Abstract Algebra – Nonabelian Semidirect Products of Order $pq$

abstract-algebragroup-theory

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.

Best Answer

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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