Groups of order $pq$ without using Sylow theorems

Question

If $|G| = pq$, $p,q$ primes, $p \gt q, q \nmid p-1 $, then how do I prove $G$ is cyclic without using Sylow's theorems?

Answer 1

^{This solution will mostly use Lagrange and the fact that |G| has so few divisors. This is mostly an example of how looking at cosets and permutations is useful. Sylow's theorem is just an example of doing that in a more general situation. Like Sylow's theorem, we gain a lot by finding fixed points of permutations.}

By Lagrange's theorem, an element of G has order 1, p, q, or pq. There is only one element of order 1. If there is an element of order pq, then G is the cyclic group generated by it. Otherwise, every non-identity element of G has order p or q, and there is at least one such element, x. Let H be the subgroup generated by x.

Case I: If x has order q, then Lagrange says that there are p cosets of H in G and x acts as a permutation on them. The order of that permutation is either 1 or q (by Lagrange again), but q > p is impossibly big, and so x leaves all the cosets gH alone. That means H is normal in G, because xgH = gH and so g⁻¹xg in H for all g in G, and H is generated by x. Let y be any element of G not contained in H. Then y normalizes H, and so conjugation by y is an automorphism of H. The automorphism group of H has order q−1, and so the order of that automorphism is a divisor of gcd(q−1, pq) = 1 by Lagrange, so conjugation by y is the identity automorphism on H. In other words, y⁻¹xy = x and xy = yx. In particular, x and y commute and xy has order pq, so G is cyclic.

Case II: If x has order p, then there are q cosets of H in G, by Lagrange. Note that xH = H, so x does not move the coset 1H. We examine two subcases based on whether it leaves any other cosets alone:

Case IIa: Suppose x moves all the other cosets. By Lagrange, those other cosets are collected into p-tuples (the "orbits" of x), and so we get that q = 1 + kp, where k is the number of orbits. This explicitly contradicts the non-divisibility hypothesis.

Case IIb: Suppose x leaves at least one more coset alone, say yH for some y not contained in H. In other words, xyH = yH, or y⁻¹xy is in H. This means that y acts by conjugation on the elements of H. However, the automorphism group of H has order p−1, and so the automorphism by y is a divisor or p−1 and a divisor of pq, but gcd(p−1, pq) = 1. Hence conjugation by y is the identity automorphism: y⁻¹xy = x and xy = yx. In particular, x and y commute and xy has order pq, so G is cyclic.

Answer 2

Once again, Burnside's book (Theory of Groups of Finite Order) contains the classification of groups of order $pq$ before it tackles Sylow's Theorems. In the Dover print of the Second Edition this is contained in Page 48 (Section 36), with Sylow Theorems not occurring until section 120 (pages 149-151).

The argument relies on Cauchy's Theorem; here's the quote. I put in brackets the modern terms for some of the ones used by Burnside.

A group of order $pq$ must contain a subgroup of order $p$ and a subgroup of order $q$. If the latter is not self-conjugate [normal] it must be one of $p$ conjugate sub-groups, which contain $p(q-1)$ distinct operations [elements] of order $q$. The remaining $p$ operations [elements] must constitute a subgroup of order $p$, which is therefore self-conjugate [normal]. A group of order $pq$ has therefore either a self-conjugate subgroup [normal subgroup] of order $p$, or one of order $q$. Take $p\lt q$, and suppose first that there is a self-conjugate [normal] subgroup $\{P\}$ [$\langle P\rangle$] of order $p$. Let $Q$ be an operation [element] of order $q$. Then:

$$\begin{align*} Q^{-1}PQ &= P^{\alpha}\\ Q^{-q}PQ^q &= P^{\alpha^q},\\ \alpha^q\equiv 1&\pmod{p},\\ \text{and therefore }\alpha\equiv 1&\pmod{p}. \end{align*}$$

