We know that if $ p$ is a prime number then $ O^p (G) $ is the smallest normal subgroup of $ G $ such that $ G/O^p (G) $ is a $ p $-group.


Now let $ G $ be a finite group of order $ p^aq^b $ where $ p $ and $ q $ are prime numbers. Is this true that $ O^p (G) \ne G$ and also $ O^q (G)\ne G $? We note that by the solvability of $ G $ we know that one of them is not equal to $ G $.


Thanks for your help



on the solvable groups of order $p^aq^b$
The alternating group $A\_4$ is a counterexample: It has order $2^2\cdot 3$, so $O^2(A\_4)$ will contain an order $3$ element. But any order $3$ element of $A\_4$ generates the whole group as a normal divisor, as is seen by playing around with permutations. So $O^2(A\_4) = A\_4$.


If you put some additional assumptions on $p$ and $q$, sometimes the Sylow theorems could give you a positive answer.


