How to prove that if $G$ is a group with $|G|= p^k*q$ with $p,q$ distinct primes, that $G$ has a subgroup of order $p^k$ I have to prove with induction on $k$ that if $G$ is a group with $|G|= p^k*q$ with $p,q$ distinct primes, that $G$ has a subgroup of order $p^k$. I already know that if $p$ is not a divisor of the center of $G$, $Z(G)$, that there is a subgroup of order $p^k$. I also know that if $G$ is abelian, G has an element of order $p$ and order q. I haven't got any further than this.

For the induction step: suppose $G$ has order $p^k q$. If $p$ does not divide the order of the center of $G$, then you're done. Else, by your second lemma, $Z(G)$ has an element of order $p$. Call the cyclic subgroup generated by this element $P$. Then $P$ is a normal (in fact, central) subgroup of $G$. The quotient $G/P$ is a group of order $p^{k-1} q$, so by induction it contains a subgroup of order $p^{k-1}$. Apply the correspondence theorem and you're done.
Make sure you understand that you can take $k=1$ in the above argument, so the base step can be chosen to be the trivial $k=0$.
