Given that $G$ is a finite group, prove that $F(G/Z(G))$ = $F(G) / Z(G)$.












2












$begingroup$


I have seen the equation in the title while reading about the Fitting subgroup, but I can't work out why those two quantities are equal.




Could anyone please explain the proof?




Thank you very much.



I know that $F(G)$ is the product of all $O_p(G)$ for all primes $p$.




Is it true that $O_p(G)/Z(G)$ = $O_p(G/Z)$ for each prime $p$?











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$endgroup$












  • $begingroup$
    What do you mean by $O_p(G)$?
    $endgroup$
    – Shaun
    Dec 22 '18 at 5:17










  • $begingroup$
    Here's a question concerning the notation $O_p(G)$.
    $endgroup$
    – Shaun
    Dec 22 '18 at 6:20
















2












$begingroup$


I have seen the equation in the title while reading about the Fitting subgroup, but I can't work out why those two quantities are equal.




Could anyone please explain the proof?




Thank you very much.



I know that $F(G)$ is the product of all $O_p(G)$ for all primes $p$.




Is it true that $O_p(G)/Z(G)$ = $O_p(G/Z)$ for each prime $p$?











share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you mean by $O_p(G)$?
    $endgroup$
    – Shaun
    Dec 22 '18 at 5:17










  • $begingroup$
    Here's a question concerning the notation $O_p(G)$.
    $endgroup$
    – Shaun
    Dec 22 '18 at 6:20














2












2








2





$begingroup$


I have seen the equation in the title while reading about the Fitting subgroup, but I can't work out why those two quantities are equal.




Could anyone please explain the proof?




Thank you very much.



I know that $F(G)$ is the product of all $O_p(G)$ for all primes $p$.




Is it true that $O_p(G)/Z(G)$ = $O_p(G/Z)$ for each prime $p$?











share|cite|improve this question











$endgroup$




I have seen the equation in the title while reading about the Fitting subgroup, but I can't work out why those two quantities are equal.




Could anyone please explain the proof?




Thank you very much.



I know that $F(G)$ is the product of all $O_p(G)$ for all primes $p$.




Is it true that $O_p(G)/Z(G)$ = $O_p(G/Z)$ for each prime $p$?








group-theory finite-groups proof-explanation






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edited Dec 29 '18 at 3:09









the_fox

2,89021537




2,89021537










asked Apr 7 '17 at 3:09









K.OK.O

114




114












  • $begingroup$
    What do you mean by $O_p(G)$?
    $endgroup$
    – Shaun
    Dec 22 '18 at 5:17










  • $begingroup$
    Here's a question concerning the notation $O_p(G)$.
    $endgroup$
    – Shaun
    Dec 22 '18 at 6:20


















  • $begingroup$
    What do you mean by $O_p(G)$?
    $endgroup$
    – Shaun
    Dec 22 '18 at 5:17










  • $begingroup$
    Here's a question concerning the notation $O_p(G)$.
    $endgroup$
    – Shaun
    Dec 22 '18 at 6:20
















$begingroup$
What do you mean by $O_p(G)$?
$endgroup$
– Shaun
Dec 22 '18 at 5:17




$begingroup$
What do you mean by $O_p(G)$?
$endgroup$
– Shaun
Dec 22 '18 at 5:17












$begingroup$
Here's a question concerning the notation $O_p(G)$.
$endgroup$
– Shaun
Dec 22 '18 at 6:20




$begingroup$
Here's a question concerning the notation $O_p(G)$.
$endgroup$
– Shaun
Dec 22 '18 at 6:20










1 Answer
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$begingroup$

The group $F(G)/Z(G)$ is a nilpotent normal subgroup of $G/Z(G)$, so $F(G)/Z(G)$ is contained in $F(G/Z(G))$. What about the other direction? Write $F(G/Z(G))=H/Z(G)$. Since $H/Z(G)$ is nilpotent (because it is the Fitting subgroup of some group), it follows that $H$ is nilpotent (do you understand why?). Then $H leq F(G)$, which proves the other direction we wanted.






share|cite|improve this answer









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    $begingroup$

    The group $F(G)/Z(G)$ is a nilpotent normal subgroup of $G/Z(G)$, so $F(G)/Z(G)$ is contained in $F(G/Z(G))$. What about the other direction? Write $F(G/Z(G))=H/Z(G)$. Since $H/Z(G)$ is nilpotent (because it is the Fitting subgroup of some group), it follows that $H$ is nilpotent (do you understand why?). Then $H leq F(G)$, which proves the other direction we wanted.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      The group $F(G)/Z(G)$ is a nilpotent normal subgroup of $G/Z(G)$, so $F(G)/Z(G)$ is contained in $F(G/Z(G))$. What about the other direction? Write $F(G/Z(G))=H/Z(G)$. Since $H/Z(G)$ is nilpotent (because it is the Fitting subgroup of some group), it follows that $H$ is nilpotent (do you understand why?). Then $H leq F(G)$, which proves the other direction we wanted.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        The group $F(G)/Z(G)$ is a nilpotent normal subgroup of $G/Z(G)$, so $F(G)/Z(G)$ is contained in $F(G/Z(G))$. What about the other direction? Write $F(G/Z(G))=H/Z(G)$. Since $H/Z(G)$ is nilpotent (because it is the Fitting subgroup of some group), it follows that $H$ is nilpotent (do you understand why?). Then $H leq F(G)$, which proves the other direction we wanted.






        share|cite|improve this answer









        $endgroup$



        The group $F(G)/Z(G)$ is a nilpotent normal subgroup of $G/Z(G)$, so $F(G)/Z(G)$ is contained in $F(G/Z(G))$. What about the other direction? Write $F(G/Z(G))=H/Z(G)$. Since $H/Z(G)$ is nilpotent (because it is the Fitting subgroup of some group), it follows that $H$ is nilpotent (do you understand why?). Then $H leq F(G)$, which proves the other direction we wanted.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 22 '18 at 5:35









        the_foxthe_fox

        2,89021537




        2,89021537






























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