# Math Help - p-sylow group abelian

1. ## p-sylow group abelian

I am having some difficulties here and would like some guidance.

Let $G$ be a group such that $|G|=p^2q^2$ where $p,q$ are primes and $p>q$ and $q$ is not a divisor of $p\pm 1$. Show that $G$ is abelian.

What I can determine so far is that $G$ has p-sylow group of order $p^2$ and a q-sylow group of order $q^2$ and that both must be abelian.

Additionally I can try to eliminate the possibilities for $s_p,s_q$ (the number of p,q-sylow groups).

What I got is that the options for $s_q$ are $1, pq$ and for $s_p$ they are $1,q^2$. but I can't get anywhere from here.

Any help appreciated.

2. Originally Posted by skyking
I am having some difficulties here and would like some guidance.

Let $G$ be a group such that $|G|=p^2q^2$ where $p,q$ are primes and $p>q$ and $q$ is not a divisor of $p\pm 1$. Show that $G$ is abelian.

What I can determine so far is that $G$ has p-sylow group of order $p^2$ and a q-sylow group of order $q^2$ and that both must be abelian.

Additionally I can try to eliminate the possibilities for $s_p,s_q$ (the number of p,q-sylow groups).

What I got is that the options for $s_q$ are $1, pq$ and for $s_p$ they are $1,q^2$. but I can't get anywhere from here.

Any help appreciated.

The given conditions give us that there are exactly one Sylow p-sbgp. and one Sylow

q-sbgp. which are thus normal.

Since they're intersection is obviously trivial their product is their own direct product, and since

the product of their orders is the order of G then G is the direct product of

abelian groups...

Tonio

3. Originally Posted by tonio
The given conditions give us that there are exactly one Sylow p-sbgp. and one Sylow

q-sbgp.
Tonio
Thanks for the reply, I understand it all besides how you get the above conclusion.

The divisors of $p^2q^2$ are $1,p,q,p^2,q^2,pq,p^2q,pq^2$.

For $S_q$ we need to have $s_q=1(modq), s_q\leq p^2$ so $p,q,q^2,p^2,p^2q,pq^2$ are all out, which leaves us with $1,pq$

Similarly for $s_p$ we need $s_p=1(modp), s_p\leq q^2$ which leaves us with $1,q^2$

How do we eliminate those other options?