# on the solvable groups of order $p^aq^b$ [closed]

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 $G$.

## closed as off-topic by Bjørn Kjos-Hanssen, Peter Mueller, abx, Derek Holt, Neil StricklandMar 10 '14 at 9:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Peter Mueller, abx, Derek Holt, Neil Strickland
If this question can be reworded to fit the rules in the help center, please edit the question.

• No, obviously not. Sym(3) is a counterexample. Sym(3) has no nontrivial quotient which is a 3-group. – Johannes Hahn Mar 9 '14 at 22:43

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.