Proving an isomorphism without proving that it is Onto I am a bit perplexed. I understood that to prove an isomorphism:


*

*Prove that the group is homomorphic whereby the operation is preserved by showing that  => $\Phi (z \space \circ \space w) = \Phi (z) . \Phi (w)$

*Prove $\Phi$ is one-to-one

*Prove $\Phi$ is onto


One book argue (and some others) stipulate that you do not need to prove thatit is onto. I quote "[if you show] that $\Phi$ is an injective homomorphism from [the first Group], then [this Group] is necessarily isomorphic to $\operatorname{im}(\Phi)$"
This last quote is in reference to a question I posted whereby I could not manage to prove that the homomorphism is or not Onto: 
https://math.stackexchange.com/a/2227115?noredirect=1
 A: It is not saying that the two groups are isomorphic.  It is just saying that the first group is isomorphic to the image of the map.  By definition, the map is onto its image but that image is not necessarily the whole of the second group, it might be a subset / subgroup.  
For example, the obvious map from $\Bbb{Z}$ to $\Bbb{Q}$ regarded as groups under addition is a homomorphism and one to one.  The image of the map is the integers within $\Bbb{Q}$ which is not all of $\Bbb{Q}$.  So, it has proved that there is a subgroup of $\Bbb{Q}$ isomorphic to $\Bbb{Z}$ but not that all of $\Bbb{Q}$ is isomorphic to $\Bbb{Z}$.  
A: Your $\Phi$ is trivially onto. The group you're isomorphic with is not $\mathrm{GL}_2(\mathbb{R})$ but the image $\Phi[C^\ast] = \operatorname{Im}(\Phi)$
which is always a subgroup for any homomorphism. 
A: Both of you are correct, to prove G is Isomorphic to H you need to show 1.,2.,3. which you have stated for some $\Phi:G\to H$.
The thing is $\operatorname{Im}\Phi\space$ is a subgroup of H and the equality $H=\operatorname{Im}\Phi$ does not necessarily hold when $\Phi$ is a homomorphism.
