Why are p-elementary groups so crucial in finite group theory? I see what makes finite p-groups such a nice thing to study: non-trivial center, lots of interesting decreasing and increasing series coming from the p-power map etc.
To me it seems natural to expect huge difficulties when generalizing a result from p-groups to all finite groups. There surely are lots of ways to "go up a little bit" in the direction of non-p-ness. The direct product with a cyclic-group of order prime to p is the smallest step on this ladder.
Here comes the question: Why are often (representation theoretic) results for general finite groups a direct consequence of the special case for p-elementary groups?
Or am I mistaken with this impression?
 A: Perhaps this is not a direct answer to your question, but one of my favorite examples of $p$-groups entering into the theory of general finite groups (except of course for Sylow theory itself!) occurs in a theorem due to Frobenius. This theorem states:

Let $P$ be a Sylow $p$-subgroup of a
finite group $G$. Then the following
are equivalent:
(i) $G$ has a normal $p$-complement.
(ii) $N_G(U)$ has a normal $p$-complement
for all $p$-subgroups $U\subseteq G$
with $U>1$.
(iii) $N_G(U)/C_G(U)$ is a $p$-group for all $p$-subgroups $U\subseteq G$.
(iv) There is no fusion in $P$.

(Here $C_G(U)$ and $N_G(U)$ denote the centralizer and normalizer of the subgroup $U\subseteq G$, respectively. The quotient $N_G(U)/C_G(U)$ makes sense, of course, since $C_G(U)$ is normal in $N_G(U)$. Also, to say that there is no fusion in $P$ is to say that any pair of elements in $P$ that are $G$-conjugate are already $P$-conjugate.)
A remarkable theorem of Thompson asserts that, if $p\neq 2$ in Frobenius' theorem, to verify that $G$ has a normal $p$-complement is tantamount to checking that $N_G(U)$ has a normal $p$-complement for just two particular $p$-subgroups $U\subseteq G$. The two subgroups $U\subseteq G$ whose normalizers need to be checked are $Z(P)$ and $J(P)$ where $P$ is a fixed Sylow $p$-subgroup of $G$ and $Z(P)$ and $J(P)$ are the center and Thompson subgroup of $P$, respectively.
