# 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? [closed]

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:

• 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? Commented Nov 27, 2014 at 1:27
• 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. Commented Nov 27, 2014 at 1:36

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 $S_3$?

• 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. Commented Nov 30, 2014 at 3:43