Given that $G$ is a finite group, prove that $F(G/Z(G))$ = $F(G) / Z(G)$.
$begingroup$
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 $p$?
group-theory finite-groups proof-explanation
edited Dec 29 '18 at 3:09
the_fox
2,69221535
asked Apr 7 '17 at 3:09
K.OK.O
114
$endgroup$
add a comment |
$begingroup$
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 $p$?
group-theory finite-groups proof-explanation
edited Dec 29 '18 at 3:09
the_fox
2,69221535
asked Apr 7 '17 at 3:09
K.OK.O
114
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $p$?
group-theory finite-groups proof-explanation
edited Dec 29 '18 at 3:09
the_fox
2,69221535
asked Apr 7 '17 at 3:09
K.OK.O
114
$endgroup$
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
group-theory finite-groups proof-explanation
edited Dec 29 '18 at 3:09
the_fox
2,69221535
asked Apr 7 '17 at 3:09
K.OK.O
114
edited Dec 29 '18 at 3:09
the_fox
2,69221535
asked Apr 7 '17 at 3:09
K.OK.O
114
edited Dec 29 '18 at 3:09
the_fox
2,69221535
edited Dec 29 '18 at 3:09
the_fox
2,69221535
edited Dec 29 '18 at 3:09
the_fox
2,69221535
2,69221535
asked Apr 7 '17 at 3:09
K.OK.O
114
asked Apr 7 '17 at 3:09
K.OK.O
114
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 22 '18 at 5:35
the_foxthe_fox
2,69221535
$endgroup$
add a comment |
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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.
answered Dec 22 '18 at 5:35
the_foxthe_fox
2,69221535
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 22 '18 at 5:35
the_foxthe_fox
2,69221535
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 22 '18 at 5:35
the_foxthe_fox
2,69221535
$endgroup$
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.
answered Dec 22 '18 at 5:35
the_foxthe_fox
2,69221535
answered Dec 22 '18 at 5:35
the_foxthe_fox
2,69221535
answered Dec 22 '18 at 5:35
the_foxthe_fox
2,69221535
answered Dec 22 '18 at 5:35
the_foxthe_fox
2,69221535
2,69221535
add a comment |
add a comment |
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$begingroup$
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