D+F 4.5.32

Given that $P \in{Syl_p(H)}$ and $P \unlhd H \unlhd K$, then $P \unlhd K$

In general, normality of groups is not a transitive relationship. This exercise establishes this transitivity when there is an added condition.

$\textit{Proof}$: Since $H \unlhd K$, $$kHk^{-1} = H \quad \forall k\in{K}$$ Since $P \subseteq H$, $$kPk^{-1} \subseteq kHk^{-1} = H$$ Since $P \unlhd H$ and $P \in{Syl_p(H)}$, then by corollary 20 p.142, $P$ is the only $p$-Sylow subgroup of $H$. Sylow $p$-subgroups are conjugate to each other (part (ii) of Sylow's Theorem). So, if we conjugate $P$, then the only possibility is that this conjugate is sent to $P$ itself. $$kPk^{-1}=P \Longrightarrow P \unlhd K \quad \blacksquare$$