Orthosymplectic Lie Superalgebra I am trying to work out a presentation for the orthosymplectic Lie superalgebra $\mathfrak{osp}(m,2n)$.  I am following Musson's book "Lie Superalgebras and Enveloping Algebras".
From what I understand, we can take as the underlying $\mathbb{Z}_2$-graded vector space $k^m\oplus k^{2n}$, ($k$ is a characteristic $0$ algebraically closed field, say), and consider the bilinear form $J=\begin{bmatrix}G & 0\\0 & H\end{bmatrix}$, where $G$ is $m\times m$ symmetric nonsingular and $H$ is $2n\times 2n$ skew-symmetric nonsingular.
Then $\mathfrak{osp}(m,2n)$ is the collection $X=\begin{bmatrix}A & B\\C & D\end{bmatrix}\in\mathfrak{gl}(m,2n)$ such that $X^TJ+JX=0$, where $X^T=\begin{bmatrix}A^t & -C^t\\B^t & D^t\end{bmatrix}$ is the supertranspose of $X$, and the lower case $t$ gives the usual transpose, e.g. $A^t$ is just the transpose of the matrix $A$.  So we want 
$$
X^TJ+JX=\begin{bmatrix}A^t & -C^t\\B^t & D^t\end{bmatrix}\begin{bmatrix}G & 0\\0 & H\end{bmatrix}+\begin{bmatrix}G & 0\\0 & H\end{bmatrix}\begin{bmatrix}A & B\\C & D\end{bmatrix}=0
$$
When I write this out, I get the equations $A^tG+GA=0$, i.e. $A\in\mathfrak{o}(m)$, $D^tH+HD=0$, i.e. $D\in\mathfrak{sp}(2n)$.  These two equations seem correct.  Finally we get $B^tG+HC=0$ which is equivalent to the other equation $GB-C^tH=0$ upon taking the transpose.
However in the book, I've seen two places where the resulting equations have $B^tG-HC=0$ instead of $B^tG+HC=0$, and these equations are used for writing a more explicit presentation for $\mathfrak{osp}(m,2n)$ so I want to understand what I'm doing incorrectly here.  Can someone help? 
ANY advice/hints/help would be greatly appreciated!

I think you might have some misunderstanding relating to supertransposition. Here's the definition of supertransposition on wiki:
https://en.wikipedia.org/wiki/Supermatrix#Supertranspose.
As you see, the supertransposition is related to the parity of the supermatrix itself, so this question might not be that easy now. Hope this will help.