In this case, $P$ and $Q$ are permutable [commute] and the group is cyclical. Suppose secondly that there is no self-conjugate [normal] subgroup of order $p$. There is then necessarily a self-conjugate [normal] subgroup $\{Q\}$ of order $q$; and if $P$ is an operation of order $p$, $$\begin{align*} P^{-1}QP &= Q^{\beta}\\ P^{-p}QP^{p} &= Q^{\beta^p}\\ \beta^p\equiv 1 &\pmod{q}. \end{align*}$$ If $q\not\equiv 1\pmod{p}$ this would involve $\beta=1$, and $\{P\}$ would be self-conjugate, contrary to supposition. Hence if the group is non-cyclical, $q\equiv 1 \pmod{p}$ and $P^{-1}QP=Q^{\beta}$, where $\beta$ is a root, other than unity, of the congruence $\beta^p\equiv 1\pmod{p}$. Between the groups defined by [$E$ is the identity] $$\begin{align*} P^p&=E, &\qquad Q^q&=E,&\qquad P^{-1}QP &= Q^{\beta},\\ \text{and }P'^p&=E, & Q'^q&=E, & P'^{-1}Q'P'&=Q^{\beta^a}, \end{align*}$$ a simple isomorphism is established by taking $P'$ and $P^a$, $Q'$ and $Q$, as corresponding operations [elements]. Hence when $q\equiv 1\pmod{p}$ there is a single type of non-cyclical group of order $pq$.

Answer 3

Let $G$ be a group of order $pq$. Then order of element should be $1,p,q,pq$.

It is sufficient to show existance of subgroups of order $p$ and $q$.

• If all elements of $G$ are of order $p$ (except identity), then consider a subgroup $H$ of order $p$ and take $y\in G\backslash H$, let $K=\langle y\rangle$.

  Now $H$ can not be normalised by $y$ in $G$, otherwise $HK$ will be an abelian subgroup of $G$ of order $p^2$, contradiction. Therefore, $yHy^{-1}$ is another conjugate subgroup of order $p$. Now number of conjugates of $H$ will be the index $[G\colon N(H)]$ of normalizer of $H$ in $G$; since there are at least two conjugates, ($H,yHy^{-1}$) so $[G\colon N(H)]>1$, we deduce that $N(H)=H$. Therefore there are exactly $q$ conjugates of $H$. The non-trivial elements in collection of conjugates of $H$ will be $(p-1)q$. Then take element $z$ of $G$ outside these counted elements, proceed further for $\langle z\rangle$. We will get $(p-1)q$ non-trivial elements in the collection of all conjuagtes of $\langle z\rangle$. After some finite steps, say $m\geq 1$, we will get all non-trivial elements of $G$ (of order $p$); they will be $m(p-1)q$ in number.

  Therefore, $m(p-1)q+1=pq$, which is not valid, since $pq-m(p-1)q$ is divisible by $q$ (here all terms are non-zero).

  Therefore, we conclude that all non-trivial elements of $G$ can-not be of same order $p$.

  Similarly, we can conclude that all non-trivial elements can not have same order $q$.

  • If $G$ has element of order $pq$ then it will be cyclic.

  • Otherwise, now we must have atleast one element of order $q$, hence a subgroup $Q$ of order $q$. This subgroup must be be unique (hence normal): if $Q_1$ is another subgroup of order $q$, then $\langle Q, Q_1\rangle \supseteq QQ_1$, so $|\langle Q,Q_1\rangle | \geq |QQ_1|=q.q/1=q^2>qp = |G|$, contradiction.

  Take a subgroup $P$ of order $p$. Now $Q \triangleleft G$, $P\leq G$, hence $PQ\leq G$; in fact this is equality - $PQ=G$ (computing orders). So $G=Q\rtimes P$. Using two basic theorems on semi-direct product of groups ( Ref. Alperin-Bell - Groups and Representations), we can conclude that $G=Q\times P$, hence it is cyclic.

