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$U_n(\mathbb{F}_p)$ is the group of upper triangular matrices of order n with diagonal entries 1 and other entries from $\mathbb{F}_p$ (equipped with matrix multiplication). $$U_n = \left \{\left ( \begin{matrix} 1 & &* \\ & 1 & \\ 0& & 1 \end{matrix}\right )_n : * \in \mathbb{F}_p \right \}$$

I need to show that given any p-group $G$, it is isomorphic to a subgroup of $U_n$ where $|G| = n$

One of the hints that I have been given is to see that there is an element in $\mathbb{F}_p^n$ which is stabilised by all elements of $G$ when they are viewed as members of $GL_n(\mathbb{F}_p)$. I have proved this but I do not know how to proceed from here.

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    The hint assumes that you can embed $G$ in ${\rm GL}_n(p)$ for some $n$. Can you do see how to do that? Once you have done that, $G$ stabilizes some $v \in V = F_p^n$ and then it stabilizes a vector in $V/F_pv$, etc. Another approach is to observe that $U_n(F_p)$ is a Sylow $p$-subgroup of ${\rm GL}_n(p)$. – 2017-02-05
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    Yes, I understand that $G$ stabilises some $v1 \in \mathbb{F}^n_p, v_2 \in \mathbb{F}^{n-1}_p$ and so on, but I do not see how to proceed from there – 2017-02-05
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    Then the matrices for the elements of $G$ are upper triangular with respect to the basis $v_1,v_2,\ldots,v_n$ (or maybe it's $v_{n},v_{n-1},\ldots,v_1$. – 2017-02-05
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    @DerekHolt This is the part I do not see why – 2017-02-06
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    If $G$ fixes $v_1$ and $v_1$ is the first basis element, then the first column of the matrix of any element of $G$ is $(1\ 0\ 0 \cdots\ 0)^T$. Now,in its action on $V/F_pv_1$, $G$ fixes a vector $v_2 + F_pv_1$, and so if we make $v_2$ the second basis element then the second column of the matrix of an element of $G$ is $(x\ 1\ 0\ \cdots\ 0)^T$ for some $x \in F_p$. Carry on like this and you see that the matrix is upper triangular. – 2017-02-06

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Let $\rho$ be the regular representation for the $p$-group $G$. We have that kernel of $\rho$ is trivial so $\rho$ is an injection of $G$ into $GL_n(p)$. Identify $G$ with its image. As $G$ is a $p$-subgroup of $GL_n(p)$, it is a subgroup of some Sylow $p$-subgroup of $GL_n(p)$. Note that $UT_n(p)$ is a Sylow $p$-subgroup of $GL_n(p)$ and all Sylow $p$-subgroups are conjugate, and so they are isomorphic. Therefore, we can identify $G$ with a subgroup of $UT_n(p)$.