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Let $G=GL(2,\mathbb R)\oplus GL(2,\mathbb R)$ and let $H=\{(A,B)\in G\mid \det(A)=\det(B)\}$. Prove that $G/H \simeq (\mathbb R^*,\times)$.

I'm guessing I should use: "Let $G$ be a group and let $H$ be a normal subgroup of $G$. The set $G/H = \{aH\mid a \in G\}$ is a group under the operation $(aH)(bH)=abH$."

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up vote 1 down vote accepted

Do you know the first isomorphism theorem? Apply that to the following group homomorphism:

$$\varphi:G\to \mathbb R^*\\(A,B)\mapsto\frac{\det A}{\det B}$$

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The quickest way is to define $\varphi:G_1\to \mathbb (R^*,\times)$ by $\varphi(A,B)=\det(A)/{\det(B)}$. Then just observe that $\ker \varphi=G_2$.

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Consider $f(A,B)=\det(AB^{-1})$ then we have $$f((A,B)(C,D))=f(AC,BD)=\det(ACD^{-1}B^{-1})=\det(AB^{-1})\det(CD^{-1})=f(A,B)f(C,D)$$ so $f$ is group homomorphism and we have $\ker f=G_2$ so by the first isomorphism theorem we have $$G_1/G_2 \cong {(\mathbb{R^*},\times})$$

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Nicely argued, Sami! – amWhy May 2 '14 at 13:56

The reason you are having trouble is that your proposed map $f: (A,B) \mapsto \det(AB)$ is not an isomorphism $G_1/G_2 \to \mathbb R^\times$. Actually, it's not even a well-defined map. The kernel of your $f$ consists of all pairs $(A,B)$ with $\det A = \det B^{-1}$, but you want a map whose kernel is the pairs $(A,B)$ with $\det A = \det B$. You can build such a map by looking at $g: (A,B) \mapsto \det(AB^{-1})$.

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Otherwise, notice that for all $(A,B) \in GL_2(\mathbb{R}) \oplus GL_2(\mathbb{R})$,

$$(A,B)= \left(\frac{\det(A)}{\det(B)} \cdot \operatorname{I}_2, \operatorname{I}_2 \right) \cdot \underset{\in H}{\underbrace{\left( \frac{\det(B)}{\det(A)} A, B \right)}}$$

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