Elegant way to show that N is a normal subgroup of G Claim: Let $G$ be the set of all real $2 \times 2$ matrices $\left( \begin{array}{cc}
a & b \\
0 & d 
\end{array} \right)$ such that $ad \not = 0$, with matrix multiplication as the operation. Let $N$ be the subset where  $a = d = 1$. Then $N$ is a normal subgroup of $G$.
Showing that $N$ is a subgroup of $G$ is easy because $\left( \begin{array}{cc}
1 & b_1 \\
0 & 1 
\end{array} \right) \left( \begin{array}{cc}
1 & b_2 \\
0 & 1 
\end{array} \right) = \left( \begin{array}{cc}
1 & b_1 + b_2 \\
0 & 1 
\end{array} \right)$. However, I cannot think of a nice way to show that $N$ is a normal subgroup. It would be simple to do out all the computations, but also tedious. Is there a nice way to do this?
 A: Note that the eigenvalues of the matrices in $N$ are all $1$. Conjugating by any matrix doesn't change the eigenvalues. Since $G$ is a group, the conjugated matrix must still be upper triangular and have its eigenvalues on the diagonal still, so the diagonal entries must still be 1.
A: You can realize it is the kernel of the map that sends $$\begin{pmatrix}a&b\\0 &c\end{pmatrix}\mapsto \begin{pmatrix} a&0 \\ 0 &c\end{pmatrix}$$
A: This is a clarification of @jgon's answer. To show $N$ is a normal subgroup we want to show that $gNg^{-1} = N$ for any $g \in G$. This is what he means by conjugation (i.e. $gNg^{-1}$). Conjugate matrices have the same eigenvalues since $$\det(gng^{-1}) = \det(g) \cdot \det (n) \cdot \det(g^{-1}) = \det(g)\cdot \det(g^{-1}) \cdot \det(n) = \det (n),$$ and the determinant of a matrix is equal to the product of its eigenvalues. Now since $G$ is a group, $gng^{-1} \in G$ so it must be upper-triangular and have eigenvalues $1$, meaning $gng^{-1} \in N$, precisely what you wanted to prove.
