Index of $SL_2(R)$ in $GL_2(R)$

Question

"Describe the left cosets of $SL_2(R)$ in $GL_2(R)$. What is the index of $SL_2(R)$ in $GL_2(R)$?" To solve the question, do I need to find the order of $SL_2(R)$ and $GL_2(R)$ first? And how would I describe the left cosets of $SL_2(R)$ in $GL_2(R)$? Thanks!

Answer 1

Let $$f: (GL_2(\mathbb{R}),\bullet)\rightarrow (\mathbb{R}\backslash\{0\},\cdot),\ \ A\mapsto f(A):=\det(A),$$ where $"\bullet"$ denotes matrix multiplication and $"\cdot"$ the usual multiplication. You can easily see that $f$ is a group homomorphism. Therefore, from the first isomorphism theorem you obtain that $$GL_2(\mathbb{R})/\mathrm{Ker}(f)\cong \mathrm{Im}(f).\ \ \ (\dagger)$$ But $$\mathrm{Ker}(f)=\{A\in GL_2(\mathbb{R}) \: \ f(A)=1\}=SL_2(\mathbb{R})$$ and also you can easily see that $f$ is surjective. Thus, from $(\dagger)$ you obtain $$GL_2(\mathbb{R})/SL_2(\mathbb{R})\cong \mathbb{R}\backslash\{0\}.$$ This implies that the index $[GL_2(\mathbb{R}):SL_2(\mathbb{R})]=\infty.$ Now, if you want to write down the left cosets of $SL_2(\mathbb{R})$ in $GL_2(\mathbb{R})$ ($SL_2(\mathbb{R}$) is a normal subgroup of $GL_2(\mathbb{R})$ therefore the left and right cosets coincide) then you can write down the following: If $A\in GL_2(\mathbb{R})$ then $$A\ SL_2(\mathbb{R})=\{A\bullet B\ :\ B\in SL_2(\mathbb{R})\}.$$

Answer 2

The question has been almost answered at MSE, e.g., see here and here, but maybe not completely. The group homomorphism $$ \det\colon GL(2,R)\rightarrow R^* $$ has kernel $SL(2,R)$, so that $GL(2,R)/SL(2,R)\cong R^*$ and the index is given by $[GL(2,R):SL(2,R)]=|R^*|$.