Groups of order pq

In this post, we will classify groups of order pq, where p and q are primes with p<q. It turns out there are only two isomorphism classes of such groups, one being a cyclic group the other being a semidirect product.

Let G be the group of order pq.

Case 1: p does not divide q-1.

By Sylow’s Third Theorem, we have n_p\equiv 1\pmod p, n_p\mid q, n_q\equiv 1\pmod q, n_q\mid p.

Since n_q\mid p, n_q=1 or p. Since p<q and n_q\equiv 1\pmod q, we conclude n_q=1. Similarly, since n_p\mid q, n_p=1 or q. Since p\nmid q-1, n_p\equiv 1\pmod p implies n_p=1.

Let P, Q be the Sylow p-subgroup and Sylow q-subgroup respectively. By Lagrange’s Theorem, P\cap Q=\{1_G\}. Thus |P\cup Q|=p+q-1. Since \displaystyle pq\geq 2q>p+q>p+q-1, there is a non-identity element in G which is not in P\cup Q. Its order has to be pq, thus G is cyclic. Therefore G\cong\mathbb{Z}_{pq}.

Case 2: p divides q-1.

From previous arguments, n_q=1 hence Q is normal. Thus QP=PQ so PQ is a subgroup of G. \displaystyle |PQ|=\frac{|P||Q|}{|P\cap Q|}=pq, thus G=PQ. \text{Aut}(Q)\cong(\mathbb{Z}/q\mathbb{Z})^*\cong\mathbb{Z}_{q-1} is cyclic, thus it has a unique subgroup P' of order p, where P'=\{x\mapsto x^i\mid i\in\mathbb{Z}_q, i^p=1\}.

Let a and b be generators for P and Q respectively. Suppose the action of a on Q by conjugation is x\mapsto x^{i_0}, where i_0^p=1. (We may conclude this since the action of a on Q by conjugation is an automorphism which has order 1 or P, thus it lies in P'.)

If i_0=1, then G=P\times Q\cong\mathbb{Z}_{pq}.

If i_0\neq 1, then \displaystyle G=PQ=\langle P,Q\rangle=\langle a,b\mid a^p=b^q=1, aba^{-1}=b^{i_0}\rangle. Choosing a different i_0 amounts to choosing a different generator a for P, and hence does not result in a new isomorphism class.


Author: mathtuition88

Math and Education Blog

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