Full question: Let $p$ be a prime and let $k$ be a positive integer. Let $G$ be a group and let $H \triangleleft G$ with $[G : H] = p^k$. Prove that there is $x \in G$ such that $x \notin H, x^2 \notin H,...,x^{p - 1} \notin H$ but $x^p \in H$.
I've started by looking at the quotient group and I know that $|G/H| = [G : H] = p^k$. So I think I must somehow show that $xH, x^2H...x^{p - 1}H$ are distinct cosets and that $x^pH = H$. I have no idea how to do this though? Thanks in advance.
On
It suffices to show that a group $G$ such that $|G|=p^k$ has an element $a$ of order $p$.
We can do this by induction. For $k=1$ $G$ is cyclic hence we are done.
Take $k>1$. Pick $a_1\in G$, such that $|a_1|=p^{k_1}$, if $k_1=k$ then we are done since $G$ would be cyclic. Otherwise $k_1<k$ hence by the induction the group $<a_1>$ has an element $b$ of order $p$.
Take $xH \in G/H$ of order p by Cauchy theorem then $x^p \in H$.