# Groups of order $pq$ without using Sylow theorems

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?

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Try counting elements of order $p$ and elements of order $q$ - there can't be any elements of order $pq$ (why?) and the subgroups of prime order are disjoint apart from the identity (why?) so the identity plus elements of order $p$ plus elements of order $q$ form the whole group. – Mark Bennet Sep 24 '11 at 8:10
just curious but, why would you not want to use the sylow theorems? – user12205 Sep 24 '11 at 11:32

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] subgruop $\{Q\}$ 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 $pq$.

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If $Q_1, Q_2$ are two subgroups of order q, then $<Q_1, Q_2> \supseteq Q_1Q_2$, and so $|<Q_1, Q_2>|\geq |Q_1Q_2|=q.q/1=q^2 >qp=|G|$, contradiction; so there is unique subgroup of order $q$, hence normal. – Marshal Kurosh Sep 29 '11 at 5:30
@MarshalKurosh: Perhaps you can contact your local medium and let Burnside know, instead of letting me know? – Arturo Magidin Sep 29 '11 at 13:20

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.

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In Case 1, Why can you claim that $x$ acts as a permutation on the set of left cosets of $H$ ? i.e. suppose $\{g_1H, \ldots g_pH\}$ are left cosets of H, what goes wrong if $xg_iH=xg_jH$ for $1 \leq i < j \leq p$ ? – the8thone Oct 14 '14 at 20:31

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 $G$.

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In your argument you have not used the fact that $p \neq q$, in fact if $p=q$, A group of order $p^2$ is not necessarily cyclic , i.e. it can be isomorphic to $\mathbb{Z}_p \times \mathbb{Z}_p$ – the8thone Oct 14 '14 at 20:18
@the8thone That fact is used in the last line. The product of two elements of order $p$ need not be of order $p^2$. – Brandon Carter Oct 14 '14 at 20:53

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.)

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(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.

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