Abelian normal subgroups of A-groups
$begingroup$
Let $G$ be a finite solvable group, where every Sylow subgroup is abelian. I want to show that if $Alhd G$ is an abelian normal subgroup, then
$$ A=(Acap Z(G))(Acap G')$$
This is easy if $A$ is a minimal normal subgroup, so I've tried to induct on the $G$-composition length of $A$. So if $L<A$ is a normal subgroup of $G$, maximal in the sense that there are no normal subgroups of $G$ strictly between $L$ and $A$, I want to show either $Ale LZ(G)$ or $Ale LG'$.
If there exists $ain A$ and $gin G$ with $[g,a]notin L$, then
$$ A= L(Acap G')= Acap LG'$$
so that case is done.
The other case is where $[G,A]le L$. Now I need to find an $ain Asetminus L$ that is central. But I have no idea how to do that.
group-theory finite-groups abelian-groups sylow-theory solvable-groups
edited Dec 27 '18 at 23:06
Shaun
10.6k113687
asked Dec 27 '18 at 22:01
HempeliciousHempelicious
142111
$endgroup$
add a comment |
$begingroup$
Let $G$ be a finite solvable group, where every Sylow subgroup is abelian. I want to show that if $Alhd G$ is an abelian normal subgroup, then
$$ A=(Acap Z(G))(Acap G')$$
This is easy if $A$ is a minimal normal subgroup, so I've tried to induct on the $G$-composition length of $A$. So if $L<A$ is a normal subgroup of $G$, maximal in the sense that there are no normal subgroups of $G$ strictly between $L$ and $A$, I want to show either $Ale LZ(G)$ or $Ale LG'$.
If there exists $ain A$ and $gin G$ with $[g,a]notin L$, then
$$ A= L(Acap G')= Acap LG'$$
so that case is done.
The other case is where $[G,A]le L$. Now I need to find an $ain Asetminus L$ that is central. But I have no idea how to do that.
group-theory finite-groups abelian-groups sylow-theory solvable-groups
edited Dec 27 '18 at 23:06
Shaun
10.6k113687
asked Dec 27 '18 at 22:01
HempeliciousHempelicious
142111
$endgroup$
add a comment |
$begingroup$
Let $G$ be a finite solvable group, where every Sylow subgroup is abelian. I want to show that if $Alhd G$ is an abelian normal subgroup, then
$$ A=(Acap Z(G))(Acap G')$$
This is easy if $A$ is a minimal normal subgroup, so I've tried to induct on the $G$-composition length of $A$. So if $L<A$ is a normal subgroup of $G$, maximal in the sense that there are no normal subgroups of $G$ strictly between $L$ and $A$, I want to show either $Ale LZ(G)$ or $Ale LG'$.
If there exists $ain A$ and $gin G$ with $[g,a]notin L$, then
$$ A= L(Acap G')= Acap LG'$$
so that case is done.
The other case is where $[G,A]le L$. Now I need to find an $ain Asetminus L$ that is central. But I have no idea how to do that.
group-theory finite-groups abelian-groups sylow-theory solvable-groups
edited Dec 27 '18 at 23:06
Shaun
10.6k113687
asked Dec 27 '18 at 22:01
HempeliciousHempelicious
142111
$endgroup$
Let $G$ be a finite solvable group, where every Sylow subgroup is abelian. I want to show that if $Alhd G$ is an abelian normal subgroup, then
$$ A=(Acap Z(G))(Acap G')$$
This is easy if $A$ is a minimal normal subgroup, so I've tried to induct on the $G$-composition length of $A$. So if $L<A$ is a normal subgroup of $G$, maximal in the sense that there are no normal subgroups of $G$ strictly between $L$ and $A$, I want to show either $Ale LZ(G)$ or $Ale LG'$.
If there exists $ain A$ and $gin G$ with $[g,a]notin L$, then
$$ A= L(Acap G')= Acap LG'$$
so that case is done.
The other case is where $[G,A]le L$. Now I need to find an $ain Asetminus L$ that is central. But I have no idea how to do that.
group-theory finite-groups abelian-groups sylow-theory solvable-groups
group-theory finite-groups abelian-groups sylow-theory solvable-groups
edited Dec 27 '18 at 23:06
Shaun
10.6k113687
asked Dec 27 '18 at 22:01
HempeliciousHempelicious
142111
edited Dec 27 '18 at 23:06
Shaun
10.6k113687
asked Dec 27 '18 at 22:01
HempeliciousHempelicious
142111
edited Dec 27 '18 at 23:06
Shaun
10.6k113687
edited Dec 27 '18 at 23:06
Shaun
10.6k113687
edited Dec 27 '18 at 23:06
Shaun
10.6k113687
10.6k113687
asked Dec 27 '18 at 22:01
HempeliciousHempelicious
142111
asked Dec 27 '18 at 22:01
HempeliciousHempelicious
142111
asked Dec 27 '18 at 22:01
HempeliciousHempelicious
142111
142111
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $A$ is the direct product of its Sylow subgroups, each of which is normal in $G$, we can suppose that $A$ is a $p$-group for some prime $p$.
