Product of Subgroups of Prime Power Order

Theorem
Let $p$ be a prime number.

Let $G$ be a group of order $p^a k$, where:
 * $a \in \Z_{>0}$ is a (strictly) positive integer
 * $p$ is not a divisor of $k$.

Let $P \le G$ be a subgroup of $G$ of order $p^a$.

Let $Q \le G$ be a subgroup of $G$ of order $p^b$, where $0 < b \le a$.

Let it be the case that $Q$ is not a subgroup of $P$.

Then $P Q$ is not a subgroup of $G$.

Proof
From Intersection of Subgroups is Subgroup, $P \cap Q$ is a subgroup of $P$.

Thus:
 * $\order {P \cap Q} = p^c$ for some $c \in \Z$ such that $0 \le c \le a$

where $\order {P \cap Q}$ denotes the order of $P \cap Q$.

We have:

$P Q$ is a subgroup of $G$.

By Lagrange's Theorem (Group Theory):
 * $\order {P Q} \divides \order G$

where $\divides$ denotes divisibility.

We have that $\order {P Q}$ is a power of $p$.

The highest power of $p$ which divides $\order G$ is $p^a$.

Thus $P Q$ could have order $p^a$ at most.

Thus:
 * $a + b - c \le a$

That is:
 * $b \le c$

But $P \cap Q$ is a subgroup of $P$.

Hence it must be the case that $b = c$.

Thus:
 * $P \cap Q = Q$

and so $Q$ is a subgroup of $P$.

This contradicts the fact that $Q$ is not a subgroup of $P$.

It must follow that $P Q$ is not a subgroup of $G$.