Exercise 4.3.28(Groups of order $pq$ as permutation groups)

Proof. Let $G$ be a non-abelian group $G$ of order $n=\mathit{𝑝𝑞}$ (where $p<q$ and $p,q$ are prime numbers).

By Cauchy’s Theorem, there exists an element $a$ of order $q$ in $G$, so $H=⟨a⟩$ is a subgroup of order $q$, and $|G:H|=p$. Since $p$ is the smallest prime dividing $n=|G|$, then $H⊴G$ by Corollary 5 (p.120).

Moreover there is an element $b$ of order $p$ in $G$, so $K=⟨b⟩$ is a subgroup of order $p$ and index $q$.

Assume for the sake of contradiction that $K$ is normal in $G$.

By Lagrange’s Theorem $|H\cap K|$ divides $p$ and $q$, so $|H\cap K|=1$ and $H\cap K=\{1\}$. Then by Proposition 13 p. 93, $|\mathit{𝐻𝐾}|=\frac{|H||K|}{|H\cap K|}=\mathit{𝑝𝑞}=|G|$, so $G=\mathit{𝐻𝐾}$. In summary

(i)

$G=\mathit{𝐻𝐾}$,

(ii)

$H\cap K=\{1\}$,

(iii)

$H⊴G,K⊴G$.

Consider the map

By (i), $u$ is surjective. If $u(h,k)=u(h^{\prime },k^{\prime })$, then $\mathit{h𝑘}=h^{\prime }{k}^{\prime }$, thus ${h{}^{\prime }}^{-1}h=k^{\prime }{k}^{-1}\in H\cap K=\{1\}$ by (ii), therefore $(h,k)=(h^{\prime },k^{\prime })$, so $u$ is injective.

Moreover, if $h\in H$ and $k\in K$, then by (iii), $\mathit{h𝑘}{h}^{-1}{k}^{-1}=(\mathit{h𝑘}{h}^{-1})k^{-1}=h(k{h}^{-1}{k}^{-1})\in H\cap K=\{1\}$, thus $\mathit{h𝑘}=\mathit{𝑘h}$. If $(h,k)\in H\times K$ and $(h^{\prime },k^{\prime })\in H\times K$, then

so $u$ is an isomorphism, and $G\simeq H\times K$. Since $H$ and $K$ are cyclic, $G$ is abelian, in contradiction with the hypothesis. This contradiction shows that $K$ is not normal in $G$.

Now consider the action of $G$ on the set $A$ of left cosets of $K$, where $|A|=|G:K|=q$. This action affords a representation ${\pi }_K:G\to S_A$, defined by

Recall that $H=⟨a⟩$, where $|a|=q$. We claim that

Indeed these left cosets of $K$ are distinct: if $a^{i}K=a^{j}K$, where $0\le i\le j<q$, then $a^{j-i}\in K\cap H=\{1\}$, therefore $a^{j-i}=1$, where $0\le j-i<q$, so $i=j$. Since there are $q=|A|$ such cosets, this proves (2).

Consider the bijection $f:[[1,q]]\to A$ given by

i.e.,

This labelling of the cosets affords a permutation representation $\phi :G\to S_q$, defined by

Then $f\circ \phi (g)={\pi }_K(g)\circ f$ for all $g\in G$, so that the following diagram is commutative

Writing ${\phi }_g=\phi (g)\in S_q$, the definitions (2) and (3) imply

Then

so $\phi :G\to S_q$ is a homomorphism.

By Theorem 3 p. 119,

Since $K$ is not normal in $G$, there is some $y\in G$ such that $\mathit{𝑦𝐾}{y}^{-1}\ne K$. Therefore $L=K\cap \mathit{𝑦𝐾}{y}^{-1}$ is a subgroup of $K$, and $L\ne K$, otherwise $K\subseteq \mathit{𝑦𝐾}{y}^{-1}$, where $|K|=|\mathit{𝑦𝐾}{y}^{-1}|$, so $\mathit{𝑦𝐾}{y}^{-1}=K$, which contradicts the definition of $y$. Since the order of $K$ is prime, this shows that $K\cap \mathit{𝑦𝐾}{y}^{-1}=\{1\}$ for some $y\in G$. Since $\ker <!--\; FUNCTION\; APPLICATION\; -->({\pi }_K)\subseteq K\cap \mathit{𝑦𝐾}{y}^{-1}=\{1\}$, we obtain

So the action of $G$ on $A$ is faithful, and ${\pi }_K:G\to S_A$ is an injective homomorphism.

If $\phi (g)=()={\mathrm{id}}_{[[1,q]]}$, then ${\pi }_K(g)=f\circ \phi (g)\circ f^{-1}={\mathrm{id}}_A$, thus $g=1$. This shows that $\phi$ is an injective homomorphism.

Put $𝜃=(123\cdots q)$. We compute ${\phi }_a$ and ${\phi }_b$.

By (5), for all $i,j\in [[1,q]]$,

Therefore

(So $𝜃$ corresponds to the element $a\in G$ in the permutation representation.)

Since ${\phi }_a=𝜃\in ⟨𝜃⟩$, ${\phi }_a\in ⟨𝜃⟩\le N_{S_n}(⟨𝜃⟩)$, so

To compute ${\phi }_b$, note that $\mathit{𝑏𝑎}{b}^{-1}\in H$, since $H=⟨a⟩⊴G$. Therefore there is some fixed integer $k$ such that

Moreover, by (2), there is some fixed integer $l$ so that $b\in a^{l}K$, thus

Then for all integers $i$, $b{a}^i{b}^{-1}={(\mathit{𝑏𝑎}{b}^{-1})}^i={(a^k)}^i=a^{\mathit{𝑘𝑖}}$. Hence for all $i,j\in [[1,q]]$

This shows that

where $[j]$ denotes the representative of the class of $j$ modulo $q$ such that $1\le [j]\le q$. (Note : the permutation group of such permutations is known as $\mathrm{AGL}(1,{𝔽}_q)\le S_q$. This group was studied by Lagrange and Galois.)

It remains to prove that ${\phi }_b\in N_{s_q}(⟨𝜃⟩)$. Using the notation $[j]$, we obtain for all $i\in [[1,q]]$,

So ${\phi }_b𝜃={𝜃}^k{\phi }_b$, and

More generally, for all integers $i$, ${\phi }_b{𝜃}^i{\phi }_b^{-1}={𝜃}^{\mathit{𝑖𝑘}}\in ⟨𝜃⟩$. This shows that ${\phi }_b$ normalizes $⟨𝜃⟩$, so

Let $\tilde{G}=\phi (G)\subseteq S_q$. Since $\phi$ is an injective homomorphism, $\tilde{G}$ is a permutation group isomorphic to $G$, and

By (6) and (7), $⟨{\phi }_a,{\phi }_b⟩\le N_{S_q}(⟨𝜃⟩)$, so

$G$ is isomorphic to a subgroup of the normalizer in $S_q$ of the cyclic group generated by the $q$-cycle $(12\dots q)$. □