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Let $\mathbf{R}^*$ be group of non-zero real numbers where the operation is multiplication. The group is 2x2 matrices with non-zero determinant.
Let $f: G \rightarrow \mathbf{R}^*$ be a function given by $f(A)= det(A)$.

My proof to show f is homomorphorism is:

We must show $f(ab)=f(a)f(b)$. Here we can see
if $A= \begin{pmatrix} a & b\\ c & d \end{pmatrix}$ and $B= \begin{pmatrix} e & f\\ g & h \end{pmatrix}$,

Then $f(AB) = f(C)$ where $C = AB = \begin{pmatrix} ae+bg & af+bh\\ ce+dg & cf+dh \end{pmatrix} $

Note $\det (C)$
$=f(C)$ $=(ae+bg)(cf+dh) - (af+bh)(ce+dg)$
$=(aecf+bgcf)+(aedh+bgdh) - (afce+bhce)-(afdg+bhdg)$
$=bgcf+aedh - bhce-afdg$

Also note that $f(A)= ad-bc$ and $f(B)=eh-fg$
then $f(A)f(B) = (ad-bc)(eh-fg)=adeh-bceh+adfg-bcfg$

Thus $f(AB)=f(A)f(B)$ and as such $f$ is a homomorphism.


Now since we showed $f$ is surjective, we need to show $f$ is injective to prove $f$ is a bijective mapping and thus isomorphic.

One aspect of $2\times2$ determinant we can use is that it produce the area of parallelogram. If we think of the properties of rotation matrices, we know their determinants all produce $1$. Thus $f$ is not injective and bijective. Meaning $f$ is not isomorphic.

Is this the correct way to go about this proof or am I making wild mis-understandings?

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$f$ is surjection, but not injection. It can be prooved very easily, giving a specific example.

$$\forall x\in\mathbb{R}^*, f\begin{pmatrix} x & 0\\ 0 & 1 \end{pmatrix}=x. f(I)=f(-I)$$

The first example shows $f$ is surjective, and the second shows $f$ is not injective.

Also, your approach to injectivity is right, but your approach to surjectivity does not show that $f$ is surjection.

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