Subgroup and/or Normal Subgroup Let $G$ be the general linear group $GL_2(\mathbb{R})$ and let $H$ be the subset of matrices $A\in G$ such that $EAE^{−1} = A$, where $E$ is the matrix $$\begin{pmatrix} 1 &1 \\ 0 & 1 \end{pmatrix}$$
Determine whether $H$ is a subgroup of $G$ and whether $H$ is a normal subgroup of $G$.
I have shown that $H$ is a subgroup of $G$ but am suspecting $H$ is not a normal subgroup of $G$. 
Let $A\in H$, $X \in G$ such that $EAE^{-1}=A$
We have: $XAX^{-1}=X(EAE^{-1})X^{-1}=(XE)A(XE)^{-1} \in H \iff X=E^{-1}$
Since $XAX^{-1}\in H$ for a specific $X$, I think that $H$ is not a normal subgroup. 
Is the analysis correct ?
 A: Let E=$\left(\begin{array}{rr}1&1\\0&1\\ \end{array}\right)$, then an element $A=\left(\begin{array}{rr}a&b\\c&d\\\end{array}\right) \in G$ commutes with $E$ iff $AE-EA=0$, i.e. when $\left(\begin{array}{cc}-c&-d+a\\0&c\\\end{array}\right)=\left(\begin{array}{rr}0&0\\0&0\\\end{array}\right)$, so $A$ is of the form $\left(\begin{array}{rr}a&b\\0&a\\\end{array}\right)$, where $a \neq 0$. To see if the group $H$ of matrices of this form we have to see if any arbitrary element $X=\left(\begin{array}{rr}x&y\\z&u\\\end{array}\right) \in G$ is such that $XAX^{-1} \in H$. We use $X^{-1}=\left(\begin{array}{rr}u&-y\\-z&x\\\end{array}\right)/\Delta$ where $\Delta$ is the determinant of $X$. Calculation gives $\left(\begin{array}{cc}%
uax-ayz-bzx&bx^{2}\\
-bz^{2}&uax-ayz+bzx\\
\end{array}\right)/\Delta
$, which in order to $\in H$ must have $z=0$ giving $\left(\begin{array}{rr}
uax&bx^{2}\\
0&uax\\
\end{array}\right)/\Delta$. This is of the required form if $xu\neq 0$, but this is just $\Delta$ which is non zero by assumption. So the normalizer $\mathcal{N}(H)$ of H is the group of upper triangular matrices of $G$ and is strictly included in $G$ so $H$ is not normal. 
