Is there an elementary reason for why $SL_2(\mathbb{F}_p)$ for $p>5$ does not embed into $SL_2(\mathbb{Z}_p[w])?$ This is an exercise from Serre's book on Galois cohomology.  
Let $p>5$ and consider the groups $SL_2(\mathbb{F}_p)$ and $SL_2(\mathbb{Z}_p[w])$ where $w$ is a primitive $p$th root of unity.  
Is there an elementary reason for why there can be no embedding $SL_2(\mathbb{F}_p) \rightarrow SL_2(\mathbb{Z}_p[w])?$  
Of course, what counts as elementary is somewhat arbitrary. So to be more precise, I would be interested in a proof / reason which uses as little heavy machinery as possible.
 A: Notice that $\mathrm{SL}_2(\mathbb F_p)$ has a cyclic subgroup $U$ of order $p$ (the upper triangular unipotent matrices) such that some conjugacy class in $\mathrm{SL}_2(\mathbb F_p)$ contains at least $(p - 1)/2$ elements of $U$.  (This can be seen already if you conjugate only by diagonal matrices.  I don't know if the intersection has precisely $(p - 1)/2$ elements, but it doesn't seem to matter.)
Here's one argument for why $\mathrm{SL}_2(\mathbb Z_p[w])$ has no such subgroup, though it's probably overpowered.  Suppose it does.  Then so does $G = \mathrm{GL}_2(\overline{\mathbb Q_p[w]})$.  Let $u$ be a generator of (the image of) $U$.  Since $u$ has finite order, (by the Jordan decomposition) it is contained in some conjugate of the diagonal subgroup $D$ of $G$, and (for convenience) we may arrange by conjugation that it actually lies in that subgroup.  By explicit computation, the intersection with $D$, hence a fortiori with $U \subseteq D$, of the conjugacy class of any power of $u$ has at most 2 elements.  This is a contradiction.
