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.

Question

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.

Answer 1

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.

Answer 2

Self-contained proof: every nilpotent group $G$ of square-free order is cyclic.

1. $G$ is abelian. For each prime divisor $p$ of $|G|$ choose an element $g_p$ of order $p$ in $G$ and define $g=\prod_p g_p$. Then (using commutativity) the order of $g$ is divisible by $p$ for every prime $p$ dividing $|G|$, and hence by $|G|$ since $|G|$ is square-free, so $G=\langle g\rangle$.

2. General case ($G$ nilpotent). Write the order as product of $n$ distinct primes and argue by induction on $n$. The case $n=1$ is clear.

Suppose $n\ge 2$. The center of $G$ is nontrivial, hence contains a subgroup $Z$ of order $p$ for some prime $p$ dividing $|G|$. So $G/Z$ has square-free order, product of $n-1$ primes. By induction, $G/Z$ is cyclic. Since $G$ has a central subgroup with cyclic quotient, $G$ is abelian (as we see by lifting a generator from $G/Z$ to $x\in G$, so $G=\langle G,x\rangle$ is abelian). So the previous case applies.

[If one is not familiar with nilpotent groups, in the context of finite groups it can be defined as "every nontrivial quotient (including the group itself!) has a nontrivial center"]