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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!

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  • $\begingroup$ 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. $\endgroup$ – Greg Martin Jan 12 at 6:32
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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 $C_G(x)$.

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  • $\begingroup$ 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? $\endgroup$ – Jyrki Lahtonen Jan 12 at 13:44
  • $\begingroup$ Oh, I think I now realize that your idea may also have been to count the size of the conjugacy class by other means :-) $\endgroup$ – Jyrki Lahtonen Jan 12 at 13:54
  • $\begingroup$ 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 $S$. $\endgroup$ – Robert Chamberlain Jan 12 at 16:05
  • $\begingroup$ That is simpler :-) $\endgroup$ – Jyrki Lahtonen Jan 12 at 18:25
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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 $n_3 = [G : N_G(P)] = 10$.

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