# Thread: Sylow group question

1. ## Sylow group question

1.Let G be a group with |G|=p^2*q,where p and q are distinct primes,show that G has a normal Sylow p-subgroups or a normal Sylow q-subgroups.

2.if |G| < 100 and G is non-abelian and simple,then show that |G|=60.

2. Originally Posted by seng
1.Let G be a group with |G|=p^2*q,where p and q are distinct primes,show that G has a normal Sylow p-subgroups or a normal Sylow q-subgroups.
let $mp+1$ and $nq+1$ be the number of Sylow p-subgroups and q-subgroups respectively. we just need to show that either

m = 0 or n = 0. so suppose, on the contrary, that m > 0 and n > 0 and consider two possible cases:

Case 1: $mp+1=q, \ nq+1=p.$ then we'll have $q>p$ and $p>q,$ which is impossible.

Case 2: $mp+1=q, \ nq+1=p^2.$ thus, since every two Sylow q-subgroups intersect in the identity element only, we'll have:

$p^2q=|G| > p^2(q-1) + p^2 = p^2q,$ which is impossible. Q.E.D.

2.if |G| < 100 and G is non-abelian and simple,then show that |G|=60.
to show this, you'll need more than just this fact that for primes p, q, groups of order $p^n, \ n >1,$ or $pq$ or $p^2q$ are not simple!

3. Actually i not really undestand how to do the second question,can you guide me to do?
I also will try my best to do...

4. Originally Posted by seng
Actually i not really undestand how to do the second question,can you guide me to do?
I also will try my best to do...
Just write out the numbers: $61,62,...,100$.
And start out canceling the ones you know are not it.
Cross out the primes because you are looking for the non-abelian ones.
Now cross out the prime powers because $p$-groups.
Now cross out the $pq$ ones.
And then the $p^2q$ ones.
And see if you can get rid of all those numbers.