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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

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closed as off-topic by Bjørn Kjos-Hanssen, Peter Mueller, abx, Derek Holt, Neil Strickland Mar 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
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  • 2
    $\begingroup$ No, obviously not. Sym(3) is a counterexample. Sym(3) has no nontrivial quotient which is a 3-group. $\endgroup$ – Johannes Hahn Mar 9 '14 at 22:43
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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.

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  • $\begingroup$ oops, had that typed up before seeing Johannes' comment. $\endgroup$ – Achim Krause Mar 9 '14 at 22:54

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