Counterexample with finite index I'm on a three part question, and have got stuck on the last part. The full question is:
Let $G$ be a group and $H$ a finite index subgroup of $G$. 
(a) If $g \in G$ show that there is a smallest positive integer $k$ such that $g^{k} \in H.$ Show that $k$ divides every integer $m$ such that $g^{m} \in H$.
(b) If $H$ is normal in $G$ show that $k$ divides $[G:H]$.
(c) Produce a counterexample to the claim that for all subgroups $H$ we have $k$ dividing $[G:H]$
I'm working on part (c). From part (b), I can see that I need $H$ to not be a normal subgroup, but I'm having a hard time coming up with a counterexample. Any help would be appreciated. 

Take $G=S_3$, $H=\langle (1\ 2)\rangle$ and $g=(2\ 3)$. The minimal $k$ such that $g^k\in H$ is $2$, and it doesn't divide the index $[G:H]=3$.  
