Homomorphism of $G=GL(n,\mathbb R)$ and $H=GL(n,\mathbb R)$ where $f(A)=(A^{-1})^T$

Question

Is this $f:G\rightarrow H$ a homomorphism.

At first look it appears to be an automorphism given

$G=GL(n,\mathbb R)$ and $H=GL(n,\mathbb R)$ where $f(A)=(A^{-1})^T$

So $f(AB)=((AB)^{-1})^T=(A^{-1})^T(B^{-1})^T=f(A)f(B)$

Would the kernel then be

ker $f = SL(n, \mathbb R):(A \in GL(n, \mathbb R)| \det(A) = 1)$.

I'm unsure how the kernel of this should be commuted. Any help would be nice thanks.

Answer 1

$A$ is in the kernel iff $f(A)=I$. But $(A^{-1})^T=I$ implies $A=(I^T)^{-1}$, which is still $=I$.

Also, $g\colon H\to G$, $X\mapsto (X^T)^{-1}$ is quite obviously an inverse homomorphism, making both isomorphisms.