Normalizer and Centralizer of Upper Triangular Matrix Consider the group $H_3(\mathbb{Z})=\left\{\begin{pmatrix}1 & a & c\\0 & 1 &
b\\0 & 0 & 1\end{pmatrix}:a, b, c\in\mathbb{Z}\right\}$.
Find $Z\left(\begin{pmatrix}1 & 1 & 0\\0 & 1 & 1\\0 & 0 &
1\end{pmatrix}\right)$.
Find $N_{H_3(\mathbb{Z})}\left(\left\{\begin{pmatrix}1 & a & 0\\0 & 1
& 0\\ 0 & 0 & 1\end{pmatrix}\right\}\right)$.
So to find the centralizer of \begin{pmatrix}1 & 1 & 0\\0 & 1 & 1\\0 & 0 &
1\end{pmatrix}
I need to find things that commute with each other right?

To find the centralizer of
$$
A = \begin{bmatrix}1 & 1 & 0\\0 & 1 & 1\\0 & 0 &
1\end{bmatrix}
$$
you need to solve the equation
$$
A X = X A,
$$
where
$$
X = \begin{bmatrix}1 & a & c\\0 & 1 & b\\0 & 0 &
1\end{bmatrix}.
$$
This will yield three linear equations which, as you will see, admit the immediate solution $a = b$.
To find the normalizer, you might first note that your subgroup 
$$
\left\{ 
\begin{bmatrix}1 & y & 0\\0 & 1 & 0\\0 & 0 &
1\end{bmatrix}
:
y \in \mathbb{Z}
\right\}
$$
is generated by
$$
B = \begin{bmatrix}1 & 1 & 0\\0 & 1 & 0\\0 & 0 &
1\end{bmatrix},
$$
so it is enough to find all $X$ as above such that
$$
B X = X B',
$$
where
$$
B' = \begin{bmatrix}1 & y & 0\\0 & 1 & 0\\0 & 0 &
1\end{bmatrix},
$$
for some $y$. You will find out that the condition is $b = 0$ here.
