Mixing
If $G$ is a $p$-group then $Phi(G)=G'G^p$
$begingroup$
Okay this problem is quite the confounding one for me.
If $G$ is a $p$-group then it follows that $Phi(G)=G'G^p$.
Where:
$Phi(G)$- Frattini subgroup (which in this case is the intersection of all subgroups of index $p$)
$G'$ commutator subgroup
- $G^p={x^p:xin G}$
I am having trouble tackling a couple sub-problems with this problem.
Why is $G'G^p$ a subgroup? In general $G^p$ isn't a subgroup, so why does $G'G^p$ become a subgroup?
While I understand that $G^psubset Phi(G)$, why would $G'G^psubseteq M_i$ for all $i$ where ${M_i}$ is a collection of subgroups of index p?
Alot of my problems, center around $G^p$ not being a subgroup in general. But even if I were to prove that $G'G^p$ is a normal subgroup, I would still have to prove (2.) which would be easy if it is true that every maximal class is conjugate to one another, but I don't know that (or maybe it isn't true).
Is there a way for me to see that $G'G^p$ is a normal subgroup and that it is contained in every subgroup of index $p$.
group-theory p-groups
edited Jan 25 at 3:04
Orat
2,88021231
asked Sep 18 '15 at 21:36
user160110user160110
1,415818
$endgroup$
add a comment |
$begingroup$
Okay this problem is quite the confounding one for me.
If $G$ is a $p$-group then it follows that $Phi(G)=G'G^p$.
Where:
$Phi(G)$- Frattini subgroup (which in this case is the intersection of all subgroups of index $p$)
$G'$ commutator subgroup
- $G^p={x^p:xin G}$
I am having trouble tackling a couple sub-problems with this problem.
Why is $G'G^p$ a subgroup? In general $G^p$ isn't a subgroup, so why does $G'G^p$ become a subgroup?
While I understand that $G^psubset Phi(G)$, why would $G'G^psubseteq M_i$ for all $i$ where ${M_i}$ is a collection of subgroups of index p?
Alot of my problems, center around $G^p$ not being a subgroup in general. But even if I were to prove that $G'G^p$ is a normal subgroup, I would still have to prove (2.) which would be easy if it is true that every maximal class is conjugate to one another, but I don't know that (or maybe it isn't true).
Is there a way for me to see that $G'G^p$ is a normal subgroup and that it is contained in every subgroup of index $p$.
group-theory p-groups
edited Jan 25 at 3:04
Orat
2,88021231
asked Sep 18 '15 at 21:36
user160110user160110
1,415818
$endgroup$
3
$begingroup$
The standard definition of $G^p$ is not the set of $p$-th powers but the subgroup of $G$ generated by the $p$-th powers. So $G^p$ is always a subgroup. But having said that, with your definition it will still work, because the images of the elements $x^p$ in $G/[G,G]$ (which is abelian) do form a subgroup.
$endgroup$
– Derek Holt
Sep 18 '15 at 22:03
$begingroup$
Oh, thanks for that.
$endgroup$
– user160110
Sep 18 '15 at 22:07
$begingroup$
But would $G'subset Phi(G)$?
$endgroup$
– user160110
Sep 18 '15 at 22:16
add a comment |
$begingroup$
Okay this problem is quite the confounding one for me.
If $G$ is a $p$-group then it follows that $Phi(G)=G'G^p$.
Where:
$Phi(G)$- Frattini subgroup (which in this case is the intersection of all subgroups of index $p$)
$G'$ commutator subgroup
- $G^p={x^p:xin G}$
I am having trouble tackling a couple sub-problems with this problem.
Why is $G'G^p$ a subgroup? In general $G^p$ isn't a subgroup, so why does $G'G^p$ become a subgroup?
While I understand that $G^psubset Phi(G)$, why would $G'G^psubseteq M_i$ for all $i$ where ${M_i}$ is a collection of subgroups of index p?
Alot of my problems, center around $G^p$ not being a subgroup in general. But even if I were to prove that $G'G^p$ is a normal subgroup, I would still have to prove (2.) which would be easy if it is true that every maximal class is conjugate to one another, but I don't know that (or maybe it isn't true).
Is there a way for me to see that $G'G^p$ is a normal subgroup and that it is contained in every subgroup of index $p$.
group-theory p-groups
edited Jan 25 at 3:04
Orat
2,88021231
asked Sep 18 '15 at 21:36
user160110user160110
1,415818
$endgroup$
Okay this problem is quite the confounding one for me.
If $G$ is a $p$-group then it follows that $Phi(G)=G'G^p$.
Where:
$Phi(G)$- Frattini subgroup (which in this case is the intersection of all subgroups of index $p$)
$G'$ commutator subgroup
- $G^p={x^p:xin G}$
I am having trouble tackling a couple sub-problems with this problem.
