Given a field $K$, $n \in \mathbb{N}$ and $J \in K^{n \times n}$, show that:
$$ S = \{A \in GL_n(K) | AJA^t = J\} $$ is a subgroup of $GL_n(K)$.
To show that S is a subgroup, we must show:
It must include the neutral element: $e \in S$
For $a,b \in S$, $S$ must include: $a \cdot b^{-1} \in S$.
1.) Showing that $e \in S$:
Since $J$ is an element over the field $K$, the multiplication has a neutral element. Therefore we choose $J = E_n$. The neutral element must also be part of the linear group over $K$, therefore if we choose $A = E_n$, we can conclude that the neutral element is in $S$.
2.) Given an $a := AJA^t = J , b := BIB^t = I$ we will show that $a \cdot b^{-1}$ will also be in $S$. The inverse of $b$ is defined as: $b^{-1} = (B^t)^{-1}I^{-1}B^{-1}$. Therefore: $$ AJA^t(B^t)^{-1}I^{-1}B^{-1} = JI^{-1} $$ But I am not sure how to proceed ...
The following method shows that $S$ is the kernel of a homomorphism defined on $GL_{n}(K)$. The kernel of a homomorphism is always a normal subgroup.
Regard $K^{n \times n}$ as a vector space (of dimension $n^2$) over $K$.
For each $A \in GL_{n}(K)$, we define a linear transformation $\phi[A]$ on the vector space $K^{n \times n}$ as follows: $$ \phi[A] J = A J A^{t}. $$ Furthermore, $\phi[A]$ is invertible, and $$ \phi[B] \phi[A] = \phi[BA]. $$ Therefore, $\phi$ is a homomorphism from group $GL_{n}(K)$ to the group of the invertible linear transformations on $K^n$.
And $S$ is the kernel of that homomorphism, hence is a subgroup of $GL_{n}(K)$, and even a normal subgroup.