Every Group Of Order Pq Is Solvable

# Every Group of Order pq is Solvable

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:

\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 $G$.

- Let:

\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. $\blacksquare$