Why is $G'G^p$ a subgroup? In general $G^p$ isn't a subgroup, so why does $G'G^p$ become a subgroup?
While I understand that $G^psubset Phi(G)$, why would $G'G^psubseteq M_i$ for all $i$ where ${M_i}$ is a collection of subgroups of index p?
Alot of my problems, center around $G^p$ not being a subgroup in general. But even if I were to prove that $G'G^p$ is a normal subgroup, I would still have to prove (2.) which would be easy if it is true that every maximal class is conjugate to one another, but I don't know that (or maybe it isn't true).
Is there a way for me to see that $G'G^p$ is a normal subgroup and that it is contained in every subgroup of index $p$.
group-theory p-groups
group-theory p-groups
edited Jan 25 at 3:04
Orat
2,88021231
asked Sep 18 '15 at 21:36
user160110user160110
1,415818
edited Jan 25 at 3:04
Orat
2,88021231
asked Sep 18 '15 at 21:36
user160110user160110
1,415818
edited Jan 25 at 3:04
Orat
2,88021231
edited Jan 25 at 3:04
Orat
2,88021231
edited Jan 25 at 3:04
Orat
2,88021231
2,88021231
asked Sep 18 '15 at 21:36
user160110user160110
1,415818
asked Sep 18 '15 at 21:36
user160110user160110
1,415818
asked Sep 18 '15 at 21:36
user160110user160110
1,415818
1,415818
3
$begingroup$
The standard definition of $G^p$ is not the set of $p$-th powers but the subgroup of $G$ generated by the $p$-th powers. So $G^p$ is always a subgroup. But having said that, with your definition it will still work, because the images of the elements $x^p$ in $G/[G,G]$ (which is abelian) do form a subgroup.
$endgroup$
– Derek Holt
Sep 18 '15 at 22:03
$begingroup$
Oh, thanks for that.
$endgroup$
– user160110
Sep 18 '15 at 22:07
$begingroup$
But would $G'subset Phi(G)$?
$endgroup$
– user160110
Sep 18 '15 at 22:16
add a comment |
3
$begingroup$
The standard definition of $G^p$ is not the set of $p$-th powers but the subgroup of $G$ generated by the $p$-th powers. So $G^p$ is always a subgroup. But having said that, with your definition it will still work, because the images of the elements $x^p$ in $G/[G,G]$ (which is abelian) do form a subgroup.
$endgroup$
– Derek Holt
Sep 18 '15 at 22:03
$begingroup$
Oh, thanks for that.
$endgroup$
– user160110
Sep 18 '15 at 22:07
$begingroup$
But would $G'subset Phi(G)$?
$endgroup$
– user160110
Sep 18 '15 at 22:16
3
3
$begingroup$
The standard definition of $G^p$ is not the set of $p$-th powers but the subgroup of $G$ generated by the $p$-th powers. So $G^p$ is always a subgroup. But having said that, with your definition it will still work, because the images of the elements $x^p$ in $G/[G,G]$ (which is abelian) do form a subgroup.
$endgroup$
– Derek Holt
Sep 18 '15 at 22:03
$begingroup$
The standard definition of $G^p$ is not the set of $p$-th powers but the subgroup of $G$ generated by the $p$-th powers. So $G^p$ is always a subgroup. But having said that, with your definition it will still work, because the images of the elements $x^p$ in $G/[G,G]$ (which is abelian) do form a subgroup.
$endgroup$
– Derek Holt
Sep 18 '15 at 22:03
$begingroup$
Oh, thanks for that.
$endgroup$
– user160110
Sep 18 '15 at 22:07
$begingroup$
Oh, thanks for that.
$endgroup$
– user160110
Sep 18 '15 at 22:07
$begingroup$
But would $G'subset Phi(G)$?
$endgroup$
– user160110
Sep 18 '15 at 22:16
$begingroup$
But would $G'subset Phi(G)$?
$endgroup$
– user160110
Sep 18 '15 at 22:16
add a comment |
1 Answer
1
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oldest
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$begingroup$
For a finite $p$-group $G$, any maximal subgroup $M<G$ is normal of index $p$.
Proof. $G$ has nontrivial center so pick a central subgroup $Z$ of order $p$. If $Zsubseteq M$ then $M/Z<G/Z$ is normal of index $p$ by induction hypothesis, and then $M<G$ is normal of index $p$. Otherwise if $Znotsubseteq M$ then the whole group $G=MZ$ normalizes $M$, and $G$ has $|Z|$ cosets of $M$ in it.
Because $G/Mcong C_p$, the $p$-power map on $G/M$ is the zero map so $G^psubseteq M$, and the quotient $G/M$ is abelian so $G'subseteq M$. Therefore $G^pG'subseteq M$ for every maximal $M$. Both $G^p$ and $G'$ are normal so we know that $G^pG'$ is normal too.
answered Sep 19 '15 at 3:50
whackawhacka
12.1k1734
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For a finite $p$-group $G$, any maximal subgroup $M<G$ is normal of index $p$.
