----- > [!proposition] Proposition. ([[classification of groups of order pq]]) > Let $G$ be a [[group]] with [[order of a group|order]] $pq$, where $p$ and $q$ are distinct [[prime number|prime numbers]] with $p<q$. Then, either $G \cong C_{pq} \ \ \text{(when }q \not{|} (p-1) \ \text{)}$ > or > $G \cong C_{p} \rtimes C_{q}$ ^c5e9a6 > [!proof]- Proof. ([[classification of groups of order pq]]) > Let $G$ have [[order of a group|order]] $pq$. Recall that WLOG $p>q$. \ [[The Sylow theorems]] tell us that $G$ has a number $n_{p}$ of [[p-Sylow subgroup]]s of [[order of a group|order]] $p$ that [[divides]] $q$ (i.e., $n_{p} \in \{ 1,q \}$) and is [[congruent]] to $1$ modulo $p$. Since $q <p$, $q \text{ mod }p=q$ so $q \not \equiv 1 \text{ mod }p$, hence $n_{p}=1$. Now the [[the Sylow theorems|second Sylow theorem]] implies that $G$ has a [[normal subgroup]] $H$ of [[order of a group|order]] $p$. The [[the Sylow theorems|first Sylow theorem]] also guarantees the existence of some number of [[p-Sylow subgroup|q-Sylow subgroups]] of [[order of a group|order]] $q$; choose one of these arbitrarily and all it $K$. We know that $H \cap K = (e)$ because by [[Lagrange's Theorem]] [[intersection of subgroups is a subgroup|the intersection of H and K]] needs to have [[order of a group|ordering]] dividing both $p$ and $q$. \ Recall from [[Frobenius product of subgroups|this characterization]] that $HK \leq G$ if and only if $HK=KH$. Since $H \trianglelefteq G$, this property is certainly satisfied. This allows us to conclude that $HK=G$, for [[cardinality of frobenius product of subgroups]] implies $|HK|=|H| \cdot |K| / \cancel{|H \cap K|}^{1}=|G|$. \ Because $HK=G$ and $H \cap K = (e)$, we [[external semi-direct product|conclude that]] $G \cong H \rtimes_{\phi} K$, where $\phi \in \hom(K, \text{Aut}(H))$. Thus, the task is now to check: which choices of $\phi$ yield non-[[group isomorphism|isomorphic groups]]? \ We know that $H$ is a [[cyclic group]] of [[order of a group|order]] $p$. So, by [[automorphism group of cyclic group]] and [[finite subgroup of multiplicative group of a field is cyclic]], we see that $\text{Aut}(H) \cong C_{p-1}$. \ What [[group homomorphism|homomorphisms]] $f$ exist from $K \cong C_{q} \to C_{p-1} \cong \text{Aut}(H)$? By [[homomorphisms preserve structure]], the image $f^{(i)}(k)$ has order [[divides|dividing]] $|C_{q}|=q$. Thus $|f^{(i)}(k)| \in \{ 1,q \}$. Moreover, by [[Lagrange's Theorem]], the image $f^{(i)}(k)$ has order [[divides|dividing]] $|C_{p-1}|=p-1$. We see that if $q$ does not [[divides|divide]] $p-1$, then we can't have $f^{(i)}(k)=q$ and therefore $f^{(i)}(k)=1$ for all $k$ — i.e., $f^{(i)}$ is the [[group homomorphism|trivial homomorphism]]. In this case the [[external semi-direct product]] multiplication 'collapses into' the [[direct product of groups|direct product]] $C_q \times C_{p}$, which [[coprime characterization for when direct product is cyclic|is cyclic of order]] $pq$. So, in this case we conclude $G \cong C_{pq}$. \ Else suppose that $q$ *does* [[divides|divide]] $p-1$ and $f:C_{q} \to C_{p-1}$ is a nontrivial homomorphis[](group%20homomorphisms%20preserve%20structure.md)erve structure]], $f(C_{q})$ is a [[cyclic subgroup]] of $C_{p-1}$ whose order [[divides]] $q$. Of course, $q$ is [[prime number|prime]] so $|f(C_{q})|=q$. In fact, [[uniqueness of subgroups of finite cyclic groups]] tell us that $f(C_{q})$ is the unique [[subgroup]] of $C_{p-1}$ with [[order of a group|order]] $q$. \ We claim that any nontrivial choice of homomorphism $g:K \to \text{Aut}(H)$ yields an [[group isomorphism|isomorphic]] [[[](group%20homomorphisms%20preserve%20structure.md)rnal semi-direct product]] using $f$. We'll apply the [[sufficiency condition for external semidirect products to be isomorphic]]: rphic#^006d2a]]\ We know $g(K)$ is the unique (cyclic) [[subgroup]] of $C_{p-1}$ of order $q$. Since $g$ and $f$ have the same image, onto which they are [[group isomorphism|isomorphic]], for all $k \in K$, $f(k_{})=g(k_{}')$ for some $k' \in K$. In particular, define $h : K \to K$, $k_{} \mapsto$ $g^{-1}\big(f(k)\big)$. We see that $f(k_{}) = g \circ h$, implying that $f$ and $g$ yield [[group isomorphism|isomorphic]] groups. This completes the classification. ^48ddcd ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```