Order of elements of the group of $3 \times 3$ upper-triangular matrices over $\mathbb{Z}/p\mathbb{Z}$ with $1$'s in the diagonal 
Suppose $p$ is an odd prime. Let $G$ be the group of $3 \times 3$ upper-triangular matrices over $\mathbb{Z}/p\mathbb{Z}$ with $1$'s down the diagonal. Show that every element of $G$ has order that divides $p$.

I know the objective is to show that for $g^{p} = 1$ for every $g \in G$, where $1$ is the identity matrix. However, I am stuck on this problem and don't know how to begin. Any help is appreciated.
 A: You can prove by induction that$$(\forall n\in\Bbb N):\begin{bmatrix}1&a&c\\0&1&b\\0&0&1\end{bmatrix}^n=\begin{bmatrix}1&na&\frac{n(n-1)}2ab+nc\\0&1&nb\\0&0&1\end{bmatrix}.$$Therefore, if $p$ is an odd prime, you have$$\begin{bmatrix}1&a&c\\0&1&b\\0&0&1\end{bmatrix}^p=\operatorname{Id}_3.$$
A: Let $N = \begin{pmatrix} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{pmatrix}$. Then you want to check that $(I + N)^p \equiv I \pmod p$.
So this follows from the binomial theorem and the fact that $N^3 = 0$.
Notice that this does not work for $p = 2$ because
$$(I + N)^2 = I + 2N + N^2 \equiv I + N^2 = I + \begin{pmatrix} 0 & 0 & ac \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. $$
A: Let
$$
g = \begin{pmatrix}
1 & a & b \\
0 & 1 & c \\
0 & 0 & 1
\end{pmatrix}
$$
be an arbitrary matrix in your group $G$. Such a matrix $g$ is what is called unipotent: It is an addition of the identity matrix $I$ with a nilpotent matrix $N$; in this case
$$
N = \begin{pmatrix}
0 & a & b \\
0 & 0 & c \\
0 & 0 & 0
\end{pmatrix}.
$$
Note that $N$ is a nil-cube matrix, i.e., $N^3 = 0$. This lets you write (because $I$ and $N$ commute with each other)
$$
g^p = (I + N)^p = \sum_{k=0}^{p}\binom{p}{k}I^{p-k}N^{k} = I + pN + \binom{p}{2}N^2 = I.
$$
