Every p-group is isomorphic to some subgroup of U(n,p) 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.
 A: 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 $U(n,p)$.
A: Presumably you mean finite $p$-groups, and in this case the answer is yes.  First we can always assume that $G$ is a subgroup of $\mathrm{GL}_m(\mathbb F_p)$ for some $m$.  Then note that the elements of $G$ are unipotent matrices, i.e., their eigenvalues are all $1$.  This is because for each $g \in G$ we have $g^{p^n} = 1$ so the minimum polynomial of $g$ divides $t^{p^n} - 1 = (t - 1)^{p^n}$.  Then we apply the Lie-Kolchin theorem.
A: Let $G$ be a finite $p$-group. Then $G$ acts faithfully and linearly on the $|G|=n$-dimensional vector space $\mathbb{F}_p[G]= \bigoplus\limits_{g \in G} \mathbb{F}_p$ by permuting coordinates, hence a monomorphism $\rho : G \to \text{GL}_n(\mathbb{F}_p)$. But $\rho(G)$ is contained in a $p$-Sylow $S$ of $\text{GL}_n(\mathbb{F}_p)$; since $p$-Sylow are conjugate, $S \simeq U(n,p)$.
