A special subset of $GL(n;\Bbb R)$

Question

Let $M(n,\Bbb R)$ denote set of all $n\times n$ real matrices. For $A\in M(n,\Bbb R)$ we denote $A^t$ as the transpose of $A$. Denote $GL(n,\Bbb R)$ as set of all invertible real $n\times n$ matrices. A matix $S\in M(n,\Bbb R)$ is said to be skew-symmetric if $S^t=-S$.

Define $$G:=\{A\in GL(n,\Bbb R)\ :\ A^tSA=S,\text{ for all skew-symmetric matrix $S$}\}.$$ It easy to show that if $A_1,A_2\in G$ then $A_1A_2$ is also in $G$.

My question: if $A\in G$, then is it true that $A^{-1}\in G$ where $A^{-1}$ denotes the multiplicative inverse of $A$ in $GL(n,\Bbb R)$?

Answer 1

If $A \in G$ and $ A^tSA=S$, then $A^tS=SA^{-1}$, hence $S=(A^t)^{-1}SA^{-1}$.

Since $(A^t)^{-1} = (A^{-1})^t$, we get $A^{-1} \in G$.

Answer 2

Yes. It stems from the fact that $(A^t)^{-1} = (A^{-1})^t$.