Let $G$ be a nilpotent group and $|G| = p_1p_2p_3$ where $p_i$ are different prime numbers. Prove that $G$ is an abelian group.
It has been a long time since I worked with algebra and I am at a loss here. I was thinking of creating the sequence:
$$ 0 \rightarrow G_1 \rightarrow G_{1,2} \rightarrow G_{1,2,3} = G $$
where $|G_1| = p_1$ and $|G_{1,2}| = p_1p_2$ and $|G_{1,2,3}| = p_1p_2p_3$. Now I think I will have to work my way from here considering their quotient groups are cyclic and I remember of a theorem where the quotient group $G/G'$ (where $G'$ is the commutator) is cyclic then the group is abelian. Probably it can be adapted. But the details of a proof escape me.
Best Answer
As you can find in the comments and related post, all Sylow subgroups of $G$ is normal. Let $P_i$ be Sylow $p_i$-subgroups of $G$ for each $i=1,2,3$. Then it can be deduced that $G$ is the direct product of $P_1,P_2$ and $P_3$. Note that a group of prime order must be cyclic and hence abelian. Since a direct product of abelian groups must be abelian, therefore we conclude that $G$ is abelian.