About direct products of Groups Let $G$ be a finite soluble group. Let $T$, $S\leq G$ such that $T$ and $S$ are nilpotent ($T=T_{p_1}\times \cdots \times T_{p_n}$) and $T^{*}= T_{p_1}^{g_1}\times\cdots\times T_{p_n}^{g_n}\leq S$, for some $g_1,\dots, g_n\in G$. Can I prove that $T\leq S$? If not, what conditions do I need?
 A: Let's restrict to the simpler case where $T$ and $S$ are $p$-groups and consider the contrapositive of your statement.

Suppose that $S,T$ are subgroups of a solvable group $G$ and $S$ and $T$ are $p$-groups.  If $T$ is not a subgroup of $S$, then $T^g$ is not a subgroup of $S$ for any $g\in G$.

You can see clearly this isn't true.  If it were, then any subgroup of $S$ would be unable to conjugate out of $S$.  Thus every single $p$-subgroup would be normal; in particular, every solvable group would be nilpotent.
So, any time $G$ is solvable but not nilpotent, there is at least one counterexample to your statement.  This implies we should maybe look at the Fitting subgroup of $G$ for a specific class of counterexamples.

Definition. The Fitting subgroup $F(G)$ of a solvable group $G$ is the largest nilpotent normal subgroup of $G$.  Equivalently, $F(G)$ is the direct product $$F(G)=\bigotimes_{p\mid |G|} O_p(G)$$ where $\displaystyle O_p(G)=\bigcap_{P\in \operatorname{Syl}_p(G)}P$.

You can see that any group $H\not \leqslant F(G)$ is going to contain a counterexample.  (This is not sufficient, however.  Subgroups of $F(G)$ can still contain counterexamples, since not every subgroup of $O_p(G)$ is necessarily normal in $G$.)
