Which are the nine Sylow 2-subgroups of $S_3 \times S_3$? What is the only Sylow 3-subgroup of $S_3 \times S_3$? And the most important... why?

You can find excellent information about the Sylow theorems in the website of the scholars John Beachy and William Blair. Here is the link:

  • $\begingroup$ What have you tried so far? Can you imagine where 'nine' would come from? What orders do elements in S3 have, and why does that matter? $\endgroup$ Commented Nov 27, 2014 at 1:27
  • $\begingroup$ I think you're a little bit confused yet - you don't know that there are 9 Sylow 2-subgroups as opposed to 1 or 3; the Sylow theorems aren't enough to give you that kind of granularity. It's much easier to work here by first understanding what the Sylow subgroups of $S_3$. The key point here is that $9=3\times 3$; see if you can independently pick subgroups of each S3. $\endgroup$ Commented Nov 27, 2014 at 1:36

2 Answers 2


Quang's method is undoubtedly the easier approach here, although there are alternative ways of attacking this problem.

Sylow $2$-subgroups have order $4$ in $S_3 \times S_3$. However, none of them can be isomorphic to $\mathbb{Z}_4$ because this would imply that there exists an element of order $4$ in $S_3 \times S_3$. Can you see why such an element cannot exist?

Thus, every Sylow $2$-subgroup will be isomorphic to the Klein-four group, $\mathbb{Z}_2 \times \mathbb{Z}_2$. Such a subgroup will be generated by two elements of order $2$ whose product is also of order $2$. An example of such a subgroup is $H = \langle (e, 12);(13,e) \rangle$. How many ways are there of constructing a similar subgroup?

As for Sylow $3$-subgroups of $S_3 \times S_3$, the Sylow theorems tell us that there are either $4$ of them, or only $1$ of them. Furthermore, each Sylow $3$-subgroup will have order $9$.

Now, it is a fact that any group of order $p^2$ is isomorphic to $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_p \times \mathbb{Z}_p$. However, I claim that it's not possible to have an element of order $9$ in $S_3 \times S_3$, so the latter must be the case.

What is special about $\mathbb{Z}_p \times \mathbb{Z}_p$? Every non-identity element has order $3$. How can we use this information to conclude that only a single Sylow $3$-subgroup exists?

Useful facts for considering the above:

  • In any direct product of groups, the order of an element $(g_1, g_2) \in G_1 \times G_2$ is the least common multiple of the orders of $g_1$ and $g_2$ in their respective groups.

  • Any permutation in $S_n$ can be written as a product of disjoint cycles, and the order of that permutation will be the least common multiple of the cycle lengths.


Sylow subgroups of $G\times H$ are also products of Sylow subgroups of $G$ and of $H$. Now what are the Sylow subgroups of $S_3$?

  • $\begingroup$ S3 Sylow 2-subgroups: <(12)>, <(13)>, <(23)>. S3 Sylow 3-subgroups: <(123)>. As Quang Hoang and Stadnicki told us, to find the Sylow p-subgroups of S3 x S3 we need to do the cartesian product of the Sylow p-subgroups of S3. As there are three Sylow 2-subgroups in S3, in S3 x S3, there are nine Sylow 2-subgroups because 3 times 3 is 9. As there is only one Sylow 3-subgroups in S3, in S3 x S3, there is only one Sylow 3-subgroup, for the same reason. $\endgroup$
    – Beginner
    Commented Nov 30, 2014 at 3:43

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