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I know that a group of order $pq$ where $p < q$ and $p \not\mid q-1$ is abelian (in particular it is cyclic), and that a group of order $p^2q^2$ where $p$ and $q$ are distinct primes, $q \not\mid p^2-1$, and $p \not\mid q^2-1$ is abelian. I have attempted to generalize this (in what seems like the obvious way) and would appreciate it if someone could confirm the validity (or lack thereof) of my "conjecture".

Let $G$ be a group with $|G|=p^nq^m$ where $p$, $q$ are distinct primes and $n, m \ge 1$. Furthermore, assume that $p \not\mid q^k-1$ and $q \not\mid p^k-1$ for $1 \le k \le m$ and $1 \le k \le n$, respectively. Then $G$ is abelian.

Proof: Every divisor of $|G|$ is of the form $p^sq^t$ with $0 \le s \le n$ and $0 \le t \le m$. If $s \ne 0$ then $p\mid p^sq^t$ so that $p^sq^t \equiv 0 \,\,\mbox{mod}\,p$, and similarly if $t \ne 0$ then $p^sq^t \equiv 0 \,\,\mbox{mod}\,q$. Therefore we will now assume that $s=0$ or $t=0$. Let $1 \le k \le n$. If $p \not\mid q^k-1$ for all $k$, then $q^k \not\equiv 1 \,\, \mbox{mod}\,p$ and if $q \not\mid p^k-1$ for all $k$ then $p^k \not\equiv 1\, \mbox{mod} \,q$. Therefore $1$ is the only divisor of $|G|$ that is congruent to $1 \,\mbox{mod}\,p$ and $1 \, \mbox{mod}\,q$.

It follows by Sylow's Third Theorem that $G$ has only one Sylow $p$-subgroup and one Sylow $q$-subgroup. In addition, each of these subgroups is normal. Clearly the order of their product is $|G|$ and they intersect trivially. Furthermore, it can be shown (see proof of Theorem 24.6 in Contemporary Abstract Algebra, Gallian, 5th ed.) that all elements in these subgroups commute. Therefore, $G$ is abelian.

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As has been pointed out, the error is the assumption that the Sylow subgroups are themselves abelian. If you want a full characterization of the "abelian numbers", you can find that and more in this answer by Pete L. Clark. – Arturo Magidin Feb 16 '12 at 4:14
up vote 2 down vote accepted

This isn't true. $G = H_5 \times H_7$, where $H_p$ is the Heisenberg group mod $p$, is a counterexample. See Keith Conrad's notes here.

I don't know what Theorem 24.6 of Gallian's book states, but groups of order $p^k$ are not guaranteed to be abelian unless $k \leq 2$.

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Why the change in counterexample? And where exactly do things go wrong with my attempted generalization? The part of the proof in Gallian that I referred to just shows that the elements of two trivially intersecting normal subgroups commute. – Alex Petzke Feb 16 '12 at 3:43
I changed the counterexample to a standard group of order $p^3$. And it is true that the elements of the two distinct normal $p$-Sylows commute with each other. However, the $p$-Sylows are not commutative themselves. – Brandon Carter Feb 16 '12 at 3:50
...hence failure in the case of an exponent greater than 3. I understand now. Thanks. – Alex Petzke Feb 16 '12 at 18:39

"The Abelian orders (or Abelian numbers): n such that every group of order n is Abelian" is at, where it gives conditions on the primes dividing $n$.

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