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$\begingroup$

Here $p_{i}$'s are distinct primes and $O(G)=p_{1}^{2}\cdots p_{n}^{2}$. Then we need to show that $G$ is Abelian iff 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 $O(G) = p_{1}.p_{2}.p_{3}.....p_{n}$ , $G$ is cyclic iff all the Sylow subgroups of $G$ are normal .I think for both the problems method is same.Is it correct or some different concepts will be used ?

Please elaborate.

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    **Hint:** Let $H$ and $K$ be a normal subgroup of $G$ with $G=HK$ and $H\cap K=\big\{1_G\big\}$. Then, $G$ is the internal direct product $H\times K$. – 2017-01-26
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    You need to show an iff. Can you show at least one direction? – 2017-01-26
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    @Batominovski ok i got it one direction now for other way ,if $G$ is Abelian how to prove all the Sylow groups of $G$ are normal ? – 2017-01-26
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    If $G$ is abelian, then every subgroup of $G$ is normal. – 2017-01-26

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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.

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    means the theorem will be false if its not cube free order? – 2017-02-17
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    Yes, then it becomes false, there are (two types of) non-abelian groups of order $p^3$. The quaternion group of order $8$ is an example. – 2017-02-17
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    And to falsify it otherwise: there exists a group of order 12 that is not abelian. Do you know one? Apparently there are non-normal Sylow subgroups there! – 2017-02-17
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    ok, i will try to figure it out . – 2017-02-17
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    @NickyHekster Your proof works, but I don't understand the need for complication. You proved by induction that $N$ is a normal subgroup, so why instead of N being the product of $P_2$ to $P_n$, why not taking $N$ to be the product of $P_1$ to $P_n$ that equals $G$, to end sooner the proof? – 2017-07-12
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    No, you need this for the induction step, that is why I take $n-1$ subgroups. And that is why you need to assume that $n \gt 1$. – 2017-07-12