Is $H$ is a normal subgroup of $G$? Yes/NO

Question

$G = GL_n(\mathbb{R})$ and $H$ is the subgroup of all matrices in $G$ with positive determinant

Is $H$ is a normal subgroup of $G$?

My attempt : Take G= $ \begin{bmatrix} 2 &0 \\ 0& 1 \end{bmatrix}$ and $H=\begin{bmatrix} 1 &2 \\ 0& 1 \end{bmatrix} $

H will not normal subgroup of G

Is its true ?

Answer 1

Yes, $H$ is normal in $G$; for

$h \in H, \; g \in G \tag 1$

we have

$$\begin{align} \det (g^{-1}hg) &= \det(g^{-1}) \det(h) \det(g)\\ & = \det(h) \det(g^{-1})\det(g) \\ &= \det(h) \det(g^{-1}g) \\ &= \det(h) \det(1_G) \\ &= \det(h) > 0; \tag 2 \end{align}$$

thus,

$g^{-1}hg \in H, \tag 3$

which shows that $H$ is normal.

Note: This demonstration apparently binds for $GL(n, F)$, where $F$ is any field which accomodates a notion of "positivity". If $F$ is not such a field, we can still assert that the subgroup of elements with $\det(h) = 1$ is normal. Thanks to Jyrki Lahtonene for pointing this out.

Answer 2

The determinants of similar matrices are the same: $\operatorname{det}A=\operatorname{det}PBP^{-1}=\operatorname{det}P\cdot \operatorname{det}B\cdot\operatorname{det}P^{-1}=\operatorname{det}P\cdot \frac1{\operatorname{det}P}\cdot \operatorname{det}B=\operatorname{det}B$. This means that $H\triangleleft G$.

Alternatively, the surjective homomorphism $h:GL_n(\Bbb R)\to\{1,-1\}$ given by the sign of the determinant, has $H$ as kernel. Thus $H$ has index $2$, hence is normal.