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We know that $(GL_{2} (\mathbb{R}), ·)$ and ($\mathbb{R}$ − {0}, ·) are a group. Prove that the function det : $GL_{2} (\mathbb{R})$ → $\mathbb{R}$ − {0} is a homomorphism. What is the kernel of this function?

For the first part I denoted $f=det:Gl_{2}(\mathbb{R})$ → $\mathbb{R}-\{0\}$

And then:

Let $A, B \in GL_{2}$

$F(A \cdot B)=det(A\cdot B)=det(A)\cdot det(B)=f(A)\cdot f(B)$

But my problem is with the kernel. I know that if $f:G \rightarrow H$

$kernel f=\{x \in G: f(x)=e_H\}$

Since my 'H' is $\mathbb{R}-\{0\}$ i don't know what is the identity element there. How can i get the kernel?

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    $\begingroup$ $\mathbb{R}-\{0\}$ must be intended as a multiplicative group, so the identity element is $1$. $\endgroup$ – Francesco Polizzi Aug 22 '17 at 20:26
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    $\begingroup$ so what i must look is for all the matrices such that det(A)=1? is that the kernel? $\endgroup$ – Lexie Walker Aug 22 '17 at 20:31
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    $\begingroup$ yes. It is called $\mathrm{SL}_2(\mathbb{R})$. See en.wikipedia.org/wiki/Special_linear_group $\endgroup$ – Francesco Polizzi Aug 22 '17 at 20:34
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The group structure on $\mathbb{R} \setminus \{0\}$ is multiplication here: you implicitly assumed so in verifying that $f$ is a homomorphism! What is the identity of the multiplicative group $\mathbb{R}\setminus \{0\}$?

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