I try to solve the following problem, wich related with Tate's theorem.
Let $P$ be a Sylow $p$-subgroup of finite group $G$ and assume that $P\subseteq K\subseteq G$. Show that $O^p(K)\subseteq K\cap O^p(G)$ and if equality holds then $A^p(K) = K\cap A^p(G)$.
I explain the definitions. Here $O^p(G)$ is the smallest normal subgroup of $G$ such that the corresponding factor group is a $p$-group. And $A^p(G) = G'O^p(G)$.
I can prove that $O^p(K)\subseteq O^p(G)\cap K$. But I stuck with the second part.
Also I recall that the inverse implication $$ A^p(K) = A^p(G)\cap K \Rightarrow O^p(K) = K\cap O^p(G) $$ is known as Tate's Theorem.
This is easier than Tate's Theorem. $O^p(K) = K \cap O^p(G)$ says that the largest quotients of $G$ and $K$ that are $p$-groups are isomorphic, in which case the largest quotients that are abelian $p$-groups are also isomorphic, which is equivalent to $A^p(K) = K \cap A^p(G)$. Here is a formal proof.
\begin{eqnarray*}O^p(K) = K \cap O^p(G) &\Rightarrow& K/O^p(K) = K/(K \cap O^p(G)) \cong\\ KO^p(G)/O^p(G) = G/O^p(G) &\Rightarrow& K/A^p(K) = K/K'O^p(K) \cong\\ G/G'O^p(G) = KA^p(G)/A^p(G) &\cong& K/(K \cap A^p(G)),\end{eqnarray*}.
Since $A^p(K) \le K \cap A^p(G)$, this implies that we have equality.