Let $G = \{A \in G L_n(\mathbb{R}) : \mathrm{det} A = 2^k$ for some $k \in Z \}$. Show that $G$ is a normal subgroup of $H = GL_n(\mathbb{R})$.

Question

Let $G = \{A \in G L_n(\mathbb{R}) : \mathrm{det} A = 2^k$ for some $k \in Z \}$. Show that $G$ is a normal subgroup of $H = GL_n(\mathbb{R})$.

I want to show that for $g \in G$; $gHg^{-1} = H$

But am unsure what property of $G$ shows this.

Should I use the inverse of the determinant to derive some characteristic of $g$ to show this holds?

Or should I use $g \in G$, $h \in H$; $ghg^{-1} = h$ and try to prove with some example that this holds?

Answer 1

You should prove that if $h\in H$ and $g\in G$, then $hgh^{-1}\in G$. That is easy, since $\det(hgh^{-1})=\det g$.