If $[G:H]=n$, is it true that $x^n\in H$ for all $x\in G$?

Solution 1:

For a counterexample, take $S_3$ and the subgroup $H = \{\rm{id},(12)\}$. This has index $3$, but $(13)^3 = (13)\not\in H$.

Solution 2:

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 $H \unlhd G$.

Solution 3:

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 $r \leq [G:H]$.