"Extension" of orthogonal group I'm looking for a Lie group $G$, subgroup of $GL(n,\Bbb{R})$, and which contains $O(n,\Bbb{R})$ as a subgroup:
$$
O(n,\Bbb{R}) \subseteq G \subseteq GL(n,\Bbb{R}).
$$
Obvious examples: the conformal group, and the special linear group. 
Now, I would like $G$ to have exactly one dimension more than $O(n,\Bbb{R})$. Is the conformal group the only possibility, or are there any others?
Thanks.
(Feel free to edit tags appropriately.)
 A: $n>2$ let $G$ be a subgroup of $GL(n)$ such that $O(n)\subset G$ and $dim(G)=dim(O(n))+1$. We denote by $G_0$ the connected component of $G$. We know that $SO(n)$ is simple, this implies that $SO(n)\subset [G_0,G_0]$. Thus $dim([G_0,G_0]\geq dim(G_0)-1$. We can't have $dim([G_0,G_0])=dim(G_0)$ since this would implies that $G$ is simple a fact which is impossible, (look at the dimension of real simple groups https://en.wikipedia.org/wiki/Simple_Lie_group). We deduce that $SO(n)$ is normal in $G_0$.
Let $A\in G_0$ not in $SO(n)$, we can write $A=UV$ where $U$ is orthogonal and $V$ a positive definite symmetric matrix (the polar decomposition of $A$), since $A$ is in the normalizer of $SO(n)$, so is $V$. We know that a positive definite symmetric matrix is diagonalizable. Suppose that there exists two eigenvectors, $c,d$ of $V$ whose norm is 1 associated to distinct eigenvalues $x$ and $y$. We can define an orthogonal map $B$ such that $B(c)=d$, there exists an orthogonal map $E$ such that $VB=EV$ since $V$ normalizes $SO(n)$. We have $VB(c)=V(d)=yd=EV(c)=E(xc)$. Since $E$ is orthogonal, we deduce that the norm of $xc$ is the norm of $yd$, this is impossible since the fact that the norm of $c$ and $d$ is $1$ implies that $x$ and $y$ have the same absolute value, this last fact implies that $x=y$ since $V$ is positive definite.
We deduce that $G_0$ is the group of conformal transformations.
A: Orthogonal transformations preserve the norm of a vector. You can take the product of that group with the one-dimensional multiplicative group that scales the vectors by nonzero real factors (a copy of $\Bbb{R} \backslash \{0\}$), in other words, the nonzero multiples of the unit matrix.
