Bijective Homomorphisms between non-isomorphic groups I am starting to study Group Theory and I'm having a problem in defining what is a isomorphism in relation to an homomorphism.    
Consider a mapping $f : G_1 \to G_2$ between two groups.
Considering I know this mapping represents an homomorphism AND a bijection.       
Question 1:
Is it true that i dont know yet if it represents an isomorphism? Do I have to check if it's inverse is also a bijection?       
I was told the groups $(\operatorname{GL}_n(\mathbb{R}) , \times )$ and $(\mathbb{R}^*, \times )$ are not isomorphic, that is, there is not any possible isomorphism between them because one is abelian whereas the other isn't.
At the same time I was told that there can be bijective mappings between these two.
Question 2 :
Does anyone know any example of bijective homomorphisms between those two?     
Thanks
 A: *

*Usually, isomorphisms for groups, rings, vector spaces, modules etc are defined to be bijective homomorphisms. However, if your definition of isomorphism $f$ is that there is another homomorphism $g$ such that $fg$ and $gf$ are identity maps, then Tobias Kildetoft's comment on your post provides a full explanation for that. That is, you can prove that the inverse mapping of a bijective homomorphism is also a homomorphism.

*It's very easy to disprove that $GL(n,\Bbb R)$ is not isomorphic to $(\Bbb R^\ast,\times)$: you just have to think of a group behavior possible in one and not in the other!
The one that came to mind for me is that there are exactly two things in $\Bbb R^\ast$ which square to 1, namely $1$ and $-1$. But in $GL(2,\Bbb R)$, for example, it is very easy to find more than two things which square to 1:
$$
\begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}1&0\\0&-1\end{bmatrix},\begin{bmatrix}-1&0\\0&1\end{bmatrix},\begin{bmatrix}-1&0\\0&-1\end{bmatrix},\begin{bmatrix}0&1\\1&0\end{bmatrix}\dots
$$
A: this is some kind of hints:
1) the inverse of a bijection is a bijection (just apply the definition, you can change the role of $f$ with $f^{-1}$, otherwise you can use the fact that bijection is an equivalence relation)
2) the bijection is a function giving an exact pairing of the elements of two sets. We do not assume NOTHING on "is this function preserving the structure and properties of the 2 sets?
a bijective homomorphism in fact PRESERVE qualities of a set, for example, if a group called "A" is isomorphic to $\mathbb{Z}_p$ with $p$ prime, we know that A is a cyclic group of order $p$ (and many other things). Just because exist a bijective homomorphism between the 2 groups.
hope this solves some of your doubts
