element not in subgoup Let $G$ be a finite group and $H$ a subgroup of $G$ of index $k$ i.e. $|G| = k* |H|$. Consider any $g \in G$\ $H$ (in the set $G$ but not in $H$). 
Now if $H$ is normal then the quotient group $G$ \ $H$ has order $k$ so by Lagrange we have that $g^k \in H$ for all $g \in G$.
My question is, if $H$ is not normal, we should find a $g \in G$\ $H$ such that $g^k \notin H$, but looking at some particular examples I couldn't find a counterexample.
 A: Let's try the smallest example of a non-normal subgroup: $G$ is the symmetric group on $\{1,2,3\}$ and $H$ is the subgroup consisting of the identity and the transposition $(12)$. Then $H$ has index $3$ in $G$ and either of the other transpositions, say $g=(13)$, has $g^3\notin H$.
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 $H \unlhd G$.

So, from the Proposition it follows that whenever you have a non-normal Sylow $p$-subgroup $P$ of $G$, the pair $(G,P)$ forms a counterexample. See the example of Andrea Blass above.
