TITLE: Show that there is a conjugation class that is contained in the subset of triangular matrices
QUESTION [6 upvotes]: The question goes like this:

Let $p$ be a prime, $G$ a subgroup of $GL(n, \mathbb{F}_p)$ such that $|G| = p^k$ for some $k \ge 1$. Show that there exists an element $g \in GL(n, \mathbb{F}_p)$ such that $G^g$ is contained in the subgroup of $GL(n, \mathbb{F}_p)$ composed of upper triangular matrices with ones in the diagonal entries.

Here $G^g = \{x^{-1} g x \mid x \in G\}$.
I think this must have something to do with the class equation, since $|G| = p^k$ must imply there is some conjugacy class with just one element. I'm not sure.
EDIT: a previous question says that there is a common eigenvector $v$ with eigenvalue 1 for all matrices in $G$, and there is a hint to use induction over $n$, applying the induction hypothesis on $\mathbb{F}_p^n / \langle v \rangle$.

REPLY [6 votes]: It's worth noting that this follows from Sylow II and the observation that the upper triangular matrices you're talking about form a Sylow subgroup. But it's possible to give an independent proof (which can be used to motivate the Sylow theorems, and which proves Sylow II for $GL_n(\mathbb{F}_p)$ at the prime $p$) as follows. We will use what I like to call the

$p$-group fixed point theorem: Let $P$ be a finite $p$-group acting on a finite set $X$, and let $X^P$ denote the set of fixed points. Then $|X^P| \equiv |X| \bmod p$. In particular, if $|X| \not \equiv 0 \bmod p$ then there is at least one fixed point.

Proof. The non-fixed points lie in orbits of size a power of $p$. $\Box$
Let's apply the PGFPT to the action of a finite $p$-group $P$ by linear transformations on a finite-dimensional vector space $V$ over $\mathbb{F}_p$, or more specifically to $V \setminus \{ 0 \}$.

Corollary: A finite $p$-group $P$ acting on a finite-dimensional vector space $V$ over $\mathbb{F}_p$ fixes at least one nonzero vector.

If we repeatedly apply the corollary the following happens. Let $v_1 \in V$ be a nonzero fixed vector (equivalently, a copy of the trivial representation of $P$). Then $P$ acts on the quotient $V / \text{span}(v_1)$ and has a nontrivial fixed vector there. Lift it to a vector $v_2 \in V$. Then $P$ acts on the quotient $V / \text{span}(v_1, v_2)$ and has a nontrivial fixed vector there. We again lift it to a vector $v_3$ in $V$. Continuing in this way, we construct a sequence of vectors $v_1, \dots v_n$ (where $\dim V = n$) such that $P$ leaves $V_k = \text{span}(v_1, \dots v_k)$ invariant and fixes $v_k \bmod V_{k-1}$, for every $k$. (This is slightly more information than saying that $P$ leaves a complete flag invariant.)
Specializing to $V = \mathbb{F}_p^n$, there is a unique element $g \in GL_n(\mathbb{F}_p)$ which sends the $v_i$ to the standard basis $e_i$, and hence which conjugates $P$ to a subgroup of $GL_n(\mathbb{F}_p)$ which leaves $E_k = \text{span}(e_1, \dots e_k)$ invariant and fixes $e_k \bmod E_{k-1}$. Now you can check that the upper triangular matrices with $1$s on the diagonal are precisely the subgroup of $GL_n(\mathbb{F}_p)$ of elements with this property. $\Box$
Essentially the same argument over an algebraically closed field $k$ shows that once you know eigenvalues and eigenvectors exist, every square matrix $M$ over $k$ is conjugate to an upper triangular matrix (whose diagonal entries must be the eigenvalues of $M$), which is a baby form of the Jordan normal form theorem and can often be substituted for it. Among other things this result lets you define the characteristic polynomial without using determinants, which I believe Axler does in Linear Algebra Done Right (he also uses a nice proof that eigenvalues and eigenvectors exist that avoids the characteristic polynomial so this isn't circular). The same argument shows that once you know that commuting matrices have a simultaneous eigenvector then commuting matrices can be simultaneously upper triangularized.
Part of the point of calling the result above a fixed point theorem - it is often used but rarely named - is to draw an analogy between the corollary above and the Lie-Kolchin theorem / the Borel fixed point theorem (Serre points this out in Finite Groups: An Introduction). The upper triangular matrices (with no restriction on the diagonal) form a Borel subgroup and there's a lot to say about these.