  (The crucial step stated in proof is existance of subgroups of order $p$ and $q$. Using theorems on semi-direct products doesn't uses Sylow's theorems.)

Answer 4

This doesn't use any Sylow, as desired.

In general, if $|G|=pq$, with $p$ and $q$ distinct primes, then $G$ has center either trivial or the whole group (see here).

If $Z(G)$ is trivial, then the Class Equation reads: $$pq=1+k_pp+k_qq \tag 1$$ where $k_i$ are the number of conjugacy classes of size $i$. Now, there are exactly $k_qq$ elements of order $p$ (they are the ones in the conjugacy classes of size $q$)$^\dagger$. Since each subgroup of order $p$ contributes $p-1$ elements of order $p$, and two subgroups of order $p$ intersect trivially, then $k_qq=m(p-1)$ for some positive integer $m$ such that $q\mid m$ (because by assumption $q\nmid p-1$). Therefore, $(1)$ yields: $$pq=1+k_pp+m'q(p-1) \tag 2$$ for some positive integer $m'$; but then $q\mid 1+k_pp$, namely $1+k_pp=nq$ 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 $n=1$ (which in turn implies $m'=1$). So, $1+k_pp=q$: but this is a contradiction, because $p\nmid q-1$ for $p>q$. So we are left with $G$ Abelian, and hence (Cauchy) cyclic.

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$^\dagger$An element lies in a conjugacy class of size $q$ (respectively, $p$) if and only if its order is $p$ (respectively, $q$). This claim follows from the Orbit-Stabilizer Theorem and the fact that, for $g\in G\setminus\{e\}$, it is $\langle g\rangle=C_G(g)$.

Answer 5

For any group $G$ and a normal subgroup $H$, $G$ acts on $H$ by conjugation as automorphisms of $H$. This gives a map from $G \to \text{Aut}(H)$ via the permutation representation with kernel $C_G(H)$. So by the First Isomorphism Theorem we have $G/C_G(H) \hookrightarrow \text{Aut}(H)$.

Now let $G$ be a group of order $pq$ as above. Clearly if $Z(G)$ is nontrivial then $G/Z(G)$ is cyclic, and thus $G$ is abelian. So we may assume that $Z(G) = \{e\}$. If every element of $G$ besides the identity has order $q$, then the size of each conjugacy class must be $p$ for every nontrivial element. Then we would have the class equation $pq = 1 + kp$ for some $k \in \mathbb{Z}$. But clearly this is impossible as $p$ divides $pq$ but not $1 + kp$. So $G$ must have an element of order $p$, say $g$. Define $H = \langle x \rangle$. Then $|G:H| = q$, and since $q$ is the smallest prime dividing $|G|$, we have that $H$ must be normal. So $N_G(H) = G$ and since $Z(G) = \{e\}$, we must have that $C_G(H) = H$. Then by the above work, $G/C_G(H) \hookrightarrow \text{Aut}(H)$. But since $H$ is cyclic, we have that $\text{Aut}(H) \simeq (\mathbb{Z}/p\mathbb{Z})^\times$ by a standard result in group theory. Since $C_G(H) = H$, $|G/C_G(H)| = q$. But $|\text{Aut}(H)|=p-1$. Since $G/C_G(H) \hookrightarrow \text{Aut}(H)$, this implies that $\text{Aut}(H)$ has a subgroup of order $q$, but this would imply that $q \mid p-1$, which is a contradiction. Hence $G$ must be abelian. From here you just need a single element of order $p$ and one of order $q$. Their product has order $pq$ and thus generates $G$.

