I just constructed the semidirect product in Lang, and I'm trying to tie some facts together. From Ash's Algebra, I know that if $p\lt q$ are distinct primes, if $q\not\equiv 1\pmod{p}$, then any group $G$ of order $pq$ is abelian.

Is the converse true, that for any primes $p\lt q$, if $q\equiv 1\pmod{p}$ then there exists a nonabelian group of order $pq$?

One example I found online is that $\mathbb{Z}_3\ltimes \mathbb{Z}_7$ is nonabelian, and here $7\equiv 1\pmod{3}$. I was considering then semidirect products $\mathbb{Z}_p\ltimes\mathbb{Z}_q$ where $q\equiv 1\pmod{p}$ and some homomorphism $\phi\colon \mathbb{Z}_p\to\operatorname{Aut}(\mathbb{Z}_q)$ I calculate that $$ (1,0)(0,1)=(1+0,\phi_0(0)+1)=(1,1) $$ and $$ (0,1)(1,0)=(0+1,\phi_{-1}(1)+0)=(1,\phi_{p-1}(1)). $$ Is it true somehow that $\phi_{p-1}(1)\neq 1$ in each case to show the group is nonabelian? I guess if it did this would imply $\phi_{p-1}$ is the trivial automorphism, so maybe there's something there? If not, is there a way to show $\mathbb{Z}_p\ltimes\mathbb{Z}_q$ is nonabelian in these cases in general? Thanks.

  • $\begingroup$ @Dylan, I have my group operation defined as $(a_1,b_1)(a_2,b_2)=(a_1a_2,\phi_{a_2^{-1}}(b_1)b_2)$. $\endgroup$ – yunone Sep 9 '11 at 23:40
  • $\begingroup$ I see. I'm used to a different convention, so I became confused. No issues now! $\endgroup$ – Dylan Moreland Sep 9 '11 at 23:49

I'm going to use a few facts from group theory, which are certainly in Lang. Let me know if I can clear anything up!

It should be clear that a non-trivial homomorphism $$ \mathbf Z/p\mathbf Z \to \operatorname{Aut}(\mathbf Z/q\mathbf Z) $$ will answer your question in the affirmative. You could use such a map to finish your calculations, since it will necessarily be an injection and so the class of $-1$ will map to an automorphism which is not the identity, which must move $1$.

Recall that $\operatorname{Aut}(\mathbf Z/q\mathbf Z)$ is isomorphic to $(\mathbf Z/q\mathbf Z)^*$. Since $q$ is prime, the latter group is cyclic; we can identify it with $\mathbf Z/(q - 1)\mathbf Z$ if we choose a primitive root mod $q$. Now, to give a non-trivial homomorphism $\mathbf Z/p\mathbf Z \to \mathbf Z/(q - 1)\mathbf Z$ is to give an element of $\mathbf Z/(q - 1)\mathbf Z$ having period $p$. I claim that you can find $p - 1$ such elements if $p$ divides $q - 1$.

In your example, $3$ is a primitive root mod $7$ and we can send $1 \in \mathbf Z/3\mathbf Z$ to the automorphism of $\mathbf Z/7\mathbf Z$ given by $a \mapsto 9a = 2a$.

  • $\begingroup$ @yunone Yes, although I think it might be hard to actually calculate $\phi_{p - 1}$ in general. Also, I messed up some numbers. One second. $\endgroup$ – Dylan Moreland Sep 10 '11 at 0:36
  • $\begingroup$ Thanks, when you have the chance, doyou mind expanding on the line "Now, to give a non-trivial homomorphism...having period $p$." How does an element of period $p$ show a nontrivial homomorphism? And why does $q\equiv 1\pmod{p}$ imply such elements exist? $\endgroup$ – yunone Sep 10 '11 at 0:42
  • $\begingroup$ @yunone Sure. You only have to specify one element because the domain is a cyclic group, so you just have to specify a valid image of $1$. The only constraint on the image in this case is that it have period (maybe I should say order) dividing $p$. To get an element of period $p$, I would review stuff like Theorem 3.5 in Keith Conrad's handout. $\endgroup$ – Dylan Moreland Sep 10 '11 at 0:50
  • $\begingroup$ There's also a discussion in section I.4 of Lang. $\endgroup$ – Dylan Moreland Sep 10 '11 at 0:54
  • $\begingroup$ Ah, I remember now. This comes from 4.3(iv). Thanks for your help. $\endgroup$ – yunone Sep 10 '11 at 0:59

Given $p,q$, with $p<q$ and $p|(q-1)$, there exist a non-abelian group of order $pq$ and it is unique up to isomorphism:

If $G$ is a non-abelian group, $|G|=pq$, then subgroup $H$ of order $q$ is normal (by Sylow theorem); let $K$ be its subgroup of order $p$. Then $HK\leq G$ (since $H\triangleleft G$) and $HK=G$ (since $|KH|=|H|.|K|/|H\cap K|=pq=|G|$). So,

$H\triangleleft G$, $K\leq G$, $HK=G$, $H\cap K=(1)$ $\implies G\cong H \rtimes K$.

Now, to get all possible (non-isomorphic) semidirect products $H$ by $K$, we have to consider homomorphisms $\phi \colon K \rightarrow Aut(H)$.

Since $K\cong \mathbb{Z}/p$, $H\cong \mathbb{Z}/q$, we know that $Aut(H)\cong \mathbb{Z}/(q-1)$.

As $p|(q-1)$, and $Aut(H)$ is cyclic group of order $q-1$, it contains unique subgroup of order $p$. So for any two non-trivial homomorphisms $\phi, \psi \colon K\rightarrow Aut(H)$, we have $\phi(K)=\psi(K)$, and $K$ is cyclic. Also, these homomorphisms are injective, since $K$ is of prime order. (Note that trivial homomorphisms will give direct product of $H$ by $K$, so $G\cong H\times K$, abelian group of order $pq$).

Therefore by following theorem semidirect products of $H$ by $K$ w.r.t. $\phi$ and $\psi$ are isomorphic:

If $K$ is a cyclic group and $\phi,\psi\colon K\rightarrow Aut(H)$ are two injective homomorphisms such that $\phi(K)=\psi(K)$, then these two homomorphisms give isomorphic semi-direct products of $H$ by $K$ (Alperin and Bell- Groups and Representations).

Now it is clear the existance and uniqueness (up to isomorphism) of non-abelian groups of order $pq$.


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