Since $G$ has abelian Sylow subgroups, the group $C cong G/C_G(A)$ of automorphisms of $A$ that is induced by conjugation by $G$ is a a $p'$-group, and it is enough to prove that there is an $a in A setminus L$ that is central in the semidirect product $A rtimes C$.
Let $b in A setminus L$. Then, since $[G,A] = [C,A] le L$, we have $C^b le CL$. Since $C$ is a $p'$-group, we can apply the Schur-Zassenhaus Theorem to deduce that $C$ and $C^b$ are conjugate in $CL$, and hence by an element $l in L$. So $bl^{-1} in A cap N_G(C) le C_G(C)$, and so $a = bl^{-1}$ is central.
answered Dec 27 '18 at 23:27
Derek HoltDerek Holt
54.6k53574
$endgroup$
$begingroup$
Thank you! The idea to focus on a p-group really makes it clear. I think this proof also works with $C$ replaced by a Hall $p'$-subgroup, and we can use the conjugacy there.
$endgroup$
– Hempelicious
Dec 28 '18 at 4:32
1
$begingroup$
Yes. But I don't think the proof above (which is extracted from a more general result that, if a finite $p'$-group $C$ acts on a finite group $A$ then $A = C_A(C)[C,A]$) uses the solvability of $G$.
$endgroup$
– Derek Holt
Dec 28 '18 at 8:43
add a comment |
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1 Answer
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$begingroup$
Since $A$ is the direct product of its Sylow subgroups, each of which is normal in $G$, we can suppose that $A$ is a $p$-group for some prime $p$.
Since $G$ has abelian Sylow subgroups, the group $C cong G/C_G(A)$ of automorphisms of $A$ that is induced by conjugation by $G$ is a a $p'$-group, and it is enough to prove that there is an $a in A setminus L$ that is central in the semidirect product $A rtimes C$.
Let $b in A setminus L$. Then, since $[G,A] = [C,A] le L$, we have $C^b le CL$. Since $C$ is a $p'$-group, we can apply the Schur-Zassenhaus Theorem to deduce that $C$ and $C^b$ are conjugate in $CL$, and hence by an element $l in L$. So $bl^{-1} in A cap N_G(C) le C_G(C)$, and so $a = bl^{-1}$ is central.
answered Dec 27 '18 at 23:27
Derek HoltDerek Holt
54.6k53574
$endgroup$
$begingroup$
Thank you! The idea to focus on a p-group really makes it clear. I think this proof also works with $C$ replaced by a Hall $p'$-subgroup, and we can use the conjugacy there.
$endgroup$
– Hempelicious
Dec 28 '18 at 4:32
1
$begingroup$
Yes. But I don't think the proof above (which is extracted from a more general result that, if a finite $p'$-group $C$ acts on a finite group $A$ then $A = C_A(C)[C,A]$) uses the solvability of $G$.
$endgroup$
– Derek Holt
Dec 28 '18 at 8:43
add a comment |
$begingroup$
Since $A$ is the direct product of its Sylow subgroups, each of which is normal in $G$, we can suppose that $A$ is a $p$-group for some prime $p$.
Since $G$ has abelian Sylow subgroups, the group $C cong G/C_G(A)$ of automorphisms of $A$ that is induced by conjugation by $G$ is a a $p'$-group, and it is enough to prove that there is an $a in A setminus L$ that is central in the semidirect product $A rtimes C$.
Let $b in A setminus L$. Then, since $[G,A] = [C,A] le L$, we have $C^b le CL$. Since $C$ is a $p'$-group, we can apply the Schur-Zassenhaus Theorem to deduce that $C$ and $C^b$ are conjugate in $CL$, and hence by an element $l in L$. So $bl^{-1} in A cap N_G(C) le C_G(C)$, and so $a = bl^{-1}$ is central.
answered Dec 27 '18 at 23:27
Derek HoltDerek Holt
54.6k53574
$endgroup$
$begingroup$
Thank you! The idea to focus on a p-group really makes it clear. I think this proof also works with $C$ replaced by a Hall $p'$-subgroup, and we can use the conjugacy there.
$endgroup$
– Hempelicious
Dec 28 '18 at 4:32
1
$begingroup$
Yes. But I don't think the proof above (which is extracted from a more general result that, if a finite $p'$-group $C$ acts on a finite group $A$ then $A = C_A(C)[C,A]$) uses the solvability of $G$.
