Prove that a subset is normal Let $H$ be the subset of $GL(2, \mathbb R)$ consisting of all matrices of the form
$$ \left[
    \begin{array}{cc}
x & 0\\
0 & x\\
    \end{array}
\right] $$
where $x \neq 0$
Prove that H is normal.
UPDATE
So if I show $\left[
    \begin{array}{cc}
h & 0\\
0 & h\\
    \end{array}
\right]$ 
commutes with everything in $ GL(2, \mathbb R)$, I can do something like $$\left[
    \begin{array}{cc}
a & b\\
c & d\\
    \end{array}
\right]\left[
    \begin{array}{cc}
h & 0\\
0 & h\\
    \end{array}
\right]\left[
    \begin{array}{cc}
a & b\\
c & d\\
    \end{array}
\right]^{-1}$$  $$= \left[
    \begin{array}{cc}
h & 0\\
0 & h\\
    \end{array}
\right]\left[
    \begin{array}{cc}
a & b\\
c & d\\
    \end{array}
\right]\left[
    \begin{array}{cc}
a & b\\
c & d\\
    \end{array}
\right]^{-1}$$
$$ = \left[
    \begin{array}{cc}
h & 0\\
0 & h\\
    \end{array}
\right] \in H$$
Yes?
 A: Incorrect, you want to show that for any $$\left[
    \begin{array}{cc}
a & b\\
c & d\\
    \end{array}
\right] \in GL(2, \mathbb R)$$
and $$\left[
    \begin{array}{cc}
h & 0\\
0 & h\\
    \end{array}
\right] \in H$$
then $$\left[
    \begin{array}{cc}
a & b\\
c & d\\
    \end{array}
\right]\left[
    \begin{array}{cc}
h & 0\\
0 & h\\
    \end{array}
\right]\left[
    \begin{array}{cc}
a & b\\
c & d\\
    \end{array}
\right]^{-1}\in H$$
To begin, I'd actually try to prove the general fact that any subgroup whose elements commute with all other elements of the group, is normal.
A: You are indeed on the wrong track. Sounds like you have shown $H$ to be a subgroup. As for proving $H$ is a normal subgroup, we pick any arbitrary element $\left[\begin{array}{cc}
a & b\\
c & d\\
    \end{array}\right] \in \text{GL}_2(\mathbb{R})$ and prove that $\left[\begin{array}{cc}
a & b\\
c & d\\
    \end{array}\right]\left[\begin{array}{cc}
x & 0\\
0 & x\\
    \end{array}\right]\left[\begin{array}{cc}
a & b\\
c & d\\
    \end{array}\right]^{-1} \in H$. It will probably be helpful to recall $$\left[\begin{array}{cc}
a & b\\
c & d\\
    \end{array}\right]^{-1} = \frac{1}{ad-bc}\left[\begin{array}{cc}
d & -b\\
-c & a\\
    \end{array}\right]$$ So simply (although a bit tiresome) calculate by hand that the product $$\left[\begin{array}{cc}
a & b\\
c & d\\
    \end{array}\right]\left[\begin{array}{cc}
x & 0\\
0 & x\\
    \end{array}\right]\left(\frac{1}{ad-bc}\left[\begin{array}{cc}
d & -b\\
-c & a\\
    \end{array}\right]\right)$$ and show it will equal some matrix of the form $\left[\begin{array}{cc}
z & 0\\
0 & z\\
    \end{array}\right]$ with $z \neq 0 $. This will complete the proof.
A: Definition
A subgroup $H$ of $G$ is normal if for each $g \in G$ and $h \in H$ $ ghg^{-1} \in H$. 
So you are off to a bad start to begin with. First you  need to show that $H$ is in fact a subgroup. It is not difficult to show closure and the existence of the inverse for each element. Then you need to show that, $$ \left({  \begin{array}{cc}
a & b\\
c & d\\
    \end{array}} \right) \left({  \begin{array}{cc}
x & 0\\
0 & x\\
    \end{array}} \right) \left({  \begin{array}{cc}
a & b\\
c & d\\
    \end{array}} \right)^{-1} \in H $$
$$ $$ 
