Let $G$ be a group of order $p^kq$ with $p$ and $q$ distinct primes and $k \geqslant 1$.
I have to prove that $G$ has a subgroup of order $p^k$ using the third isomorphism theorem and induction on $k$.
I have tried to play around a bit with definitions and theorems but I can't seem to figure it out.
I found some help on other posts but none use the third isomorphism theorem.
Thanks in advance!
The case $k=1$ follows from Cauchy's theorem. Now, we shall use induction hypothesis to prove the required. Let $|G| = p^kq$ for $k>1$. Consider the class equation of $G$: $$|G|=|Z(G)|+\sum |G:C_G(g_i)|$$ where $g_i$ are representatives of the non-singleton conjugacy classes of $G$.
Case $1$ : $|Z(G)|=1$
Considering the class equation modulo $p$, we would need $\sum |G:C_G(g_i)| \equiv -1 \pmod{p}$. This forces that there exists some $g_i$ for which $|G:C_G(g_i)|=q$. Then, $|C_G(g_i)|=p^k$ is a subgroup of the required form.
Case $2$ : $|Z(G)| = p^t$ for $0<t \leqslant k$
In this case, by induction hypothesis, we know that $G/Z(G)$ has a subgroup of order $p^{k-t}$. Using the Lattice Isomorphism Theorem, this implies that $G$ has a subgroup of order $p^{k-t} \cdot p^t = p^k$ as required.
Case $3$ : $|Z(G)| = p^tq$ for $0 \leqslant t < k$
We know from induction hypothesis that $Z(G)$ must have a subgroup $H$ of order $p^t$. Moreover, $H$ is a normal subgroup of $G$. We have $|G/H|=p^{k-t}q$ and the quotient group, again by induction hypothesis, must have a subgroup of order $p^{k-t}$. Similar to the previous case, the Lattice Isomorphism Theorem gives a subgroup of order $p^k$.
Case $4$ : $|Z(G)|=p^kq$
This implies that $G=Z(G)$ i.e. $G$ is abelian. By Cauchy's theorem, there exists an element of order $p$ in $G$. Let the cyclic subgroup of order $p$ generated by this element be $H$. As $G$ is abelian, $H$ is normal. Thus, $G/H$ has a subgroup of order $p^{k-1}q$ and once again, we can finish off using the Lattice Isomorphism Theorem.