TITLE: Every p-group is isomorphic to some subgroup of U(n,p)
QUESTION [5 upvotes]: Let $U(n,p)$ be the group of upper diagonal matrices with elements from $\mathbb{F}_p$ and determinant $1$. Then prove/disprove that every $p$-group is isomorphic to some subgroup of $U(n,p)$.
My ideas: We can go about by induction, since if $G = A \times B$, then we can append the matrices along the diagonal, to get a corresponding matrix for G.
Another idea is that we can consider the Jordan decomposition of the permutation matrices.

REPLY [4 votes]: One way is the following. First note that just because of order considerations, $U(n, p)$ is a $p$-Sylow subgroup of $\operatorname{GL}(n, p)$.
Now let $G$ be a $p$-group of order $n = p^{k}$. Consider a faithful permutation representation of $G$: the regular representation will do (see Cayley's theorem), so you may consider $G$ as a subgroup of the symmetric group $S_{n}$. Now $S_{n}$ can be regarded as the subgroup of $\operatorname{GL}(n, p)$ given by the permutation matrices.
Composing the underlying maps, you now have that $G$ is isomorphic to a subgroup $G_{1}$ of $\operatorname{GL}(n, p)$. Now use Sylow's theorem to get that $G_{1}$ is conjugate to (and thus isomorphic to) a subgroup of $U(n,p)$.