$G$ finite metabelian group, $P$ a Sylow $p$-subgroup. Show that $P'$ is abelian and normal in $G$

Question

Let $G$ be a finite metabelian group and let $P$ be a Sylow $p$-subgroup of $G$. I have to observe that the derived subgroup $P'$ is abelian and normal in $G$.

Since $G'$ is abelian, the subgroup $P' \leq G'$ is abelian too, and it's normal in $G'$. How can I obtain the normality in the entire $G$?

Answer 1

For $g \in G$, we have $[P,P]^g = [P^g,P^g]$. Now $P$ and $P^g$ are Sylow $p$-subgroups of $PG'$, and hence they are conjugate by an element of $G'$. So there exists $h \in G'$ with $P^g=P^h$ and hence $$[P,P]^g = [P^g,P^g] = [P^h,P^h] = [P,P]^h = [P,P]$$ because $G'$ is abelian.

Answer 2

A slightly different proof runs along the following lines. First some generalities: if a subgroup $H$ of $G$ contains the commutator subgroup $G'$, it must be normal (proof: $H/G'$ is normal in the abelian group $G/G'$). So if $P \in Syl_p(G)$, then $PG' \unlhd G$. Of course $P$ is a Sylow $p$-subgroup of $PG'$, hence the Frattini Argument yields $G=N_G(P)PG'=N_G(P)G'$. So in general the following holds.

Proposition Let $G$ be a finite group and $P \in Syl_p(G)$, then $G=N_G(P)G'$.

Now observe that $P' \text{ char } P \unlhd N_G(P)$, so $P' \unlhd N_G(P)$. Assume from this point on that $G'$ is abelian (that is, $G$ is metabelian), then $P' \unlhd G'$. The proposition now implies $P' \unlhd G$.
Finally we mention an additional observation: note that $P \cap G'$ is the unique Sylow $p$-subgroup of the abelian $G'$. Hence $P \cap G' \text{ char } G' \unlhd G$, whence $P \cap G' \unlhd G$.

Note Using Dedekind's Modular Law and induction one can actually prove this.

Proposition Let $G$ be a finite group, $P \in Syl_p(G)$, and $i$ a non-negative integer, then $G=N_G(P \cap G^{(i)})G^{(i+1)}$.

Corollary Let $G$ be a finite solvable group of derived length $n$, $P \in Syl_p(G)$, then $P^{(n-1)}$ and $P \cap G^{(n-1)}$ are both abelian normal subgroups of $G$.