How to show that the mapping $\varphi : GL(2,\mathbb{R})\rightarrow GL(2,\mathbb{R})$ is isomorphic

Question

Question: Show that the map $\varphi : GL(2,\mathbb{R})\rightarrow GL(2,\mathbb{R})$ sending a matrix $A$ to the inverse of its transposed matrix $(A^t)^{-1}$ is an isomorphism of groups.

I came across this question and wasn't entirely sure how to prove it. I'm currently going through a textbook that hasn't quite gone through something like this and came across this question. How would you show that this is an isomorphism of groups?

Answer 1

The kernel is $\{A: (A^t)^{-1}=I\}=\{I\}$.

It is also surjective, since for any $A\in\operatorname{GL}(2,\Bbb R), \varphi((A^t)^{-1})=A$. (Here we use that $\operatorname{det}(A^t)=\operatorname{det}(A)$ and $\operatorname{det}(A^{-1})=1/\operatorname{det}(A)$.)

But you need to show it's a homomorphism: $\varphi(AB)=((AB)^t)^{-1}=(B^tA^t)^{-1}=(A^t)^{-1}(B^t)^{-1}=\varphi(A)\varphi(B)$ (I refer you to @JensSchwaiger's answer.)

Answer 2

Note that 1) $(AB)^t=B^t A^t$ and 2) $(AB)^{-1}=B^{-1} A^{-1}$. Thus $((AB)^t)^{-1}=(B^t A^t)^{-1}=(A^t)^{-1} (B^t)^{-1}$.