Group of Order p^2 q has Normal Sylow p-Subgroup

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Theorem

Let $p$ and $q$ be prime numbers such that $p \ne q$.

Let $G$ be a group of order $p^2 q$.


Then $G$ has a normal Sylow $p$-subgroup.


Proof

Let $n_p$ denote the number of Sylow $p$-subgroups in $G$.


From the Fourth Sylow Theorem:

$n_p \equiv 1 \pmod p$

From the Fifth Sylow Theorem:

$n_p \divides p^2 q$

where $\divides$ denotes divisibility.


Thus $n_p \in \set {1, q}$.


Suppose $p > q$.

Then:

$q \not \equiv 1 \pmod p$

and so $n_p \ne q$.

Hence $n_p = 1$.

From Sylow p-Subgroup is Unique iff Normal it follows that this Sylow $p$-subgroup is normal.

$\Box$


Suppose $q > p$.

Then $n_p$ could equal $q$ if $q \equiv 1 \pmod p$.

Let $n_q$ denote the number of Sylow $q$-subgroups in $G$.

From the Fourth Sylow Theorem:

$n_q \equiv 1 \pmod q$

and so $n_q$ is not a multiple of $q$.

From the Fifth Sylow Theorem:

$n_q \divides p^2 q$

It follows that:

$n_q \in \set {1, p, p^2}$

If $n_q = 1$ then it follows from Sylow p-Subgroup is Unique iff Normal that this Sylow $q$-subgroup is normal.

$\Box$


As $p \not \equiv 1 \pmod q$, it follows that $n_q \ne p$.

The final possibility to be explored is when $n_q = p^2$.

Then:

\(\ds p^2\) \(\equiv\) \(\ds 1\) \(\ds \pmod q\)
\(\ds \leadsto \ \ \) \(\ds q\) \(\divides\) \(\ds p^2 - 1\)
\(\ds \leadsto \ \ \) \(\ds q\) \(\divides\) \(\ds \paren {p + 1} \paren {p - 1}\) Difference of Two Squares
\(\ds \leadsto \ \ \) \(\ds q\) \(=\) \(\ds p + 1\) as $q > p$

But then $p$ and $q$ are consecutive prime numbers, and so:

$p = 2, q = 3$

This gives the specific case:

$\order G = 2^2 \times 3$

and the result follows from Normal Sylow p-Subgroups in Group of Order 12.

$\blacksquare$


Sources

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