Group of order $pq$ with $p\not\mid (q-1)$

Question

Let $p, q$ be prime numbers, with $p<q$. If $G$ is a group of order $pq$ and $p\not\mid (q-1)$, then $G\cong \mathbb{Z}/pq\mathbb{Z}$.

The standard way to prove this fact is using Sylow theorems, but I'm looking for an alternative proof.

My attempt of (alternative) proof: if we show that there exist $H, K$ normal subgroups of $G$ such that $|H|=p$ and $|K|=q$, we are done because $G\cong H\times K$. By Cauchy theorem, there exists a subgroup $K$ of order $q$ and by a well know lemma $K$ is a normal subgroup because $[G:K]=p$ and $p$ is the smallest prime such that $p\mid |G|$. Now it remains to find $H$ using the hypothesis $p\not\mid (q-1)$, any idea?

Answer 1

I will write an elemantary proof by using a lemma:

lemma: number of the elements of order $p$ equivalent to $-1$ mod $p$.

Claim$1$: The subgroup of order $q$ is uniqe.

Otherwise, we have $H,K$ with order $q$ and since $H\cap K=1$ and $q^2>pq$ which is imposible.

Now, Assume $G$ is noncyclic which means all nonidentity elements have order $p$ or $q$. By claim $1$, we have $pq-q$ elements of order $p$ and by lemma

$$pq-q\equiv -q \equiv -1$$ which implies $p|q-1$ which is a contradiction.

Note: In that solution, claim 1 is essantial which is more stronger than normality.

Answer 2

By Cauchy's theorem (or by a different argument), there is a subgroup $H$ of order $p$. Even if $H$ is not normal, the fact that $K$ is already implies that $G$ is a semi-direct product of $H$ and $K$. So the only thing left to do is to examine the morphisms $H \to Aut(K)$ and prove that there is only the trivial one. Thus the semi-direct product must be direct.

Answer 3

By Cauchy theorem, there are $H\cong\Bbb Z_p$ and $K\cong\Bbb Z_q$ subgroups of $G$. Moreover: $H\cap K=\{1\}$, $HK=G$ (because $HK\subseteq G$ and $|HK|=pq$) and, for counting reason, $K\unlhd G$. Therefore, $G\cong \Bbb Z_p\ltimes \Bbb Z_q$. Now, consider any action $\Bbb Z_p\curvearrowright \Bbb Z_q$; $\operatorname{Fix}(j)$ is a subgroup of $\Bbb Z_q$, for every $j\in\Bbb Z_p$. A nontrivial action requires that, for some $j\in\Bbb Z_p$, $\operatorname{Fix}(j)$ is trivial and hence (Lagrange): $$\sum_{j=0}^{p-1}\left|\operatorname{Fix}(j)\right|=l+(p-l)q \tag1$$ for some integer $l$, with $0<l<p$ (recall that $\operatorname{Fix}(0)=\Bbb Z_q$). By Burnside's (counting) lemma, $p$ divides the RHS of $(1)$, namely $p\mid pq-l(q-1)$, and hence, if $p\nmid q-1$, necessarily $p\mid l$: contradiction, because $p>l$. Therefore, if $p\nmid q-1$, $\Bbb Z_p\curvearrowright \Bbb Z_q$ just trivially, and hence $G\cong \Bbb Z_p\ltimes\Bbb Z_q=$ $\Bbb Z_p\times\Bbb Z_q\cong$ $\Bbb Z_{pq}$.

Answer 4

By contradiction, let's suppose $G$ nonabelian. Then $G$ is centerless, because $|Z(G)|=p,q$ would require that the noncentral elements have centralizer of order $pq$, a contradiction. By counting reason, a subgroup $K$ of order $q$ is unique and hence normal. Now there are at least two arguments to conclude.

Proof #1. Since $K$ is normal, it is the union of conjugacy classes. Then necessarily $q=1+kp$, for some positive integer $k$, namely $p\mid q-1$: contradiction. Therefore $G$ is abelian, and hence cyclic (the product of two elements of order $p$ and $q$ has order $pq$).

Proof #2. Since $G$ is centerless, the class equation reads: $$pq=1+kp+lq \tag 1$$ where $k$ and $l$ are the (positive) number of the conjugacy classes of size $p$ and $q$, respectively. Now, there are exactly $lq$ elements of order $p$ (they are the ones in the conjugacy classes of size $q$). Since each subgroup of order $p$ contributes $p-1$ elements of order $p$, and two subgroups of order $p$ intersect trivially, then $lq=m(p-1)$ for some positive integer $m$ such that $q\mid m$ (because $q> p-1$). Therefore, $(1)$ yields: $$pq=1+kp+m'q(p-1) \tag 2$$ for some positive integer $m'(=m/q)$; but then $q\mid 1+kp$, namely $1+kp=nq$ for some positive integer $n$, which replaced in $(2)$ yields: $$p=n+m'(p-1) \tag 3$$ In order for $m'$ to be a positive integer, it must be $n=1$, whence $1+kp=q$, namely $p\mid q-1$: contradiction. Therefore $G$ is abelian, and hence cyclic (the product of two elements of order $p$ and $q$ has order $pq$).