Proof. $G$ has nontrivial center so pick a central subgroup $Z$ of order $p$. If $Zsubseteq M$ then $M/Z<G/Z$ is normal of index $p$ by induction hypothesis, and then $M<G$ is normal of index $p$. Otherwise if $Znotsubseteq M$ then the whole group $G=MZ$ normalizes $M$, and $G$ has $|Z|$ cosets of $M$ in it.
Because $G/Mcong C_p$, the $p$-power map on $G/M$ is the zero map so $G^psubseteq M$, and the quotient $G/M$ is abelian so $G'subseteq M$. Therefore $G^pG'subseteq M$ for every maximal $M$. Both $G^p$ and $G'$ are normal so we know that $G^pG'$ is normal too.
answered Sep 19 '15 at 3:50
whackawhacka
12.1k1734
$endgroup$
add a comment |
$begingroup$
For a finite $p$-group $G$, any maximal subgroup $M<G$ is normal of index $p$.
Proof. $G$ has nontrivial center so pick a central subgroup $Z$ of order $p$. If $Zsubseteq M$ then $M/Z<G/Z$ is normal of index $p$ by induction hypothesis, and then $M<G$ is normal of index $p$. Otherwise if $Znotsubseteq M$ then the whole group $G=MZ$ normalizes $M$, and $G$ has $|Z|$ cosets of $M$ in it.
Because $G/Mcong C_p$, the $p$-power map on $G/M$ is the zero map so $G^psubseteq M$, and the quotient $G/M$ is abelian so $G'subseteq M$. Therefore $G^pG'subseteq M$ for every maximal $M$. Both $G^p$ and $G'$ are normal so we know that $G^pG'$ is normal too.
answered Sep 19 '15 at 3:50
whackawhacka
12.1k1734
$endgroup$
add a comment |
$begingroup$
For a finite $p$-group $G$, any maximal subgroup $M<G$ is normal of index $p$.
Proof. $G$ has nontrivial center so pick a central subgroup $Z$ of order $p$. If $Zsubseteq M$ then $M/Z<G/Z$ is normal of index $p$ by induction hypothesis, and then $M<G$ is normal of index $p$. Otherwise if $Znotsubseteq M$ then the whole group $G=MZ$ normalizes $M$, and $G$ has $|Z|$ cosets of $M$ in it.
Because $G/Mcong C_p$, the $p$-power map on $G/M$ is the zero map so $G^psubseteq M$, and the quotient $G/M$ is abelian so $G'subseteq M$. Therefore $G^pG'subseteq M$ for every maximal $M$. Both $G^p$ and $G'$ are normal so we know that $G^pG'$ is normal too.
answered Sep 19 '15 at 3:50
whackawhacka
12.1k1734
$endgroup$
For a finite $p$-group $G$, any maximal subgroup $M<G$ is normal of index $p$.
Proof. $G$ has nontrivial center so pick a central subgroup $Z$ of order $p$. If $Zsubseteq M$ then $M/Z<G/Z$ is normal of index $p$ by induction hypothesis, and then $M<G$ is normal of index $p$. Otherwise if $Znotsubseteq M$ then the whole group $G=MZ$ normalizes $M$, and $G$ has $|Z|$ cosets of $M$ in it.
Because $G/Mcong C_p$, the $p$-power map on $G/M$ is the zero map so $G^psubseteq M$, and the quotient $G/M$ is abelian so $G'subseteq M$. Therefore $G^pG'subseteq M$ for every maximal $M$. Both $G^p$ and $G'$ are normal so we know that $G^pG'$ is normal too.
answered Sep 19 '15 at 3:50
whackawhacka
12.1k1734
answered Sep 19 '15 at 3:50
whackawhacka
12.1k1734
answered Sep 19 '15 at 3:50
whackawhacka
12.1k1734
answered Sep 19 '15 at 3:50
whackawhacka
12.1k1734
12.1k1734
add a comment |
add a comment |
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3
$begingroup$
The standard definition of $G^p$ is not the set of $p$-th powers but the subgroup of $G$ generated by the $p$-th powers. So $G^p$ is always a subgroup. But having said that, with your definition it will still work, because the images of the elements $x^p$ in $G/[G,G]$ (which is abelian) do form a subgroup.
$endgroup$
– Derek Holt
Sep 18 '15 at 22:03
$begingroup$
Oh, thanks for that.
$endgroup$
– user160110
Sep 18 '15 at 22:07
$begingroup$
But would $G'subset Phi(G)$?
$endgroup$
– user160110
Sep 18 '15 at 22:16