When $H$ is a subgroup of $G$, we can define the focal subgroup of $H$ as $$H^\ast:=\langle h^{-1}h'\mid h,h'\in H, h'=h^g, g\in G\rangle.$$ I'm confused about the proposition "$H^\ast$ is a normal subgroup of $H$."
I understood the property $[H,H]<H^\ast<H$ and $[H,H]\lhd H$ , but I doubt it is not a sufficient condition of the normality. (In general, $K<H<G, K\lhd G\Longrightarrow H\lhd G$ is false.) I want to know the proof of this normality.
If $[H,H] \le H^* \le H$, then for every $x \in H^*$ and $h \in H$, we have $hxh^{-1}x^{-1} \in [H,H] \le H^*$, so $hxh^{-1} \in xH^* = H^*$. Therefore $H^*$ is normal.
Note that this argument generalizes to any situation where $K \le H \le G$, $K \lhd G$, and $G/K$ is abelian - we may conclude that $H \lhd G$.