# Criterion for a group to be abelian

Let $G$ be a group of order $pq$ , where $p,q$ are primes and $p<q$ . Show that if $p \nmid (q-1)$ (i.e $q \not \equiv 1\mod p )$ then $G$ is abelian.

I was able to use Sylow's theorems to prove that the group is cyclic and hence abelian (Sketch of what I did: the number of $p$-Sylow subgroups is $1$ or $q$ but since $q \not \equiv 1\mod p$ there is one $p$-Sylow subgroup and it is normal , similarly I showed that the number of $q$-Sylow subgroups is $1$ and also normal I then took their direct product, of the sylow subgroups that is, both are cyclic since they are of prime orders I then showed that the multiplication of generators generates the direct product which is isomorphic to $G$ it follows that G is abelian )

Besides the fact that the above proof looks very messy, I'm looking for a proof that does not use any Sylow theorem.

-
What do you know about semidirect products? – mixedmath Mar 25 '13 at 14:00
@mixedmath I've never heard of them, I just looked them up on Wikipedia. They seem similar to internal direct products but only requiring one of the subgroups to be normal. – user10444 Mar 25 '13 at 14:04

Yes, this can be done without Sylow as follows:

By Cauchy's theorem, the group has a subgroup of order $p$ and one of order $q$. If $p < q$ then the subgroup of order $q$ has index $p$ which is the smallest prime dividing the order of the group, so it is normal (see various proofs of this at Normal subgroup of prime index).

Now we know that the group is a semidirect product of $C_p$ with $C_q$ and since there is no non-trivial homomorphism from $C_p$ to $\rm{Aut}(C_q)$, since the latter has order $q-1$ which $p$ does not divide, this semidirect product is in fact direct, and we are done.

Just saw that you were not familiar with semidirect products, so here is a way without them:

We want to show that not only is the subgroup of order $q$ normal, it is in fact central. To see this, let $H$ be the $q$-Sylow subgroup (that normality implies uniqueness for Sylow subgroups is a lot easier to prove than the general theorems. For a generalization with an elementary proof, see Rotman introduction to theory of groups exercise).

We know from the N/C theorem that $G/C_G(H)$ is isomorphic to a subgroup of $\rm{Aut}(H)$ But we also know that since $H\leq C_G(H)$ we have that $|G/C_G(H)|$ divides $p$, and since $p$ does not divide the order of $\rm{Aut}(H)$ the only possibility is that $C_G(H) = G$ so $H$ is central.

Now let $K$ be any subgroup of order $p$. Since $|HK| = |G|$ we see that any element of $G$ can be written uniquely as $hk$ with $h\in H$ and $k\in K$ and since $H$ is central, elements of this form commute, so the group is abelian.

Added: The N/C theorem is the following: Let $G$ be a group and $H$ a subgroup of $G$. Then $C_G(H)$ is a normal subgroup of $N_G(H)$ and $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $\rm{Aut}(H)$. The proof is by defining a homomorphism from $N_G(H)$ to $\rm{Aut}(H)$ by letting any $g\in N_G(H)$ go to the action on $H$ by conjugation and noting that this homomorphism has exactly $C_G(H)$ as its kernel.

-
thank you for the proof, I am actually unfamiliar with most of the theorems besides Cauchy's theorem but you posted sources which is great, I'll understand them and get back to this. – user10444 Mar 25 '13 at 14:28
@user10444 I added a short remark on the N/C theorem. – Tobias Kildetoft Mar 25 '13 at 14:42

By Cauchy's Theorem, $G$ contains subgroups of order $p$ and $q$. Let $x,y\in G$ such that $\langle x\rangle\cong\mathbb{Z}_p,\langle y\rangle\cong\mathbb{Z}_q$. Since $\mathbb{Z}_p\not\leq Aut(\mathbb{Z}_q)$, so $x$ centralises $y$, i.e., $x^{-1}yx=y$, and hence $x,y$ commute. Then $|\langle x,y\rangle|=pq$, and so $G=\langle x,y\rangle$ is abelian.

-
Is the derived group the commutator subgroup (the elements of the form $xyx^{-1}y^{-1}$ )? Also isn't there a missing order in the first case? – user10444 Mar 25 '13 at 14:37
Yes, the derived group is an alternative name for the commutator group. And the missing case you meant $|G'|=pq$? – Easy Mar 25 '13 at 14:40
@Easy Indeed the case $|G'| = pq$ is the tricky one (showing that it cannot happen, that is), since once you show that the group is not perfect, the rest follows. – Tobias Kildetoft Mar 25 '13 at 14:43
@TobiasKildetoft, you are right, I have just amended my method. – Easy Mar 25 '13 at 14:58