1. ## p-sylow group abelian

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

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

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

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

What I got is that the options for $\displaystyle s_q$ are $\displaystyle 1, pq$ and for $\displaystyle s_p$ they are $\displaystyle 1,q^2$. 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 $\displaystyle G$ be a group such that $\displaystyle |G|=p^2q^2$ where $\displaystyle p,q$ are primes and $\displaystyle p>q$ and $\displaystyle q$ is not a divisor of $\displaystyle p\pm 1$. Show that $\displaystyle G$ is abelian.

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

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

What I got is that the options for $\displaystyle s_q$ are $\displaystyle 1, pq$ and for $\displaystyle s_p$ they are $\displaystyle 1,q^2$. 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 $\displaystyle p^2q^2$ are $\displaystyle 1,p,q,p^2,q^2,pq,p^2q,pq^2$.

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

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

How do we eliminate those other options?