Order of the stabiliser of a vector

Question

$\overrightarrow{e}_1=(1\space 0 \space 0\space 0\space 0\space 0)^T$, $\overrightarrow{e}_2=(0\space 1 \space 0\space 0\space 0\space 0)^T$,... $\overrightarrow{e}_6=(0\space 0 \space 0\space 0\space 0\space 1)^T$ is the canonical base of the vector space $\mathbb{R}^6$.

Define an action of the symmetric group $S_6$ on $\mathbb{R}^6$ by "permutation of the coordinates" as follows: For a permutation $\sigma$ in $S_6$ and a vector $\lambda_1\overrightarrow{e}_1 + ... \lambda_6\overrightarrow{e}_6$ in $\mathbb{R}^6$ we set $\sigma \cdot (\lambda_1\overrightarrow{e}_1 + ... \lambda_6\overrightarrow{e}_6)= \lambda_1\overrightarrow{e}_{\sigma(1)} + ... \lambda_6\overrightarrow{e}_{\sigma(6)}$

For this action, what is the order of the stabiliser of the vector $(1\space 2\space2\space 3\space 3\space3 )^T$ ?

What I know:

I think the group has order $6!$ and the stabiliser of the vector is a $6 \times 6$ matrix because the stabiliser $\cdot$ the vector has to give the vector. But I don't know how to compute the order of a matrix.

Answer 1

We are looking for the permutations $\sigma$ such that $$\sigma \cdot (1\space 2\space2\space 3\space 3\space3 )^T = (1\space 2\space2\space 3\space 3\space3 )^T$$ that is $$1\overrightarrow{e}_{\sigma(1)} + 2\overrightarrow{e}_{\sigma(2)} +2\overrightarrow{e}_{\sigma(3)} + 3\overrightarrow{e}_{\sigma(4)}+ 3\overrightarrow{e}_{\sigma(5)}+ 3\overrightarrow{e}_{\sigma(6)} = (1\space 2\space2\space 3\space 3\space3 )^T$$ So, we can look at our options now. We must have $\sigma(1) =1$ since $\overrightarrow{e}_{\sigma(1)}$ is the only term with coefficient $1$. Similarly, we can have $\sigma(2) = 2 \text{ or }3$ and $\sigma(3) = 2 \text{ or }3$. And finally, we can have $\sigma(4) = 4 \text{ or } 5 \text{ or } 6$, $\sigma(5) = 4 \text{ or } 5 \text{ or } 6$ and $\sigma(6) = 4 \text{ or } 5 \text{ or } 6$ (Of course since $\sigma$ is a permutation, the choices for these are not independent from each other). Can you find the corresponding stabilizer elements in $S_6$ now?

Also note that stabilizer is not a $6 \times 6$ matrix since the action here is defined as $S_6 \times \mathbb{R}^6 \to \mathbb{R}^6$ so stabilizer of an element in $\mathbb{R}^6$ must be a subgroup of $S_6$.

Answer 2

Don't think of the $S_{6}$ as a group of matrices, think of it as a group of permutations; so for example the permutation $(i,j) \in S_{6}$ acts by sending $e_{i} \mapsto e_{j}$, $e_{j} \mapsto e_{i}$, and $e_{k} \mapsto e_{k}$ for $k \neq i,j$.

Then it is clear that the stabilizer group of the vector $(1,2,2,3,3,3)$ contains the permutation $(2,3)$, and also contains all permutations on $\{4,5,6\}$; this gives a permutation group $S_{\{4,5,6\}} \simeq S_{3}$, which is generated by the permutations $(4,5)$, $(4,5,6)$.

So you need to compute the order of $\langle (2,3), (4,5), (4,5,6) \rangle$. The easiest way to do this is to recognize that it is isomorphic to $\langle (1,2) \rangle \times S_{\{4,5,6\}} \simeq \mathbb{Z}_{2} \times S_{3}$ which has order 12.