I need some help to conclude a proof or find a more enlightening proof.
Let us first define the orthogonal group wrt a quadratic form $q:\mathbb{R}^n\to \mathbb{R}$, i.e.
$$O(q)=\{h\in Gl(n,\mathbb{R})| q(hx)=q(x), \forall x\in \mathbb{R}\}$$
We can represent $q$ by a symmetric $A\in Mat(n,\mathbb{R})$, explicitely $q(x)=x^TAx$. Now we can equivalently say that $O(q)=\{h\in Gl(n,\mathbb{R})| h^T Ah=A\}$.
In question is now how to prove that $O(q)$ is compact iff $q$ is either positive or negative definite (obviously we can replace $q$ by $A$ in the statement and switch between them).
My attempts so far: If $q$ is indefinite we have some $x\in \mathbb{R} \backslash 0$ satisfying $q(x)=0$. Then define for any $n\in \mathbb{N}$ the matrices $g_n\in GL(n,\mathbb{R})$ to be in first column the vector $nx$ and zero otherwise. Of course now $g_n^TAg_n=0$ and by setting $h_n=id+g_n$ we get $h_n^TAh_n=A$ for the unbounded sequence $h_n\in O(q)$. This shows that $O(q)$ cannot be compact.
For the converse, $O(q)$ is obviously closed by the given equation. For the boundedness notice that the claim is true for the canonical positive and negative definite matrices ($id$ and $-id$). I tried now to conclude by generalizing this to arbitrary definite matrices. We can find by basis transformations (for bilinear forms) a matrix $g_0\in Gl(n,\mathbb{R})$ such that $g_0^TAg_0=id$ (case $-id$ is treated analogeously), i.e. assume $A=g_0^{-T}g_0^{-1}$.
Now the condition we want can be led to the $id$-case by $$h^T Ah=A \Leftrightarrow h^Tg_0^{-T}g_0^{-1}h=g_0^{-T}g_0^{-1} \Leftrightarrow g_0^Th^Tg_0^{-T}g_0^{-1}hg_0=id$$
Now we know that the group $g_0^{-1}O(q)g_0$ is bounded. How can I deduce that the conjugate of a bounded group is bounded? Is this an easy fact?
Also my proof seems rather messy, is there an alternative more beautiful or shorter proof? I also kindly encourage you to mention any corrections and clarifications that wander your minds.