# Distinct $3$-Sylow subgroups of $S_6$ intersect trivially

Let $$S$$ and $$T$$ be distinct $$3$$-Sylow subgroups of the symmetric group $$S_6$$. Prove that $$S$$ and $$T$$ intersect trivially.

Here are my thoughts so far:

I figured the Sylow Theorems could give us some insight here. Let $$G = S_6$$. Then $$G$$ has order $$6! = 720 = 2^4 \cdot 3^2 \cdot 5$$. It follows that any $$3$$-Sylow subgroup of $$G$$ must have order $$9$$, and that, denoting $$n_3$$ by the number of $$3$$-Sylow subgroups of $$G$$, $$n_3 | 80$$ and $$n_3 \equiv 1$$ (mod $$3$$) $$\Rightarrow$$ $$n_3 = 1, 4, 10, 16, 40$$.

I'm not sure how to proceed from here. There's a long list of possibilities of possible orders for the $$5$$-Sylow and $$2$$-Sylow subgroups 0f $$G$$ -- so it doesn't seem like we can get away with a counting argument here, showing that if $$3$$-Sylow subgroups of $$G$$ intersect non-trivially, we end up with more than $$720$$ elements in our group, contradicting the order of $$G$$. How can I reach a contradiction?

I appreciate all the help. Thanks!

• Remark: there are certainly at least $10$ Sylow-$3$ subgroups, namely the ones generated by $\{(1\ 2\ 3),(4\ 5\ 6)\}$ and its conjugates of the same shape. – Greg Martin Jan 12 at 6:32

Sketch proof:

$$S$$ and $$T$$ are abelian (justify how you like).

If $$S\cap T$$ is non-trivial, then for $$1\ne x\in S\cap T$$ you have $$C_G(x)\ge ST$$.

Pick any $$S$$ and $$x$$ you like and show that this cannot hold as $$S\trianglelefteq C_G(x)$$ and is therefore the only Sylow $$3$$ subgroup of $$C_G(x)$$.

• The idea that $S$ and $T$ are abelian, and thus contained in the centalizer of $x$ is IMO the key (+1). It's not entirely clear to me why we should have $S\unlhd C_G(x)$ though. I would have done the endgame using the observation that as $C_G(x)$ has at least two Sylow $3$-subgroups, it must have at least four, and consequently we would have $|C_G(x)|\ge36$ leaving the conjugacy class too small. Looks like I missed something simpler? – Jyrki Lahtonen Jan 12 at 13:44
• Oh, I think I now realize that your idea may also have been to count the size of the conjugacy class by other means :-) – Jyrki Lahtonen Jan 12 at 13:54
• I actually just like to avoid long counting arguments. As $S'=\langle (1,2,3),(4,5,6)\rangle$ is a Sylow 3-subgroup of $G$, there is some $g\in G$ with $S^g=S'$, so $S'\cap T^{g^{-1}}$ is non-trivial and we may as well assume $S=S'$. Now $x$ is either a 3-cycle or a product of two 3-cycles. In either case if $h\le C_G(x)$ we have that $h$ preserves the partition $\{\{1,2,3\},\{4,5,6\}\}$ and therefore normalises $S$. – Robert Chamberlain Jan 12 at 16:05
• That is simpler :-) – Jyrki Lahtonen Jan 12 at 18:25

You can easily calculate that in $$G$$ the number of elements of order $$3$$ is $$80$$.

An example of a $$3$$-Sylow subgroup of $$G$$ is $$P = \langle (123), (456) \rangle$$, so any $$3$$-Sylow contains a total of $$8$$ elements of order $$3$$.

Hence if you can show that the the number of $$3$$-Sylow subgroups is $$10$$, the intersection of any two of them must be trivial: otherwise we wouldn't get enough elements of order $$3$$.

You can calculate that $$N_G(P) = (S_3 \times S_3)\langle \sigma \rangle$$, where $$\sigma = (1,4)(2,5)(3,6)$$ (involution swapping the sets $$\{1,2,3\}$$ and $$\{4,5,6\}$$). So $$N_G(P)$$ has order $$2^3 \cdot 3^2$$, hence $$n_3 = [G : N_G(P)] = 10$$.