Product of Subgroups of Prime Power Order

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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:

\(\ds \order {P Q}\) \(=\) \(\ds \frac {\order P \order Q} {\order {P \cap Q} }\) Order of Subgroup Product
\(\ds \) \(=\) \(\ds \frac {p^a p^b} {p^c}\)
\(\ds \) \(=\) \(\ds p^{a + b - c}\)


Aiming for a contradiction, suppose $P Q$ 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 $G$.

$\blacksquare$


Sources