Every Group of Order p^2q is Solvable

Every Group of Order p^2q is Solvable

 Proposition 1: Let $G$ be a group of order $p^2q$ where $p$ and $q$ are primes. Then $G$ is solvable.
• Proof: Let $G$ be a group of order $n: = p^2q$ and let $n_p$ denote the number of Sylow $p$-subgroups of $G$, and let $n_q$ denote the number of Sylow $q$-subgroups of $G$. There are three cases to consider.
• Case 1: Suppose that $p = q$. Then $G$ is a group of order $p^3$. So $G$ is a $p$-group. From the result on the Every p-Group is Solvable page we have that $G$ is solvable.
• Case 2: Suppose that $p > q$. Since $G$ is a finite group and $p$ is a prime such that $p \mid n = p^2q$ and $p \nmid q$ 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 a prime we have that either $n_p = 1$ or $n_p = q$. If $n_p = q$ then the above congruence implies that $q \equiv 1 \pmod p$, i.e., $p \mid q - 1$ which is a contradiction since $q - 1 > 0$ and $p > q$. Thus $n_p = 1$. So there is only one Sylow $p$-subgroup of $G$, call it $G_1$. By Lagrange's Theorem the possible orders of a subgroup of $G$ are $1$, $p$, $p^2$, $q$, $pq$, and $p^2q$. So $G_1$ has order $p^2$
• So $\{ 0 \} = G_0 \leq G_1 \leq G_2 = G$ is such that $G_0$ is (trivially) normal in $G_1$ and $G_1$ is normal in $G_2$, while $|G_1/G_0| = p^2$ and from the the result on the Every Group of Order p^2 is Abelian page we have that $G_1/G_0$ is abelian. Also, $|G_2/G_1| = q$ and since $q$ is a prime we have that $|G_2/G_1|$ is cyclic and thus abelian. So $G$ is solvable.
• Case 3: Suppose that $p < q$. Since $G$ is a finite group and $q$ is a prime such that $q \mid n = p^2q$ and $q \nmid p^2$ we have by The Third Sylow Theorem that:
(2)
\begin{align} \quad n_q & \equiv 1 \pmod q \\ \quad n_q & \mid p^2 \end{align}
• The only positive divisors of $p^2$ are $1$, $p$, and $p^2$. So $n_q$ is either $1$, $p$, or $p^2$. Note that if $n_q = p$ then from the above congruence we have that $p \equiv 1 \pmod q$. So $q \mid p - 1$ which cannot happen since $p -1 \neq 0$ and $p < q$.
• Thus either $n_q = 1$ or $n_q = p^2$.
• If $n_q = 1$ then similarly from above, then let $G_1$ be the unique Sylow $q$-subgroup of $G$. Then by Lagrange's Theorem it has order $q$. Since it is the only subgroup of $G$ with order $q$, it is normal in $G$. So the chain $\{ 0 \} = G_0 \leq G_1 \leq G_2 = G$ is such that $G_0$ is normal in $G_1$ and $G_1$ is normal in $G_2$. The composition factors are $G_1/G_0 \cong G_1$ which is abelian since it has prime order $q$, and $G_2/G_1$ which has order $p^2$ - which is abelian since every group of order $p^2$ is abelian. So $G$ is solvable.
• If $n_q = p^2$. Then $G$ has $p^2$ Sylow $q$-subgroups, i.e., $G$ has $p^2$ subgroups of order $q$. Every element except for the identity in each subgroup has order $q$. So $G$ has $p^2 \cdot (q - 1)$ elements of order $q$. But $p^2 \cdot (q - 1) = p^2q - p^2 = n - p^2$. So $G$ has $p^2$ elements not of order $q$.
• Since $p \neq q$, these $p^2$ elements must lie in a Sylow $p$-subgroup as the intersection of a Sylow $p$-subgroup and a Sylow $q$-subgroup with $p \neq q$ is trivial. Let $G_1$ denote this unique group of order $p^2$. Then it is normal in $G$, and $\{ 0 \} = G_0 \leq G_1 \leq G_2 = G$ is a chain of subgroups with $G_0$ being (trivially) normal in $G_1$, $G_1$ being normal in $G_2$, and $|G_1/G_0| = p^2$, $|G_2/G_1| = q$, both of which must then be abelian. So $G$ is solvable.