Show there exists a non-abelian group of order $pq$ without semi-direct product

Question

Let $p,q$ be primes with $p<q$ and $p \mid q-1$ . Show that there is a non-abelian group of order $pq$

By Sylow's theorems, we have $n_p=1 $ or $q$, and $n_q=1$.

However, if $n_p=1$, then both Sylow-p subgroup $P$ and Sylow-q subgroup $Q$ are unique and normal. By some work, we can see that $G$ is abelian.

So in order to show that there exists a non-abelian group of order $pq$, we must have $n_p=q$.

Thus, there are $q$ number of Sylow-p subgroups. Moreover, since $\gcd(q,p)=1$, for any Sylow-p subgroup $P$, we must have $P \cap Q = \{ e\}$.

From here, I'm a bit stuck on showing that $G$ is non-abelian.

I've seen answers that require the use of semi-direct product. But I wonder if there is a way without using it.

Thanks!

Answer 1

$\mathbb Z_q$ is a finite field, so its multiplicative group $\mathbb Z_q^{\times}$ is a cyclic group of order $q-1$. Since $p\mid q-1=|\mathbb Z_q^{\times}|$, there is a unique subgroup $P\leq \mathbb Z_q^{\times}$ satisfying $|P|=p$. Now consider the group of linear polynomial functions $G=\{ax+b\in \mathbb Z_q[x]\;|\;a\in P, b\in \mathbb Z_q\}$ under the operation of composition. By counting the number of choices for $a$ and $b$, derive that $|G|=pq$. Also, the linear function $t(x)=x+1\in G$ does not commute with $s(x)=ax\in G$ provided $a\neq 1$. There must exist such an $a\in P$, since $|P|=p>1$.

Answer 2

You can use a semi-direct product without saying that you are doing so, by a matrix trick. Namely, if $\mathbb Z/p$ is to (secretly?!?) act on $\mathbb Z/q$, for $x\in \mathbb Z/p$ and $y\in \mathbb Z/q$, heuristically map to $\pmatrix{x & y \cr 0 & 1}$... with the understanding that $xy$ means $y^x$, meaning exponentiation in $\mathbb Z/q$ by $x$ mod $p$. Of course, this is not well defined without hypotheses on $p,q$.

Answer 3

It is a theorem that $$\langle a,b\mid a^p,b^q,aba^{-1}b^{-r}\rangle $$ is a presentation of a non-abelian group of order $pq$ when $p\mid q-1, r\ne1\pmod q$ and $r^p\equiv 1\mod q$.

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It's easy to see that its order is at most $pq$; as well as that it's non-abelian.

What remains is to show the order is $\ge pq$. So, say $a^wb^x=a^yb^z$. Then $a^{w-y}=b^{z-x}\implies w\equiv y\pmod p$ and $z\equiv x\bmod q$, because $\langle a\rangle \cap \langle b\rangle =\{e\}$. Thus we see that the elements $a^wb^x\,,0\le w\lt p,\,0\le x\lt q$ are all distinct.

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Without the condition $r^p\equiv 1\pmod q$, it will turn out that $b=e$, and the group is then $\Bbb Z_p$.