Counterexample that $a\in G$, $a^n\notin H$, for $H$ a subgroup of finite index $n$ in $G$.

Question

Let $G$ be a group and $H$ a subgroup of finite index $n$. Give a counterexample that $a\in G$, $a^n\notin H$ (although I can prove that there exists $k\in\{1,2,\dots,n\}$ such that $a^k\in H$).

Really do not know how to construct the counterexample...

Thanks.

Answer 1

Let $S_4$ be a symmetric group with $4$ letters and $H=\langle (1,2,3),(1,2)\rangle$. It is not hard to see that $[G:H]=4$.

Now, consider $(2,3,4)\in S_4$. Obviously, we have $(2,3,4)^4=(2,3,4)\notin H$.

Answer 2

I would suggest you to go through following exercise :

If $H\unlhd G$ and $[G:H]=n$ then $g^n\in H$ for all $g\in G$.

Now, If you see the proof of this, you would see that you can not relax normality of $H$.

Can you find some group with a subgroup which is not Normal?

Will that be sufficient?