y,e)$ or $(e,x),(e,y)$, then the subgroup generated is one of the factors of the direct product, hence has order 6 and is isomorphic to $S_3$.