Consider the action of the permutation group $S_3$ on the vector space $M =\lbrace (\lambda_1, \lambda_2, \lambda_3)\in \mathbb{R}^3, \lambda_1+\lambda_2+\lambda_3=0\rbrace$.
The problem I have is the following: represent the action of $S_3$ on $\lambda = (1,2,-3) \in M $ on the plane $\mathbb{R}^2$ (wich we identifiy with M).
The hints I've been given is first to choose an orthonormal basis of M and then to draw the two axis of symmetry wich represent the two generators of $S_3$ namely the permutations $(1,2)$ and $(2,3)$.
My questions are:
why do we have to choose an orthonormal basis of M, is it not enought to work with an arbitrary basis ?
what are the equations of the two axis of symmetry in $\mathbb{R}^2$ that we use to illustrate the image of $\lambda $ under the action of $S_3$ ?
I guess in general you don't have to choose an orthonormal basis. You could still choose any basis and identify $M$ with $\Bbb{R}^2$ in such a way. Maybe though, since all permutation matrices have determinant $\pm 1$, and are thus orthogonal, it is nice to choose an orthonormal basis so that they remain orthogonal transformations in the new basis.
This will depend on the basis you choose for $M$. In $\Bbb{R}^3$, the action of $(1,2)$ is the reflection over the plane $y=x$. This plane intersects $M$ in the line spanned by the vector $(1,1,-2)$. So the action of $(1,2)$ on $M$ is the reflection over this line. After choosing a basis, you can try to express this line in those coordinates.
Similarly, the permutation $(2,3)$ acts as reflection over the plane $y=z$, which intersects $M$ in the line spanned by $(-2,1,1)$. Again, you'll want to find an equation for this line in the coordinates corresponding to the basis you choose for $M$.