Here $p_{i}$’s are distinct primes and $O(G) = p_{1}^{2} \dotsm p_{n}^{2}$. Then we need to show that $G$ is abelian if and only if all the Sylow subgroups of $G$ are normal.
How to solve this question?
Also a little modification to the above leads to a different scenario as $O(G) = p_{1} \cdot p_{2} \cdot p_{3} \cdot\dots\cdot p_{n}$, $G$ is cyclic if and only if all the Sylow subgroups of $G$ are normal. I think for both problems the method is same. Is it correct or will some different concepts be used?
Please elaborate.
We can prove something slightly stronger.
Theorem Let $G$ be a finite group of cube-free order. Then $G$ is abelian if and only if $P \unlhd G$ for all $P \in Syl_p(G)$.
Proof If $G$ is abelian, obviously every subgroup is normal and there is nothing to prove. So assume that all Sylow subgroups are normal. If $G$ is a $p$-group, then since $|G|$ is cube-free, it has order $p$ or $p^2$ and groups of these orders are abelian. So we can assume that $|G|$ has at least to two different prime factors: $|G|=p_1^{a_1}p_2^{a_2} \cdots p_n^{a_n}$, where $n \gt 1$, and $a_i \in \{1,2\}$. Let $P_i \in Syl_{p_i}(G)$. Since for all $i$, $P_i \lhd G$, $N=P_2P_3 \cdots P_n \lhd G$. By induction on $n$, $N$ is hence an abelian normal subgroup of $G$. By Lagrange's Theorem we have $P_1 \cap N =1$, and $G=P_1N$. But this implies that $G \cong P_1 \times N$. Since $P_1$ is of course abelian, it follows that $G$ is abelian.