Embedding of symmetric groups into orthogonal groups Let $\Sigma_n$ be the symmetric group of order $n$. Let $O(N)$ be the orthogonal group acting on $\mathbb{R}^N$. 
Question: Given a fixed $N$, how to find the maximal $n$ such that $\Sigma_n$ can be embedded as a subgroup of $O(N)$? I have tried examples and guess the answer is $n=N+1$. But I cannot prove...
 A: Consider the $N$ subspace $x_1 + \dots + x_{N+1} = 0$ which is orthogonal to the vector $\vec{1} =\underbrace{(1,\dots, 1)}_{N+1} \in \mathbb{R}^{N+1} $
Then $S_{N+1}$ acts on this subspace and embeds into $O(N+1)$ and yet $\vec{1}$ is preserved, so this action embeds into $O(N)$.  This shows $n \geq N+1$.  This is the same as @hardmath's answer.

Showing $n \leq N+1$ is less easy and I am not sure how to do it.  Let $v \in \mathbb{R}^N$.  Then $S_{N+2}$ permutes the $(N+2)!$ vectors 
$\{ gv: g \in S_{N+2}\}\subset \mathbb{R}^N$.  This doesn't seem quite right...
The $(N+2)$-cycle $a =(1,2,\dots, N+1, N+2)$ can act on the vector over and over giving a kind of chain:
$$ v_1 \to v_2 \to \dots \to v_{N+2} \to v_1$$
Since $a^{N+2}=1$, this symmetry rotates the $(N+2)$-gon generated by these vectors, and the panel it lives in.

This still seems perfectly fine to me... yet it can't be the case.  E.g. Why can't we draw a tetrahedron on a sheet of paper so that it preserves all the symmetry it has in 3-dimensions?  So... how come $S_4$ can't embed into $O(2)$, the rotation group in the plane $\mathbb{R}^2$? I think $S_4$ has a 4-cycle - preserving a square embedded in the unit circle.  It also has a 3-cycle - preserving an equilateral triangle in the same circle.  The only symmetry to preserve both shapes.
Also, it's instructive to visualize the order-4 symmetry of the tetrahedron corresponding to the action of the 4-cycle (1,2,3,4).
How to show $S_5$ can't be embedded into rotations in $SO(3)$ preserving the unit sphere $S^2$?  In fact, the symmetries of the icosahedron - isomorphic to the alternating group $A_5$ -  comes close.
