If $G$ is order $p^2q$, where $p$, $q$ are primes, prove that either a Sylow $p$-subgroup or a Sylow $q$-subgroup must be normal in G

Question

If $G$ is order $p^2q$, where $p$, $q$ are primes, prove that either a Sylow $p$-subgroup or a Sylow $q$-subgroup must be normal in $G$.

Answer 1

(Sylow). Let $G$ be a finite group of order $p^km$ with $p\nmid m$, and $r$ be the number of Sylow $p$-subgroups of $G$. Then (1) every Sylow $p$-subgroup is conjugate to one Sylow $p$-subgroup. (2) $r\mid m$ and $r\equiv1\pmod p$.

In this case, the number of Sylow $p$-subgroup of $G$, say $r_1$, satisfies $r_1\mid q$ and $r_1\equiv1\pmod p$, but $q$ is a prime, so $r_1=1$ or $r_1=q$. If $r_1=1$, then there is unique Sylow $p$-subgroup, and it is normal. Otherwise $q=r_1\equiv1\pmod p$, and let $r_2$ be the number of Sylow $q$-subgroup of $G$. Still we have $r_2\mid p^2$ and $r_2\equiv1\pmod q$. If $r_2=p^2$, then $r_2\equiv1\pmod q$ implies $p\equiv\pm1\pmod q$. This contradicts $q\equiv1\pmod p$. Similarly $r_2\ne p$, so we have $r_2=1$, as desired.

Answer 2

Case $(1)$ If $p \gt q$, then a Sylow $p$ subgroup of order $p^2$ would have index $q$ which is the smallest prime dividing $|G|$ , hence the Sylow $p$ subgroup would be normal.

Case $(2)$ If $q\gt p^2$ , then the only allowed no. of Sylow $q$ subgroups is $1$ since $qk+1\gt p^2$ for $p\in \mathbb{N}$. Hence the Sylow $q$ subgroup is normal.

Case $(3)$ If $p\lt q\lt p^2$.

Let us assume that the neither the Sylow $p$ subgroup nor Sylow $q$ subgroup are unique.

Then the no. of Sylow $q$ subgroup must be $p^2$ since the divisors of $p^2$ are $1,p,p^2$.

Any two of these $p^2$ Sylow $q$ subgroups must have trivial intersection , since the non-trivial elements of a group of prime order will generate the whole group.

Hence total no. of elements in all of these Sylow $q$ subgroups$=p^2(q-1)+1 \quad (*)$

Now, let $P_1, P_2$ be two different Sylow $p$ subgroups.

So we must have ,

$P_1\setminus P_2 \neq \emptyset$

Now, a Sylow $p$ subgroup and Sylow $q$ subgroup will intersect trivially since order of every element in the intersection will divide $p^2$ and $q$ which are relatively prime.

Adding the elements of $P_2$ (except the identity) to the total counting of elements of Sylow $q$ subgroups in $(*)$ , we must have total no. of elements

$=p^2q-p^2+1+p^2-1=p^2q$

But how to accomodate the elements of $P_1\setminus P_2$ in a group of order $p^2q$ with the above counting of elements.

So our assumotion is wrong, one of the Sylow subgroups must be unique and hence normal.

Answer 3

This is a supplement to user735332's answer.

First notice that $q\equiv 1\pmod p$ implies $p<q$.

Then consider the case $r_2=p^2$, sure it has the contradiction if $p\equiv 1\pmod q$, since it implies $p=1$, which contradicts that $p$ is prime. However, if $p\equiv -1\pmod q$, then $q\equiv 1\pmod p$ can not lead to contradiction, since $p=2$ and $q=3$ is a counterexample. Although that, the contradiction still holds except such case, since $p\equiv -1\pmod q$ and $q\equiv 1\pmod p$ implies $p=q-1$, $p$ and $q$ are both prime, so we just need to check that the case $p=2$ and $q=3$ still leads to contradiction. For the case $p=2$ and $q=3$, we have $r_1=3$ and $r_2=4$, which mean that there are $4$ subgroup with order $3$ and $3$ subgroup with order $4$, inplies that there are at least $1+4*(3-1)+3*(4-1)=18$ elements in $G$, which contradicts that order of $G$ is $12$. Hence, we done the contradiction for the case $r_2=p^2$.