The (real) general linear group is defined $GL(n)=\{A \in \mathbb{R}^{n\times n} \mid \operatorname{det}(A) \neq 0\}$. It is a matrix Lie group. Let $J$ be a constant $n$-by-$n$ real matrix. The so-called $J$-orthogonal group is defined $O_J(n)=\{A \in GL(n) \mid A^T J A = J\}$. I'm pretty sure $O_J(n)$ is a matrix Lie group regardless of what $J$ is. The corresponding Lie algebra is $o_J(n) = \{B \in \mathbb{R}^n \mid B^T J + JB = 0\}.$ For $B \in o_J(n)$, the book I am reading defines the Cayley transform of $B$ as
$$A = (I-\alpha B)^{-1}(I+\alpha B). $$
I'm assuming $\alpha$ is some real constant. The book claims that $A$ is $J$-orthogonal, that is, $A^T J A = J$. In the book, it is unclear whether they claim this for $n=4$ and $J = \operatorname{diag}(1, -1, -1, -1)$, or for all real square matrices $J$. I think I have verified that $A$ is $J$-orthogonal if $J$ is symmetric and $J^2 = I$ (which includes $J = \operatorname{diag}(1, -1, -1, -1)$. I guess I have two questions: first, is this correct? And second, are weaker assumptions on $J$ possible?
EDIT: The two factors defining $A$ above commute. If you use this fact and mimic the proof in the answer below, you can show that no assumptions on $J$ are necessary: $J$ can be any $n$-by-$n$ real matrix.
To get some idea how this works, let us first assume only that $J$ is involutory, i.e., $J^2 = I$. Then $A$ is $J$-orthogonal if and only if $AJA^T = J$. We check that condition:
$$(I - \alpha B)^{-1} (I + \alpha B) J (I + \alpha B)^T (I - \alpha B)^{-T} = J$$
and see that it is equivalent to
$$(I + \alpha B) J (I + \alpha B)^T = (I - \alpha B) J (I - \alpha B)^T.$$
This is equivalent to
$$J + \alpha (BJ + JB^T) + \alpha^2 B J B^T = J - \alpha (BJ + JB^T) + \alpha^2 B J B^T,$$
which is equivalent to
$$\alpha (BJ + JB^T) = 0.$$
Assuming that $\alpha \ne 0$, this is equivalent to
\begin{equation} BJ + JB^T = 0.\tag{*} \end{equation}
If we premultiply and postmultiply $(*)$ with $J$, and keeping in mind that $J^2 = I$, the above is equivalent to
$$0 = J(BJ + JB^T)J = JB + B^TJ,$$
which is equivalent to saying that $B \in o_J(n)$.
Of course, for $\alpha = 0$, any $B$ gives as $J$-orthogonal matrix $A = I$, with no conditions on $J$.
Now, for the general case, let us also lose the involutority of $J$. Then we have to observe the condition $A^T J A = J$. This is equivalent to
$$(I + \alpha B)^T (I - \alpha B)^{-T} J (I - \alpha B)^{-1} (I + \alpha B) = J,$$
which is equivalent to
$$(I - \alpha B)^{-T} J (I - \alpha B)^{-1} = (I + \alpha B)^{-T} J (I + \alpha B)^{-1}.$$
Since $J$ is nonsingular, we see that this is equivalent to
$$\left( (I - \alpha B) J^{-1} (I - \alpha B)^T \right)^{-1} = \left( (I + \alpha B) J^{-1} (I + \alpha B)^T \right)^{-1}.$$
In other words,
$$(I - \alpha B) J^{-1} (I - \alpha B)^T = (I + \alpha B) J^{-1} (I + \alpha B)^T.$$
We have this above, with the only difference that we now have $J^{-1}$, instead of $J$. So, using the same steps, we obtain the following equivalent of $(*)$:
$$BJ^{-1} + J^{-1}B^T = 0.$$
Again, premultiplying and postmultiplying with $J$, we see that this is equivalent to
$$JB + B^T J = 0,$$
which means that the above is equivalent to $B \in o_{J^{\color{red}{-1}}}(n)$.
In other words, the first part of this proof is just an obvious special case ($J = J^{-1}$). This makes sense, since we've used involutority twice, which cancels itself out.