$endgroup$
– Derek Holt
Dec 28 '18 at 8:43
add a comment |
$begingroup$
Since $A$ is the direct product of its Sylow subgroups, each of which is normal in $G$, we can suppose that $A$ is a $p$-group for some prime $p$.
Since $G$ has abelian Sylow subgroups, the group $C cong G/C_G(A)$ of automorphisms of $A$ that is induced by conjugation by $G$ is a a $p'$-group, and it is enough to prove that there is an $a in A setminus L$ that is central in the semidirect product $A rtimes C$.
Let $b in A setminus L$. Then, since $[G,A] = [C,A] le L$, we have $C^b le CL$. Since $C$ is a $p'$-group, we can apply the Schur-Zassenhaus Theorem to deduce that $C$ and $C^b$ are conjugate in $CL$, and hence by an element $l in L$. So $bl^{-1} in A cap N_G(C) le C_G(C)$, and so $a = bl^{-1}$ is central.
answered Dec 27 '18 at 23:27
Derek HoltDerek Holt
54.6k53574
$endgroup$
Since $A$ is the direct product of its Sylow subgroups, each of which is normal in $G$, we can suppose that $A$ is a $p$-group for some prime $p$.
Since $G$ has abelian Sylow subgroups, the group $C cong G/C_G(A)$ of automorphisms of $A$ that is induced by conjugation by $G$ is a a $p'$-group, and it is enough to prove that there is an $a in A setminus L$ that is central in the semidirect product $A rtimes C$.
Let $b in A setminus L$. Then, since $[G,A] = [C,A] le L$, we have $C^b le CL$. Since $C$ is a $p'$-group, we can apply the Schur-Zassenhaus Theorem to deduce that $C$ and $C^b$ are conjugate in $CL$, and hence by an element $l in L$. So $bl^{-1} in A cap N_G(C) le C_G(C)$, and so $a = bl^{-1}$ is central.
answered Dec 27 '18 at 23:27
Derek HoltDerek Holt
54.6k53574
answered Dec 27 '18 at 23:27
Derek HoltDerek Holt
54.6k53574
answered Dec 27 '18 at 23:27
Derek HoltDerek Holt
54.6k53574
answered Dec 27 '18 at 23:27
Derek HoltDerek Holt
54.6k53574
54.6k53574
$begingroup$
Thank you! The idea to focus on a p-group really makes it clear. I think this proof also works with $C$ replaced by a Hall $p'$-subgroup, and we can use the conjugacy there.
$endgroup$
– Hempelicious
Dec 28 '18 at 4:32
1
$begingroup$
Yes. But I don't think the proof above (which is extracted from a more general result that, if a finite $p'$-group $C$ acts on a finite group $A$ then $A = C_A(C)[C,A]$) uses the solvability of $G$.
$endgroup$
– Derek Holt
Dec 28 '18 at 8:43
add a comment |
$begingroup$
Thank you! The idea to focus on a p-group really makes it clear. I think this proof also works with $C$ replaced by a Hall $p'$-subgroup, and we can use the conjugacy there.
$endgroup$
– Hempelicious
Dec 28 '18 at 4:32
1
$begingroup$
Yes. But I don't think the proof above (which is extracted from a more general result that, if a finite $p'$-group $C$ acts on a finite group $A$ then $A = C_A(C)[C,A]$) uses the solvability of $G$.
$endgroup$
– Derek Holt
Dec 28 '18 at 8:43
$begingroup$
Thank you! The idea to focus on a p-group really makes it clear. I think this proof also works with $C$ replaced by a Hall $p'$-subgroup, and we can use the conjugacy there.
$endgroup$
– Hempelicious
Dec 28 '18 at 4:32
$begingroup$
Thank you! The idea to focus on a p-group really makes it clear. I think this proof also works with $C$ replaced by a Hall $p'$-subgroup, and we can use the conjugacy there.
$endgroup$
– Hempelicious
Dec 28 '18 at 4:32
1
1
$begingroup$
Yes. But I don't think the proof above (which is extracted from a more general result that, if a finite $p'$-group $C$ acts on a finite group $A$ then $A = C_A(C)[C,A]$) uses the solvability of $G$.
$endgroup$
– Derek Holt
Dec 28 '18 at 8:43
$begingroup$
Yes. But I don't think the proof above (which is extracted from a more general result that, if a finite $p'$-group $C$ acts on a finite group $A$ then $A = C_A(C)[C,A]$) uses the solvability of $G$.
$endgroup$
– Derek Holt
Dec 28 '18 at 8:43
add a comment |
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