A group of order $p^2q$ will be abelian

This problem is not homework but, I was stuck to it when I reviewed the Sylow theorems and problems. I am really interested of finding a test in which we can examine whether a finite group of certain order is abelian or not. It tells:

$G$ is a finite group of order $p^2q$ wherein $p$ and $q$ are distinct primes such that $p^2 \not\equiv 1$ (mod $q$) and $q \not\equiv 1$ (mod $p$). Then $G$ is an abelian group.

We know that $n_p=1+kp$ and it must divide $p^2q$. So, $1+kp|q$ and because of $q≢1$ (mod $p$), we get $n_p=1$. This means that we have a unique $p$-sylow of $G$, for example $P$, in the group and so is normal and ofcourse isomorphic to $\mathbb Z_{p^2}$ or $\mathbb Z_p×Z_p$. What should I do next? Thanks for helping me.

Solution 1:

Since $ q\nmid p^{2}-1$ therefore $n_{q} = 1+kq \neq p,\: p^{2}$. So we also have a normal $q$ -Sylow subgroup. So

• $G \cong \mathbb{Z}_{p^2} \times \mathbb{Z}_{q}$

• $G \cong \mathbb{Z}_{p} \times \mathbb{Z}_{p} \times \mathbb{Z}_{q}$.

Hence $G$ is abelian.

Solution 2:

@Babak, your problem is a special case of the following.

A positive integer $n=p_1^{a_1}. .. p_t^{a_t}$, $p_i$ distinct, is said to have good factorization if and only if $p_i^{k} \not\equiv 1$ mod $p_j$ for all integers $i$, $j$ and $k$, where $1 \leqslant k \leqslant a_i$.

Theorem The groups of order n are all abelian if and only if n is cube-free and has good factorization.

See for example the nice paper Nilpotent and Solvable Numbers by J. Pakianathan and K. Shankar