If $K = \langle K \cap Z(S) : S \in \mbox{Syl}_p(G)$ and $K$ is elementary abelian normal $p$-subgroup, then $K = [K,G]C_K(G)$. Let $G$ be a finite group and $p$ a prime dividing the order of $G$. Also let $K$  be an elementary abelian normal $p$-subgroup of $G$ such that
$$
 K = \langle K \cap Z(S) : S \in \mbox{Syl}_p(G) \rangle.
$$
Then $K = [K, G]C_K(G)$.
Can you please help me solving this?
Of course $[K, G]C_G(G) \le K$ is clear, for the other direction I have absolutely no clue...
 A: I think that you are intended to use the theorem of Gaschütz that immediately precedes these exercises. But it does not seem to be easy to see exactly how to bring that in. There may be a more natural approach to the one below.
Note that the hypotheses and conclusion of this question both concern only $K$ the induced conjugation action of $G/K$ on $K$. The conclusion concerns $[K,G] = \langle k^{-1}k^g \mid k \in K, g \in G \rangle$, $C_K(G) = \{k \in K \mid k^g=g\,\forall g \in G \}$, and the hypothesis involves $K \cap Z(S) = \{k \in K \mid k^g=k\,\forall g \in S \}$, where $S \in {\rm Syl}_p(G)$. So $[K,G]$, $C_K(G)$ and $C_K(S)$ are the same for any extension of $K$ by $G/K$ in which the conjugation action opf $G$ on $K$ is the same as in the given group $G$.
So we can assume that the extension is split. That is $G = K \rtimes G/K$ with the action of $G/K$ on $N$ the same as the action in the actual given group $G$. So I will assume that $G=KH$ with $H$ a complement to $K$ in $G$.
Let $P = KQ \in {\rm Syl}_p(G)$, where $Q \in {\rm Syl}_p(H)$. The hypothesis $K = \langle K \cap Z(S): S \in {\rm Syl}_p(G) \rangle$ implies that $K = (K \cap Z(P))[G,K]$ so, since $K$ is elementary abelian, there is a subgroup $L$ of $K \cap Z(P)$ with $K = L \times [G,K]$.
So now $QL$ is a complement of $[G,K]$ in $P$ and hence, by Gaschütz's Theorem, there is a complement $R$ of $[G,K]$ in $G$. Let $L' = R \cap K$. Then $L' \le Z(R)$ and hence $L' \le Z(G)$, so $L \le C_K(G)$. Also, $K = L' \times [G,K]$, so $K=C_K(G)[G,K]$ as claimed.
