Problem. A powerful $2$-generated $p$-group is metacyclic.
My book gives a hint: "prove that $G'$ is cyclic."
Using this hint, I can solve the problem, but I'm in troubles to prove it. My attempt is to write $G = \langle a,b \rangle$ and show that $G' = \langle [a,b] \rangle$. I found a previous question which may be helpul:
Lemma. If $N$ is powerful embedding in $G$, $S \subset G$ and $N = \langle S^{G} \rangle$ then $N = \langle S \rangle$.
I tried to use induction over the order of $ S $, but I can't solve.
Using this lemma, I only needs to show that $G' = \langle [a,b]^{G} \rangle$.
So, my problems are: prove that $G' = \langle [a,b]^{G} \rangle$ and prove the lemma.
Thanks for the advance!
Suppose $G = \langle x, y\rangle$. First note that if $[ x, y] = 1$ the result holds so suppose $[x,y]\neq1$. Since we know that $[x,y,y]=[x,y]^{-1}[x,y]^y$. We can see that the derived group $G^{\prime}$ is generated by $\langle[x,y]^G\rangle$.Since $G$ is powerful it follows that $G^{\prime}$ is powerfully embedded in $G$ - suppose $[G, G, G, G] = 1$ then $[G,G]^p\ge[G^p,G]\ge[G,G,G]$. Hence $G^\prime=\langle[x,y]\rangle$. Choose $r$ maximal such that $[x,y]\in G^{p^r}$, then $r\ge1$.
Since $G$ is powerful, there existes $u\in G$ such that $[x,y]=u^{}p^r$. Let $N=\langle u\rangle$, a cyclec normal subgroup of $G$ there existes $v\in G$ such that $G=\langle u,v\rangle$ and it follows $G/N$ is cyclic.