I'm trying to solve this problem:
Let $G$ be a finite solvable group all of whose Sylow subgroup are abelian. Show that $Z(G) \cap G' = 1$.
I tried in this way:
induction over $|G|$. If $|G|=1$ it's trivial. Let $|G| >1 $ and $M \unlhd G$ minimal subgroup of $G$. We know it is normale becouse $G$ is solvable. Let now $P/M \leq G/M$ a Sylow subgroup of $G/M$. Then $P \leq G$ is a Sylow subgroup of $G$ (i'm not sure of this!), and so it's abelian. So all Sylow subgroup of $G/M$ are abelian, and by induction $Z(G/M) \cap (G/M)' = M/M=1_{G/M}$. Now, let $g \in Z(G) \cap G'$. then $gM \in Z(G/M) \cap (G/M)'$, and then $g \in M$. So $ Z(G) \cap G' \leq M$. But $Z(G) \cap G' \unlhd G$, and by minimality of $M$ we conclude that $Z(G) \cap G' = 1$.
Let us call such groups $A$-groups. The claim is trivially true if $G$ is a $p$-group, because then $G$ is abelian by assumption, and hence satisfies $G'=1$. The proof of the claim goes by induction on the group order. Assume that the result holds for all $A$-groups of order less than the order of $G$, and assume that the result is false for $G$. Then $H:=Z(G)\cap G'$ is a non-trivial normal subgroup of $G$. Let $N$ be any non-trivial normal subgroup of $G$, and $\theta$ be any homomorphism of $G$ whose kernel is $N$. Then $\theta(G)$ is again an $A$-group, with $(\theta(G))'=\theta(G')$ and $\theta(Z(G))\subseteq Z(\theta(G))$. Since $|\theta(G)|<|G|$, by the induction hypothesis we have $Z(\theta(G))\cap (\theta(G))'=1$. Thus we have $$ \theta(H)\le \theta(G')\cap \theta(Z(G))=\theta(N), $$ and $\theta(N)$ is the identity of $\theta(G)$. hence $H\le N$, and so $H$ is the unique minimal normal subgroup of $G$. Since $G$ is solvable, $|G'|<|G|$. Now the induction hypothesis applied for $G'$ implies that $G''=1$. Suppose that we have $G''\neq 1$. Then $G''$ is a normal subgroup of $G$ with $H\le G''$. We also have $H=G'\cap Z(G)\subseteq Z(G')$, so that $H\subseteq G''\cap Z(G')=1$, a contradiction. Thus $G''=1$ and $G'$ is abelian. Then every Sylow subgroup of $G'$ is a characteristic subgroup of $G'$, and so also of $G$. Hence every Sylow subgroup of $G'$ contains $H$, which is of order $p$. It follows that $|G'|=p^k$ for some $k\in \Bbb N$.
Claim: $G'$ is cyclic. We refer the reader to the paper of D.R. Taunt, On $A$-groups, $1947$.
So let $G'=\langle x\rangle$ with $|G'|=p^k$. Then $H=\langle x^{p^{k-1}}\rangle $. Suppose that $k=1$. Then $G'=H\le Z(G)$, so that $G$ is abelian and $G'=1$, a contradiction. Hence we have $k>1$. Let $y\in G$ be an element of order $r$ with $(r,p)=1$. Then $y^{-1}xy=x^{\ell}$ for some $\ell$. Since $x^{p^{k-1}}\in Z(G)$ we have $$ x^{p^{k-1}}=y^{-1} x^{p^{k-1}}y=x^{\ell p^{k-1}}. $$ Hence $\ell p^{k-1}\equiv p^{k-1}\bmod p^k$ and $\ell\equiv 1 \bmod p$. We also have $x^{\ell^r}=y^{-r}xy^r=x$, since $y^r=1$. Hence $\ell^r\equiv 1\bmod p^k$. So we have $$ (\ell-1)(\ell^{r-1}+\cdots +\ell+1)\equiv 0\bmod p^k. $$ Since $\ell\equiv 1 \bmod p$, each of $\ell, \ell^2,\ldots ,\ell^{r-1}$ are congruent $1$ modulo $p$, so that $$ \ell^{r-1}+\cdots +\ell+1\equiv r \not\equiv 0 \bmod p, $$ since $(p,r)=1$. Thus $\ell-1\equiv 0 \bmod p^k$, and $y^{-1}xy=x$. In other words, every element $y\in G$ whose order is prime to $p$ commutes with $x$, as does every $z\in G$ which is of order a power of $p$. But any element $w\in G$ can be written as $yz$ in this way, so that every element of $G$ commutes with $x$. Thus $x\in Z(G)$ and hence $G'\le Z(G)$, which again implies that $G$ is abelian - contradiction. Hence our initial hypothesis that the result does not hold for $G$ is not true, and the induction is finished.