On $p$-subgroup of the general linear group over a vector space over $\mathbb{Z}_p$ 
Suppose $V$ is a finite-dimensional vector space over $\mathbb{Z}_p$ and $G$ is a $p$-subgroup of $GL(V)$. Then there exists such non-zero vector $v\in V$ that $gv=v$ for all $g\in G$. Moreover, we can choose a basis in $V$ in such manner that all matricies in $G$ will be unitriangular.

How do I tackle this one?
 A: $V$ is finite dimensional over $\mathbb{Z}_p$, hence it is isomorphic to 
$\mathbb{Z}_p\oplus \mathbb{Z}_p\oplus \cdots \oplus \mathbb{Z}_p$. With addition, it is elementary abelian $p$-group. 
We will write $V=\mathbb{Z}_p\oplus \mathbb{Z}_p\oplus \cdots \oplus \mathbb{Z}_p$, with $k$ copies of $\mathbb{Z}_p$, so $|V|=p^k$.
$G$ is a $p$-subgroup of $GL(V)$. $GL(V)$ (and hence any subgroup of it) acts on $V$ naturally:
$$GL(V)\times V\rightarrow V, \,\,\,\, (T,v)\mapsto T(v).$$
So $G$ is a $p$-group, acting on another $p$-group $V$. Since every $T\in G$ fixes the identity element $(0,0,\cdots,0)$, $T$ will permute the remaining elements-i.e. of $V\setminus\{0\}$:
$$G\times (V\setminus\{0\}) \rightarrow (V\setminus\{0\}), \,\,\,\, (T,v)\mapsto T(v).$$
Consider this action (on $V\setminus \{0\}$). Since $|G|$=$|Orbit(v)|.|Stab(v)|$, $|Orbit(v)|$ divides $|G|$ (which is prime power), so every orbit has size a power of $p$, possibly $p^0=1$.
If all the orbits have size a positive power of $p$, then $V\setminus\{0\}$, is disjoint union of orbits where all orbit sizes are  divisible by $p$,  i.e. $p$ also divides $|V\setminus \{0\}|=p^k-1$, a contradiction. 
Thus, there is a singleton orbit, say $\{v_1\}$ in $V\setminus \{0\}$ under $G$. What this means? This means, $v_1\neq 0$ and $T(v_1)=v_1$ for all $T\in G$.
Now $G$ leaves invariant the subspace $\langle v_1\rangle$, hence $G$ acts on quotient space $\overline{V}=V/\langle v_1\rangle$:
$$G\times \overline{V}\rightarrow\overline{V},\,\,\,\, T(w+\langle v_1\rangle)=T(w)+\langle v_1\rangle $$ 
By same process, $\overline{V}$ contains a non-zero vector say $v_2+\langle v_1\rangle$ such that $$T(v_2+\langle v_1\rangle)=v_2+\langle v_1\rangle\,\,\,\, \forall T\in G$$
This means 
$$T(v_2)+\langle v_1\rangle=v_2+\langle v_1\rangle \Longrightarrow T(v_2)=v_2+\lambda v_1.$$
Think now, if $V$ was of dimension $2$, then $\{v_1,v_2\}$ would be a basis ($v_2+\langle v_1\rangle$ is non-zero in quotient means $v_2\notin \langle v_1\rangle$, which means $v_1,v_2$ are independent.) Then how the matric of $T\in G$ w.r.t. this basis would look like? Since 
$$T(v_1)=v_1, \,\,\,\, T(v_2)=\lambda v_1+v_2$$
hence 
$$[T]=
\begin{bmatrix}
1 & \lambda\\
0 & 1
\end{bmatrix}.
$$
This is something you expected, but in dimension $2$ only. You may try to generalize now this.
A: Let $\dim(V) = n$ and $|G| = p^{k}$. There are $p^n -1$ non-zero vectors in $V$.  Now consider the orbits of $G$ on $V$.  Every element of $V$ fixes $0$, so the other $p^n -1$ vectors are partitioned by these orbits.  
Consider an orbit $O$, and look at the action induced by $G$ on $O$.  The kernel of this action must be a subgroup $H \leq G$, so since $G$ has prime power order order, $|H| = p^{i}$ for some $0 \leq i \leq k$.  The orbit stabilizer theorem tells us that $|O| = |G|/|H| = p^{k-i}$, this means that $|O|$ is either $1$ or a power of $p$ (I do know that $1=p^{0}$ but I just want to emphasize this).
Okay, now back to the orbit partition on $V$: since $p$ does not divide $p^{n}-1$, we cannot have $V \setminus \{0\}$ completely partitioned into nontrivial orbits having size a power of $p$.  Therefore we must have fixed vectors.
Okay to address the second part: once you pick a vector $v_{1} \in V$ that is fixed by $G$, you can look at the quotient space $V^{1} = V/\langle v_{1} \rangle$, which is again finite-dimensional. So you can pick another fixed vector $v_{2} \in V^{\prime}$, etc... until you have vectors $v_{1}, \ldots, v_{n}$.  This will give you a nested sequence of subspaces 
$$\langle v_{1} \rangle \leq \langle v_{1}, v_{2} \rangle \leq \cdots \leq \langle v_{1}, \ldots, v_{n} \rangle$$
that are each stabilized by $G$.  Thus this basis will give your representation as a unitriangular matrix.
A: Nice answers by others! (+1) Here's a hint for a different approach assuming basic Sylow theory has been covered:


*

*Any $p$-subgroup is contained in a Sylow $p$-subgroup.

*The unitriangular matrices form a Sylow $p$-subgroup.

*Therefore the given $p$-subgroup is conjugate to a subgroup of unitriangular matrices.