Answer 6

(We only consider the complex representation) Suppose that $G$ is non-Abelian. Then there is an irred repr. $\rho$ s.t. it is $d$-dim'l, $d>1$. By dimension theorem, we get $d=p,q$, or $pq$, but we know that $d$ must be $p$; otherwise by the property that $\sum_{\chi'\in\text{Irr}(G)}d_\chi^2=|G|$, we will get $d^2>pq$ because of the condition $p<q$. So the dimension of the irred repr's can only be $1$ or $p$, and hence we get $$ mp^2+n=pq, $$ where $m,n$ stand for the number of $p$-dim'l and $1$-dim'l irred repr's, respectively. Furthermore, let $\mathcal{L}(G)$ be the set of all $1$-dim'l characters of $G$. Then we have $$ n=|\mathcal{L}(G)|=|\text{Irr}(G/[G,G])|=\frac{|G|}{|[G,G]|}, $$ where $[G,G]$ is the commutator subgroup of $G$. Since $G$ is non-Abelian, $[G,G]$ is nontrivial. So $n=1,p$, or $q$.

[Case 1]: If $n=1$, then $mp^2=pq-1$. So we get $pq\equiv1\;(\text{mod}\,p^2)$. However, since $q\equiv 1\;(\text{mod}\,p^2)$, we get $$ pq\equiv p\equiv 1 (\text{mod}\,p^2), $$ which is a contradiction.

[Case 2]: If $n=p$, then $\displaystyle m=\frac{q-1}{p}\notin\Bbb N$ because $p\nmid(q-1)$. Contradiction.

[Case 3]: If $n=q$, then $\displaystyle m=\frac{(p-1)q}{p^2}\notin\Bbb N$, still a contradiction.

Therefore, in conclusion, we find $G$ must be Abelian, and hence, by the fundamental theorem of Abelian groups, $G\cong \Bbb Z_{pq}$ or $\Bbb Z_p\times \Bbb Z_q$. So in any case $G$ is a cyclic group.

Answer 7

If $|G|=pq, p,q$ prime, $q\lt p,q\nmid p-1$, then $G$ is cyclic.

By Cauchy we have elements of orders $p,q$. The cyclic subgroup of order $p$ is normal, since its index is the smallest prime dividing $|G|$. We have two cyclic subgroups of orders $p,q$. Their intersection is trivial and their product is the whole group. So we have a semidirect product $G\cong\mathbb Z_p\rtimes\mathbb Z_q$. That $q\nmid p-1$ means the action is trivial, or we just have the direct product $\mathbb Z_p\times\mathbb Z_q\cong\mathbb Z_{pq}$.

Answer 8

This uses group action arguments, only, hence no one Sylow's theorem.

By (a corollary of) Cauchy's theorem, there are subgroups $H\cong C_p$ and $K\cong C_q$. Now, $H$ must be unique, as two of them (say $H$ and $\tilde H$) would intersect trivially and hence would give rise to the subset $H\tilde H\subseteq G$, of size $p^2>pq$, a contradiction. Therefore, $H\unlhd G$ and, since $HK=G$ and $H\cap K=\{1\}$, then $G\cong K\ltimes H$.

Now, under the $K$-action on $H$ by automorphisms, for every $k\in K$ the integer $\left|\operatorname{Fix}(k)\right|$ is $1$ or $p$ (by Lagrange's theorem, being $\operatorname{Fix}(k)$ a subgroup of $C_p$), and hence: \begin{alignat}{1} \sum_{k\in K}\left|\operatorname{Fix}(k)\right| &= l+(q-l)p \\ &= pq-l(p-1) \\ \tag1 \end{alignat} for some integer $l$, $0\le l\le q$. By Burnside's (counting) lemma, $q$ divides the RHS of $(1)$; so, if $l\ne 0$, since by assumption $q\nmid p-1$, then necessarily $q\mid l$ and hence $q\le l$. But $l\le q$, so finally $l=q$, which replaced in $(1)$ yields: $$\text{nr. of orbits} =1$$ a contradiction, because $K$ cannot act transitively on $H$ of order $p>q=|K|$. Therefore, $l=0$ and hence there are $p$ singleton orbits, namely the action is trivial and $G\cong$ $K\ltimes H=$ $K\times H\cong$ $C_q\times C_p\cong$ $C_{qp}$.