Every Group Of Order Pq Is Solvable

Table of Contents

Proposition 1: Let

$G$

be a group of order

$pq$

where

$p$

and

$q$

are primes. Then

$G$

is solvable.

Proof: Let $$G$$ be a group of order $$pq$$ for some primes $$p$$ and $$q$$ .

Case 1: Suppose that $$p = q$$ . Then $$G$$ is a group of order $$p^2$$ . From the result on the Every Group of Order p^2 is Abelian we have that $$G$$ is an abelian group. But every abelian group is a solvable group and so $$G$$ is solvable.

Case 2: Suppose that $p \neq q$ . Without loss of generality, assume that $$p > q$$ .

Let $$n_p$$ denote the number of Sylow $$p$$ -subgroups of $$G$$ and write $$n = |G| = pq$$ . Since $$G$$ is a finite group, $p \mid n = pq$ and $p \nmid q$ (since $$p$$ and $$q$$ are distinct primes from $$p > q$$ ) we have by The Third Sylow Theorem on The Sylow Theorems page that:

(1)

\begin{align} \quad n_p &\equiv 1 \pmod p \\ \quad n_p &\mid q \end{align}

Since $n_p \mid q$ and since $$q$$ is prime we have that either $$n_p = 1$$ or $$n_p = q$$ . If $$n_p = q$$ then from the above congruence we have that $q \equiv 1 \pmod p$ , i.e., $p \mid (q - 1)$ . So $p \leq q - 1$ - a contradiction since $$p > q$$ . Thus we must have that $$n_p = 1$$ .

So $$G$$ has only one Sylow $$p$$ -subgroup. Let $$G_1$$ denote this Sylow $$p$$ -subgroup. By Lagrange's Theorem, the only possible orders of subgroups of $$G$$ are $$1$$ , $$p$$ , $$q$$ , and $$pq$$ . So the Sylow $$p$$ -subgroup $$G_1$$ must have order $$p$$ .

Since $$G$$ is a finite group and $$G_1$$ is the only subgroup of $$G$$ with order $$p$$ we have by the result on the Subgroups of Finite Groups with Unique Order are Normal Subgroups that $$G_1$$ is a normal subgroup of $$G$$ .

(2)

\begin{align} \quad \{ e \} = G_0 \leq G_1 \leq G_2 = G \quad (*) \end{align}

Then $$G_0$$ is trivially a normal subgroup in $$G_1$$ and $$G_1$$ is a normal subgroup of $$G$$ from the remarks made above. Furthermore, the composition factors of the above chain are $G_1/G_0 = G_1/\{ e \} \cong G_1$ , which is abelian since $$G_1$$ is a group of prime order $$p$$ (and is thus cyclic $\Rightarrow$ abelian). On the otherhand, we see that $$|G_2/G_1| = |G/G_1| = |G|/|G_1| = pq/p = q$$ . So $$G_2/G_1$$ is a group of prime order $$q$$ and is thus also cyclic and hence abelian.

So the chain at $$(*)$$ is such that $$G_i$$ is a normal subgroup of $G_{i+1}$ for all $0 \leq i \leq 1$ and $G_{i+1}/G_i$ is abelian for all $0 \leq i \leq 1$ . Thus $$G$$ is a solvable group. $\blacksquare$

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