2.6.4

Let $$ GL_2(\mathbb{R}) $$ be the subgroup of invertible real matrices in the group $$ GL_2(\mathbb{C}) $$. Consider the matrix $$ A = \begin{bmatrix} 1 & 0 \\ i & 1 \\ \end{bmatrix} \in GL_2(\mathbb{C})$$, with inverse $$ A^{-1} = \begin{bmatrix} 1 & 0 \\ -i & 1 \\ \end{bmatrix} $$, then we claim the left coset $$A*GL_2(\mathbb{R}) $$ is not equal to the right coset $$ GL_2(\mathbb{R})*A $$.

Suppose for contradiction that the cosets were equal, then given any invertible real matrix B, AB is in the left coset, so there must exist some B'A in the right coset such that AB = B'A, where B' is also an invertible real matrix. Then, we have $$B' = ABA^{-1} $$, and this holds for every invertible real matrix B. Now let $$B=\begin{bmatrix} 1 & 1 \\ 0 & 1 \\ \end{bmatrix}

$$, then we can explicitly compute that $$B'=ABA^{-1}=\begin{bmatrix} 1-i & 1 \\ 1 & 1+i \\ \end{bmatrix} \notin GL_2(\mathbb{R}) $$, giving us a contradiction. Thus the left and right cosets of $$GL_2(\mathbb{R}) $$ in $$GL_2(\mathbb{C}) $$ are not necessarily equal.