A Hall system of $G$ is a set $\Sigma$ of Hall subgroups of $G$ satisfying the following two properties:
-For each $\pi$ divisor if $|G|$, $\Sigma$ contains excatly one Hall $\pi$-subgroup $G_{\pi}$.
-If $H,K \in \Sigma$, then $HK=KH$
Let $G$ a group, $U \leq G$ and $\Sigma$ a Hall system of $G$. In general $\Sigma \cap U$ is not a Hall system of $U$. But if $U$ is a normally embedded subgroup of $G$ this statement is true, why?
Thanks!
I think that we have to prove:
- That $G_{\pi} \cap U \in Hall_{\pi}(G)$, and that
- That $(U \cap G_{\pi_1})(U \cap G_{\pi_2})=(U \cap G_{\pi_2})(U \cap G_{\pi_1})$.
I have only been capable to see that if $G_{\pi} \perp U$ so $G_{\pi}U$ is a subgroup, $\frac{|G_{\pi}U|}{|G_{\pi}|}=\frac{|U|}{|G_{\pi} \cap U|}$ and $G_{\pi} \cap U$ is a $Hall_{\pi}(G)$, then I proove two statements. But I don't have this hypothesis.