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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 ?

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Not every pair of order $2$ elements generates a $2$-Sylow subgroup. For example $((1 \ 2), 1)$ and $((2 \ 3), 1)$ generate $S_3 \times 1$.

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  • $\begingroup$ Thanks, have another question - Is the technique I used correct only for abelian groups ? $\endgroup$ – daPollak Jan 6 '15 at 17:19
  • $\begingroup$ I don't think so, I think you just have to modify it. In general you're just counting generating sets in the whole group and generating sets in a sylow subgroup and dividing the get the number of slyow's. This will always work as long as you can do the counts correctly. In your case you just overcounted on the first number. $\endgroup$ – Jim Jan 6 '15 at 18:47
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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.

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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 $S_3$.

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