If $G\le GL_n(\mathbb{Z}/p\mathbb{Z})$ and $\exists P$ s.t. $\forall M \in G, P^{-1}MP$ is an upper unitriangular matrix, then $G$ is a p-group

Question

Question:

Let $p$ be a prime number and $ n \in \mathbb{N}$. Let $G$ be a finite subgroup of $GL_n(\mathbb{Z}/p\mathbb{Z})$.
Suppose that $ \exists P \in GL_n(\mathbb{Z}/p\mathbb{Z}) $ such that $\forall M \in G$ we have that $P^{-1}MP$ is an upper triangular matrix with $1$ on the diagonal.
Show that $G$ is a $p$-group. Explain your answer.

I didn't success to find a satisfying answer to this question.
In my best "attempt" I wanted to first use the fact that the set $U_n $ of all upper triangular with "1" on the diagonal is unipotent matrix.
Basically this means that $ \forall M \in U_n \Rightarrow (M-I)^n=0$ but I do not understand how from here I can continue and conclude.

Thank you for your help.

Answer 1

The order of an element of a group must divide the order of that group by Lagrange's Theorem. Here $|UT_n(\Bbb Z/p\Bbb Z)|$ is a power of $p$. Use the Fundamental Theorem of Arithmetic. Conjugacy is an isomorphism. Isomorphisms preserve order.

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We have

$$\begin{pmatrix} 1 & * & * & \dots & *\\ 0 & 1 & * & \dots & *\\ 0 & 0 & \ddots & \ddots & *\\ \vdots & & & & \\ 0 & 0 & 0 & \dots & 1 \end{pmatrix}$$

as the typical matrix of $UT_n(\Bbb Z/p\Bbb Z)$.

Answer 2

Note: $T_n$ is the set of all $n \times n$ upper triangular matrix with $1$ on the diagonal.

1-
$|T_n|= p^{\frac{n(n-1)}{2}}$.
Indeed for each matrix of $T_n$ we have $n$ rows. At the first row there is $0$ element that we can place , as the $n-1$ (from the left) first elements on the first line are $0$ and the $n$-th has to be $1$ by definition. On the second row we can choose $1$ element, on the third $2$,... following the same logic at the end we get that in each matrix of $T_n$ we have that there is $1+2+...+n-1= \frac{n(n-1)}{2}$ free emplacement where we can put, at each one of this element $p$ different number (are we are in $\mathbb{Z} / p \mathbb{Z} $ ).
Hence there is $p^{\frac{n(n-1)}{2}}$ different possible matrix in $T_n$.

2-
Hence $T_n$ is a $p$-group. Now what we have to prove is that $G$ is a subgroup of $T_n$ and then by property $G$ will be too a $p$-group.
Remind that $G = \left \{ M \in GL_n( \mathbb{Z} / p \mathbb{Z}) : P^{-1}MP \in T_n \right \}$ and $G$ is a sub group. We note $ \eta $ this morphism by conjuguate $ \eta (M)= P^{-1}MP$ and this is an automorphisme from $G$ to $G \subseteq GL_n( \mathbb{Z} / p \mathbb{Z})$ (by his definition) so $|\eta(G)|=|G|$.
Thus by theorem $G$ is a sub group of $GL_n( \mathbb{Z} / p \mathbb{Z})$ and by lagranges theorem he must divide $ |GL_n( \mathbb{Z} / p \mathbb{Z})|$ hence it is a multiple of $p$ and so it is too a p-order.
Q.E.D.

Is it correct? I will be happy to read any of your comment.