Every group of prime power order is a subgroup of a group of unipotent upper-triangular matrices

Statement
Suppose $$G$$ is a fact about::group of prime power order, i.e., a group of order $$m = p^n$$ for some prime $$p$$. Then, $$G$$ is isomorphic to a subgroup of $$U_m(p)$$, where $$U_m(p)$$ is the group of upper-triangular $$m \times m$$ matrices with $$1$$s on the diagonal, over the prime field $$F_p$$.

Related facts

 * Every finite group is a subgroup of a finite simple group
 * Every finite group is a subgroup of a finite complete group
 * Every group of prime power order is a subgroup of an iterated wreath product of groups of order p

Facts used

 * uses::Cayley's theorem
 * uses::Sylow implies order-dominating: The domination part of Sylow's theorem, which states that given a $$p$$-subgroup and a $$p$$-Sylow subgroup, some conjugate of the $$p$$-subgroup lies inside the $$p$$-Sylow subgroup.

Proof
Given: A group $$G$$ of order $$m = p^n$$ for some natural number $$n$$.

To prove: $$G$$ can be embedded as a subgroup of $$U_m(p)$$, the group of upper triangular unipotent $$m \times m$$ matrices over the field of $$p$$ elements.

Proof: By Cayley's theorem (fact (1)), $$G$$ is a subgroup of the symmetric group on $$m$$ elements. This, in turn, is a subgroup of the general linear group $$GL_m(p)$$, under the embedding that sends each permutation to its corresponding permutation matrix. Thus, $$G$$ embeds as a $$p$$-subgroup of $$GL_m(p)$$.

Now, the group $$U_m(p)$$ is a $$p$$-Sylow subgroup of $$GL_m(p)$$, so by fact (2), some conjugate of $$G$$ lies inside $$U_m(p)$$. Since this conjugate subgroup is in particular isomorphic to $$G$$, we obtain an embedding of $$G$$ as a subgroup of $$U_m(p)$$.