# 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.

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Use the \not\equiv symbol for producing $\not\equiv$ – user9413 Jun 4 '12 at 8:43
I am fairly sure this is a duplicate, but from my iPhone it will be hard to find out. – Asaf Karagila Jun 4 '12 at 8:56
I am also quite sure it's a duplicate, but even on my normal computer I'm not sure how I would find it. – Tara B Jun 4 '12 at 9:08

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.

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All this provided that the OP knows the result: If $H$ and $K$ are two normal subgroups with trivial intersection, then they centralize each other. Of course, it is possible to arrive at this also by studying the automorphism groups of the possible Sylow subgroups, and observing that they don't have subgroups of certain orders. – Jyrki Lahtonen Jun 4 '12 at 16:28
@JyrkiLahtonen: Thanks for your comment. Honestly, what Chandrasekhar did was more simple that I had expected. He did it completely. I could not solve the problem cause someone told me that $Aut(G)$ and the $N/C$ sould be applied through solving. This is what you mentioned. Thanks for your considreation. – Babak S. Jun 5 '12 at 4:23
@Babak, Chandrasekhar is using the result discussed in this question. For full solution you need to include that argument. – Jyrki Lahtonen Jun 5 '12 at 7:45

@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

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