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

This article gives the statement, and possibly proof, of an embeddability theorem: a result that states that any group of a certain kind can be embedded in a group of a more restricted kind.
View a complete list of embeddability theorems

## Statement

Suppose $G$ is a 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 $F_p$.

## Facts used

• Cayley's theorem
• 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 $U_m(p)$.