If $[G:H]=n$, is it true that $x^n\in H$ for all $x\in G$? 
Let $G$ be a group and $H$ a subgroup with $[G:H]=n$. Is it true that $x^n\in H$ for all $x\in G$?

Remarks. The answer is positive whenever $H$ is normal, e.g., for $n=2$. In general, by using the normal core of $H$, one can find an $m\ge 1$ such that $x^m\in H$ for all $x\in G$.
 A: As the answer given by Tobias shows, this is not true in general. However, it is possible to say some things even when $H$ is not a normal subgroup of $G$.

Let $H \leq G$ be a subgroup. For all $x \in G$, there exists $1 \leq r \leq [G:H]$ such that $x^r \in H$.

Proof: Let $r \geq 1$ be the smallest positive integer such that $x^r \in H$. Then $x, x^2, \ldots, x^r$ are in distinct cosets of $H$ and thus $r \leq [G:H]$.
A: For a counterexample, take $S_3$ and the subgroup $H = \{\rm{id},(12)\}$. This has index $3$, but $(13)^3 = (13)\not\in H$.
A: Trying to get a whole series of counterexamples, I came up with the following, which shows you how to construct these.

Proposition. Let $H$ be a non-trivial subgroup of the finite group $G$, with $n = [G:H]$. Assume that $\gcd(|H|,n)=1$. Then the following are equivalent.
  (a) For all $g \in G$: $g^n \in H$.
  (b) $H \unlhd G$.

Proof (b)$\Rightarrow$(a) is trivial by Lagrange's Theorem. So let us prove (a)$\Rightarrow$(b) (Sketch) We are going to use induction on $|G|$. To start the induction, we argue that $\operatorname{core}_G(H) \neq \{1\}$. For suppose $\operatorname{core}_G(H) =\{1\}$ and pick $g\in G$ and $h\in H$. By the assumption (a) $(g^{-1}hg)^n=g^{-1}h^ng  \in H$, so $h^n \in H^{g^{-1}}$. We conclude that $h^n \in \operatorname{core}_G(H)$, hence $h^n=1$ and the order of $h$ must divide $n$. But the order also divides $|H|$ and since $\gcd(|H|,n)=1$, we conclude $h=1$. But $h$ was arbitrary, so $H$ must be trivial, which contradicts the assumption. If $H$ is normal there is nothing to prove, so we can safely assume that $\operatorname{core}_G(H)$ is a proper subgroup of $H$. Now write $\bar{G}$ for $G/\operatorname{core}_G(H)$ and $\bar {H}$ for $H/\operatorname{core}_G(H)$, then $\bar {G}$  and $\bar {H}$ satisfy all the conditions of the proposition. By induction we get $\bar {H} \unlhd \bar {G}$, and this implies $H \unlhd G$.
A: 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 $g^n \in G$.
Proof. Choose subgroups $H_i$ with $S=H_0 \lhd H_1\lhd \dots \lhd H_r=G$, and put index $[H_i:H_{i-1}]=n_i$, $i=1, \dots , r$. Then $n=n_1\cdots n_r$. If $g \in G$, since $H_{r-1} \lhd H_r$, $g^{n_r} \in H_{r-1}$. Since $H_{r-2} \lhd H_{r-1}$, $(g^{n_r})^{n_{r-1}}=g^{n_rn_{r-1}} \in H_{r-2}$. 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 $g^n \in G$ for all $g \in G$. For example, take $G$ a $p$-group and $S$ any *non-normal* subgroup. Smallest example: $G=D_4= \langle a,b:a^4=b^2=1,bab=a^{-1}\rangle$, with $S=\{1,b\}$, which has index $4$.
