If $H$ is a subgroup of $G$ of finite index $n$, then under what condition $g^n\in H$ for all $g\in G$ A consequence of Lagrange's theorem in finite group theory is that $g^{|G|}=e$ for all $g\in G$.
I wonder whether this be generalized along these lines:

If $H$ is a subgroup of $G$ of finite index $n$, then $g^n\in H$ for all $g\in G$. $\qquad(\star)$

This is true when $H$ is normal, but not in general, as this counter-example shows:

$G=S_4$, $H=S_3$, $g=(234)$: $[G:H]=4$ but $g^4=g\notin H$.

A previous answer says that, when $\gcd(|H|,n)=1$, $(\star)$ happens iff $H$ is normal.
Can $(\star)$ be true when $H$ is not normal? What other conditions on $G$, $H$, and $n$ ensure $(\star)$ ? 

Here is series of groups where the property holds, even if the subgroup is not normal. A subgroup $S$ of a group $G$ is called subnormal (one writes: $S \lhd \lhd G$) if there exists a chain of subgroups $H_i$ of $G$, with $S=H_0$, $H_{i-1} \lhd H_i$, for $i=1, \dots, r$ and $G=H_r$.
Proposition. Let $S \lhd \lhd G$ with index$[G:S]=n$. Then for all $g \in G$ it holds that $g^n \in S$.

Proof Choose subgroups $H_i$ with $S=H_0 \lhd H_1 \dots \lhd H_r=G$, and put index$[H_i:H_{i-1}]=n_i$, $i=1, \dots , r$. Then $n=n_1 \dot n_2 \dots n_r$. If $g \in G$, since $H_{r-1} \lhd H_r$, $g^{n_r} \in H_{r-1}$. Since $H_{r-2} \lhd H_{r-1}$, $(g^{n_r})^{n_{r-1}}=g^{n_r.n_{r-1}} \in H_{r-2}$. Now continue this argument till $S$ is reached.$\square$

Since nilpotent groups are characterized by the fact that all subgroups are subnormal (this can be proved by the “normalizers grow” principle and induction), the proposition allows for a series of examples of a group $G$ and non-normal subgroup $S$, with index$[G:S]=n$, such that $g^n \in S$ for all $g \in G$. For example, take $G$ a $p$-group and $S$ any non-normal subgroup. Smallest example: $G=D_4=<a,b:a^4=b^2=1,bab=a^{-1}>$, with $S=\{1,b\}$, which has index 4. See also the remark of Derek Holt above!
